Cubic regularization of Newton method and its global performance

Abstract

In this paper, we provide theoretical analysis for a cubic regularization of Newton method as applied to unconstrained minimization
problem. For this scheme, we prove general local convergence results. However, the main contribution of the paper is related
to global worst-case complexity bounds for different problem classes including some nonconvex cases. It is shown that the
search direction can be computed by standard linear algebra technique.

Full text

Mathematical Programming manuscript No.
(will be inserted by the editor)
Yu. Nesterov, B. Polyak
Cubic regularization of Newton method
and its global performance
Received: June, 2004 / . . . / Final version: January 2006
Abstract. In this paper, we provide theoretical analysis for a cubic regularization of Newton
method as applied to unconstrained minimization problem. For this scheme, we prove general
local convergence results. However, the main contribution of the paper is related to global
worst-case complexity bounds for different problem classes including some nonconvex cases. It
is shown that the search direction can be computed by standard linear algebra technique.
Keywords: General nonlinear optimization, unconstrained optimization, New-
ton method, trust-region methods, global complexity bounds, global rate of con-
vergence.
1. Introduction
Motivation. Starting from seminal papers by Bennet [1] and Kantorovich [6],
the Newton method turned into an imp ortant tool for numerous applied prob-
lems. In the simplest case of unconstrained minimization of a multivariate func-
tion,
min
x∈R
n
f(x),
the standard Newton scheme looks as follows:
x
k+1
= x
k
− [f
00
(x
k
)]
−1
f
0
(x
k
).
Despite to its very natural motivation, this scheme has several hidden drawbacks.
First of all, it may happen that at current test point the Hessian is degenerate;
in this case the method is not well-defined. Secondly, it may happen that this
scheme diverges or converges to a saddle point or even to a point of local maxi-
mum. In the last fifty years the number of different suggestions for improving the
Yurii Nesterov: Center for Operations Research and Econometrics (CORE), Catholic Uni-
versity of Louvain (UCL), 34 voie du Roman Pays, 1348 Louvain-la-Neuve, Belgium; e-mail:
nesterov@core.ucl.ac.be.
B.T.Polyak: Institute of Control Science, Profsojuznaya 65, Moscow 117997, Russia; e-mail:
boris@ipu.rssi.ru.
The research results presented in this paper have been supported by a grant “Action de
recherche concert`e ARC 04/09-315” from the “Direction de la recherche scientifique - Com-
munaut`e fran¸caise de Belgique”. The scientific responsibility rests with the authors.
Mathematics Subject Classification (1991): 49M15, 49M37, 58C15, 90C25, 90C30.

2 Yu. Nesterov, B. Polyak
scheme was extremely large. The reader can consult a 1000-item bibliography in
the recent exhaustive covering of the field [2]. However, most of them combine
in different ways the following ideas.
– Levenberg-Marquardt regularization. As suggested in [7,8], if f
00
(x) is not pos-
itive definite, let us regularize it with a unit matrix. Namely, use −G
−1
f
0
(x)
with G = f
00
(x) + γI Â 0 in order to perform the step:
x
k+1
= x
k
− [f
00
(x
k
) + γI]
−1
f
0
(x
k
).
This strategy sometimes is considered as a way to mix Newton’s method with
the gradient method.
– Line search. Since we are interested in a minimization, it looks reasonable to
allow a certain step size h
k
> 0:
x
k+1
= x
k
− h
k
[f
00
(x
k
)]
−1
f
0
(x
k
),
(this is a damped Newton method [12]). This can help to form a monotone
sequence of function values: f(x
k+1
) ≤ f(x
k
).
– Trust-region approach [5,4,3,2]. In accordance to this approach, at point x
k
we have to form its neighborhood, where the second-order approximation of
the function is reliable. This is a trust region ∆(x
k
), for instance ∆(x
k
) =
{x : ||x − x
k
|| ≤ ²} with some ² > 0. Then the next point x
k+1
is chosen as
a solution to the following auxiliary problem:
min
x∈∆(x
k
)
[hf
0
(x
k
), x − x
k
i +
1
2
hf
00
(x
k
)(x −x
k
), x − x
k
i].
Note that for ∆(x
k
) ≡ R
n
, this is exactly the standard Newton step.
We would encourage a reader to look in [2] for different combinations and
implementations of the above ideas. Here we only mention that despite to a
huge variety of the results, there still exist open theoretical questions in this
field. And, in our opinion, the most important group of questions is related to
the worst-case guarantees for global behavior of the second-order schemes.
Indeed, as far as we know, up to now there are very few results on the
global p erformance of Newton method. One example is an easy class of smooth
strongly convex functions where we can get a rate of convergence for a damped
Newton method [11,10]. However the number of iterations required is hard to
compare with that for the gradient method. In fact, up to now the relations
between the gradient method and the Newton method have not been clarified.
Of course, the requirements for the applicability of these methods are different
(e.g. smoothness assumptions are more strong for Newton’s method) as well as
computational burden (necessity to compute second derivatives, store matrices
and solve linear equations at each iteration of Newton’s method). However, there
exist numerous problems, where computation of the Hessian is not much harder
than computation of the gradient, and the iteration costs of both methods are
comparable. Quite often, one reads opinion that in such situations the Newton
method is good at the final stage of the minimization process, but it is better to

Cubic regularization of Newton method and its global performance 3
use the gradient method for the first iterations. Here we dispute this position:
we show that theoretically, a properly chosen Newton-type scheme outperforms
the gradient scheme (taking into account only the number of iterations) in all
situations under consideration.
In this paper we propose a modification of Newton method, which is con-
structed in a similar way to well-known gradient mapping [9]. Assume that func-
tion f has a Lipschitz continuous gradient:
kf
0
(x) −f
0
(y)k ≤ Dky − xk, ∀x, y ∈ R
n
.
Suppose we need to solve the problem
min
x∈Q
f(x),
where Q is a closed convex set. Then we can choose the next point x
k+1
in our
sequence as a solution of the following auxiliary problem:
min
y ∈Q
ξ
1,x
k
(y), ξ
1,x
k
(y) = f (x
k
) + hf
0
(x
k
), y − x
k
i +
1
2
Dky − x
k
k
2
. (1.1)
Convergence of this scheme follows from the fact that ξ
1,x
k
(y) is an upper first-
order approximation of the objective function, that is ξ
1,x
k
(y) ≥ f (y) ∀y ∈ R
n
(see, for example, [10], Section 2.2.4, for details). If Q ≡ R
n
, then the rule (1.1)
results in a usual gradient scheme:
x
k+1
= x
k
−
1
D
f
0
(x
k
).
Note that we can do similar thing with the second-order approximation. Indeed,
assume that the Hessian of our objective function is Lipschitz continuous:
kf
00
(x) −f
00
(y)k ≤ Lkx − yk, ∀x, y ∈ R
n
.
Then, it is easy to see that the auxiliary function
ξ
2,x
(y) = f (x) + hf
0
(x), y − xi +
1
2
hf
00
(x)(y − x), y − xi +
L
6
ky − xk
3
will be an upper second-order approximation for our objective function:
f(y) ≤ ξ
2,x
(y) ∀y ∈ R
n
.
Thus, we can try to find the next point in our second-order scheme from the
following auxiliary minimization problem:
x
k+1
∈ Arg min
y
ξ
2,x
k
(y) (1.2)
(here Argmin refers to a global minimizer). This is exactly the approach we
analyze in this paper; we call it cubic regularization of Newton’s method. Note
that problem (1.2) is non-convex and it can have local minima. However, our
approach is implementable since this problem is equivalent to minimizing an
explicitly written convex function of one variable.

4 Yu. Nesterov, B. Polyak
Contents. In Section 2 we introduce cubic regularization and present its main
properties. In Section 3 we analyze the general convergence of the process. We
prove that under very mild assumptions all limit points of the process satisfy
necessary second-order optimality condition. In this general setting we get a rate
of convergence for the norms of the gradients, which is better than the rate en-
sured by the gradient scheme. We prove also the lo cal quadratic convergence of
the process. In Section 4 we give the global complexity results of our scheme for
different problem classes. We show that in all situations the global rate of con-
vergence is surprisingly fast (like O(
1
k
2
) for star-convex functions, where k is the
iteration counter). Moreover, under rather weak non-degeneracy assumptions,
we have local sup er-linear convergence either of the order
4
3
or
3
2
. We show that
this happens even if the Hessian is degenerate at the solution set. In Section 5 we
show how to compute a solution to the cubic regularization problem and discuss
some efficient strategies for estimating the Lipschitz constant for the Hessian.
We conclude the paper by a short discussion presented in Section 6.
Notation. In what follows we denote by h·, ·i the standard inner product in R
n
:
hx, yi =
n
X
i=1
x
(i)
y
(i)
, x, y ∈ R
n
,
and by kxk the standard Euclidean norm:
kxk = hx, xi
1/2
.
For a symmetric n × n matrix H, its spectrum is denoted by {λ
i
(H)}
n
i=1
. We
assume that the eigenvalues are numbered in decreasing order:
λ
1
(H) ≥ . . . ≥ λ
n
(H).
Hence, we write H º 0 if and only if λ
n
(H) ≥ 0. In what follows, for a matrix
A we use the standard spectral matrix norm:
kAk = λ
1
(AA
T
)
1/2
.
Finally, I denotes a unit n × n matrix.
Acknowledgement. We are very thankful to anonymous referees for their nu-
merous comments on the initial version of the paper. Indeed, it may be too
ambitious to derive from our purely theoretical results any conclusion on the
practical efficiency of corresponding algorithmic implementations. However, the
authors do believe that the developed theory could pave a way for future progress
in computational practice.
2. Cubic regularization of quadratic approximation
Let F ⊆ R
n
be a closed convex set with non-empty interior. Consider a twice
differentiable function f(x), x ∈ F. Let x
0
∈ int F be a starting point of our

Cubic regularization of Newton method and its global performance 5
iterative schemes. We assume that the set F is large enough: It contains at least
the level set
L(f(x
0
)) ≡ {x ∈ R
n
: f(x) ≤ f(x
0
)}
in its interior. Moreover, in this paper we always assume the following.
Assumption 1 The Hessian of function f is Lipschitz continuous on F:
kf
00
(x) −f
00
(y)k ≤ Lkx − yk, ∀x, y ∈ F. (2.1)
for some L > 0.
For the sake of completeness, let us present the following trivial consequences
of our assumption (compare with [12, Section 3]).
Lemma 1. For any x and y from F we have
kf
0
(y) − f
0
(x) −f
00
(x)(y − x)k ≤
1
2
Lky − xk
2
, (2.2)
|f(y) − f(x) − hf
0
(x), y − xi −
1
2
hf
00
(x)(y − x), y − xi| ≤
L
6
ky − xk
3
.
(2.3)
Proof. Indeed,
kf
0
(y) − f
0
(x) −f
00
(x)(y − x)k = k
1
R
0
[f
00
(x + τ (y − x)) −f
00
(x)](y − x)dτk
≤ Lky − xk
2
1
R
0
τdτ =
1
2
Lky − xk
2
.
Therefore,
|f(y) − f(x) − hf
0
(x), y − xi −
1
2
hf
00
(x)(y − x), y − xi|
= |
1
R
0
hf
0
(x + λ(y − x)) − f
0
(x) −λf
00
(x)(y − x), y − xidλ|
≤
1
2
Lky − xk
3
1
R
0
λ
2
dλ =
L
6
ky − xk
3
.
ut
Let M be a positive parameter. Define a modified Newton step using the
following cubic regularization of quadratic approximation of function f(x):
T
M
(x) ∈ Arg min
y
£
hf
0
(x), y − xi +
1
2
hf
00
(x)(y − x), y − xi +
M
6
ky − xk
3
¤
,
(2.4)
where ”Arg” indicates that T
M
(x) is chosen from the set of global minima of
corresponding minimization problem. We postpone discussion of the complexity
of finding this point up to Section 5.1.
Note that point T
M
(x) satisfies the following system of nonlinear equations:
f
0
(x) + f
00
(x)(y − x) +
1
2
Mky − xk· (y − x) = 0.
(2.5)

6 Yu. Nesterov, B. Polyak
Denote r
M
(x) = kx −T
M
(x)k. Taking in (2.5) y = T
M
(x) and multiplying it by
T
M
(x) −x we get equation
hf
0
(x), T
M
(x) −xi + hf
00
(x)(T
M
(x) − x), T
M
(x) −xi +
1
2
Mr
3
M
(x) = 0.
(2.6)
In our analysis of the process (3.3), we need the following fact.
Proposition 1. For any x ∈ F we have
f
00
(x) +
1
2
Mr
M
(x)I º 0. (2.7)
This statement follows from Theorem 10, which will be proved later in Section
5.1. Now let us present the main properties of the mapping T
M
(A).
Lemma 2. For any x ∈ F, f(x) ≤ f(x
0
), we have the following relation:
hf
0
(x), x − T
M
(x)i ≥ 0. (2.8)
If M ≥
2
3
L and x ∈ int F, then T
M
(x) ∈ L(f(x)) ⊆ F.
Proof. Indeed, multiplying (2.7) by x − T
M
(x) twice, we get
hf
00
(x)(T
M
(x) − x), T
M
(x) −xi +
1
2
Mr
3
M
(x) ≥ 0.
Therefore (2.8) follows from (2.6).
Further, let M ≥
2
3
L. Assume that T
M
(x) 6∈ F. Then r
M
(x) > 0. Consider
the following points:
y
α
= x + α(T
M
(x) −x ), α ∈ [0, 1].
Since y(0) ∈ int F, the value
¯α : y
¯α
∈ ∂F
is well defined. In accordance to our assumption, ¯α < 1 and y
α
∈ F for all
α ∈ [0, ¯α]. Therefore, using (2.3), relation (2.6) and inequality (2.8), we get
f(y
α
) ≤ f(x) + hf
0
(x), y
α
− xi+
1
2
hf
00
(x)(y
α
− x), y
α
− xi+
α
3
L
6
r
3
M
(x)
≤ f(x) + hf
0
(x), y
α
− xi+
1
2
hf
00
(x)(y
α
− x), y
α
− xi+
α
3
M
4
r
3
M
(x)
= f(x) + (α −
α
2
2
)hf
0
(x), T
M
(x) −xi −
α
2
(1−α)
4
Mr
3
M
(x)
≤ f(x) −
α
2
(1−α)
4
Mr
3
M
(x).
Thus, f(y(¯α)) < f (x). Therefore y(¯α) ∈ int L(f(x)) ⊆ int F. That is a con-
tradiction. Hence, T
M
(x) ∈ F and using the same arguments we prove that
f(T
M
(x)) ≤ f(x). ut
Lemma 3. If T
M
(x) ∈ F, then
kf
0
(T
M
(x))k ≤
1
2
(L + M )r
2
M
(x). (2.9)

Cubic regularization of Newton method and its global performance 7
Proof. From equation (2.5), we get
kf
0
(x) + f
00
(x)(T
M
(x) − x)k =
1
2
Mr
2
M
(x).
On the other hand, in view of (2.2), we have
kf
0
(T
M
(x)) −f
0
(x) −f
00
(x)(T
M
(x) −x)k ≤
1
2
Lr
2
M
(x).
Combining these two relations, we obtain inequality (2.9). ut
Define
¯
f
M
(x) = min
y
£
f(x) + hf
0
(x), y − xi +
1
2
hf
00
(x)(y − x), y − xi +
M
6
ky − xk
3
¤
.
Lemma 4. For any x ∈ F we have
¯
f
M
(x) ≤ min
y ∈F
£
f(y) +
L+M
6
ky − xk
3
¤
,
(2.10)
f(x) −
¯
f
M
(x) ≥
M
12
r
3
M
(x).
(2.11)
Moreover, if M ≥ L, then T
M
(x) ∈ F and
f(T
M
(x)) ≤
¯
f
M
(x). (2.12)
Proof. Indeed, using the lower bound in (2.3), for any y ∈ F we have
f(x) + hf
0
(x), y − xi +
1
2
hf
00
(x)(y − x), y − xi ≤ f (y) +
L
6
ky − xk
3
.
and inequality in (2.10) follows from the definition of
¯
f
M
(x).
Further, in view of definition of point T
def
= T
M
(x), relation (2.6) and in-
equality (2.8), we have
f(x) −
¯
f
M
(x) = hf
0
(x), x − T i −
1
2
hf
00
(x)(T −x), T − xi−
M
6
r
3
M
(x)
=
1
2
hf
0
(x), x − T i +
M
12
r
3
M
(x) ≥
M
12
r
3
M
(x).
Finally, if M ≥ L, then T
M
(x) ∈ F in view of Lemma 2. Therefore, we get
inequality (2.12) from the upper bound in (2.3). ut
3. General convergence results
In this paper the main problem of interest is:
min
x∈R
n
f(x), (3.1)
where the objective function f(x) satisfies Assumption 1. Recall that the nec-
essary conditions for a point x
∗
to be a local solution to problem (3.1) are as
follows:
f
0
(x
∗
) = 0, f
00
(x
∗
) º 0. (3.2)

8 Yu. Nesterov, B. Polyak
Therefore, for an arbitrary x ∈ F we can introduce the following measure of
local optimality:
µ
M
(x) = max
n
q
2
L+M
kf
0
(x)k, −
2
2L+M
λ
n
(f
00
(x))
o
,
where M is a positive parameter. It is clear that for any x from F the measure
µ
M
(x) is non-negative and it vanishes only at the points satisfying conditions
(3.2). The analytical form of this measure can be justified by the following result.
Lemma 5. For any x ∈ F we have µ
M
(T
M
(x)) ≤ r
M
(x).
Proof. The proof follows immediately from inequality (2.9) and relation (2.7)
since
f
00
(T
M
(x)) ≥ f
00
(x) −Lr
M
(x)I ≥ −(
1
2
M + L)r
M
(x)I.
ut
Let L
0
∈ (0, L] be a positive parameter. Consider the following regularized
Newton scheme.
Cubic regularization of Newton method
Initialization: Choose x
0
∈ R
n
.
Iteration k, (k ≥ 0):
1. Find M
k
∈ [L
0
, 2L] such that
f(T
M
k
(x
k
)) ≤
¯
f
M
k
(x
k
).
2. Set x
k+1
= T
M
k
(x
k
).
(3.3)
Since
¯
f
M
(x) ≤ f(x), this process is monotone:
f(x
k+1
) ≤ f(x
k
).
If the constant L is known, we can take M
k
≡ L in Step 1 of this scheme.
In the opposite case, it is possible to apply a simple search procedure; we will
discuss its complexity later in Section 5.2. Now let us make the following simple
observation.
Theorem 1. Let the sequence {x
i
} be generated by method (3.3). Assume that
the objective function f(x) is bounded below:
f(x) ≥ f
∗
∀x ∈ F.

Cubic regularization of Newton method and its global performance 9
Then
∞
P
i=0
r
3
M
i
(x
i
) ≤
12
L
0
(f(x
0
)−f
∗
). Moreover, lim
i→∞
µ
L
(x
i
) = 0 and for any k ≥ 1
we have
min
1≤i≤k
µ
L
(x
i
) ≤
8
3
·
³
3(f(x
0
)−f
∗
)
2k·L
0
´
1/3
.
(3.4)
Proof. In view of inequality (2.11), we have
f(x
0
) −f
∗
≥
k−1
P
i=0
[f(x
i
) −f(x
i+1
) ≥
k−1
P
i=0
M
i
12
r
3
M
i
(x
i
) ≥
L
0
12
r
3
M
i
(x
i
).
It remains to use the statement of Lemma 5 and the upper bound on M
k
in
(3.3):
r
M
i
(x
i
) ≥ µ
M
i
(x
i+1
) ≥
3
4
µ
L
(x
i+1
).
ut
Note that inequality (3.4) implies that
min
1≤i≤k
kf
0
(x
i
)k ≤ O(k
−2/3
).
It is well known that for gradient scheme a possible level of the right-hand side
in this inequality is of the order O
¡
k
−1/2
¢
(see, for example, [10], inequality
(1.2.13)).
Theorem 1 helps to get the convergence results in many different situations.
We mention only one of them.
Theorem 2. Let sequence {x
i
} be generated by method (3.3). For some i ≥ 0,
assume the set L(f(x
i
)) be bounded. Then there exists a limit
lim
i→∞
f(x
i
) = f
∗
.
The set X
∗
of the limit points of this sequence is non-empty. Moreover, this is
a connected set, such that for any x
∗
∈ X
∗
we have
f(x
∗
) = f
∗
, f
0
(x
∗
) = 0, f
00
(x
∗
) º 0.
Proof. The proof of this theorem can be derived from Theorem 1 in a standard
way. ut
Let us describe now the behavior of the process (3.3) in a neighborhood of a
non-degenerate stationary point, which is not a point of local minimum.
Lemma 6. Let ¯x ∈ int F be a non-degenerate saddle point or a point of local
maximum of function f(x):
f
0
(¯x) = 0, λ
n
(f
00
(¯x)) < 0.
Then there exist constants ², δ > 0 such that whenever a point x
i
appears to be
in a set Q = {x : kx − ¯xk ≤ ², f(x) ≥ f(¯x)} (for instance, if x
i
= ¯x), then the
next point x
i+1
leaves the set Q:
f(x
i+1
) ≤ f(¯x) −δ.

10 Yu. Nesterov, B. Polyak
Proof. Let for some d with kdk = 1, and for some ¯τ > 0 we have
hf
00
(¯x)d, di ≡ −σ < 0, ¯x ± ¯τ d ∈ F.
Define ² = min
©
σ
2L
, ¯τ
ª
and δ =
σ
6
²
2
. Then, in view of inequality (2.10), upper
bound on M
i
, and inequality (2.3), for |τ| ≤ ¯τ we get the following estimate
f(x
i+1
) ≤ f(¯x + τd) +
L
2
k¯x + τd − x
i
k
3
≤ f(¯x) −στ
2
+
L
6
|τ|
3
+
L
2
£
²
2
+ 2τhd, ¯x − x
i
i + τ
2
¤
3/2
.
Since we are free in the choice of the sign of τ, we can guarantee that
f(x
i+1
) ≤ f(¯x) −στ
2
+
L
6
|τ|
3
+
L
2
£
²
2
+ τ
2
¤
3/2
, |τ| ≤ ¯τ.
Let us choose τ = ² ≤ ¯τ . Then
f(x
i+1
) ≤ f(¯x) −στ
2
+
5L
3
τ
3
≤ f (¯x) − στ
2
+
5L
3
·
σ
2L
· τ
2
= f (¯x) −
1
6
στ
2
.
Since the process (3.3) is monotone with respect to objective function, it will
never come again in Q. ut
Consider now the behavior of the regularized Newton scheme (3.3) in a neigh-
borhood of a non-degenerate local minimum. It appears that in such a situation
assumption L
0
> 0 is not necessary anymore. Let us analyze a relaxed version
of (3.3):
x
k+1
= T
M
k
(x
k
), k ≥ 0 (3.5)
where M
k
∈ (0, 2L]. Denote
δ
k
=
Lkf
0
(x
k
)k
λ
2
n
(f
00
(x
k
))
.
Theorem 3. Let f
00
(x
0
) Â 0 and δ
0
≤
1
4
. Let points {x
k
} be generated by (3.5).
Then:
1. For all k ≥ 0, the values δ
k
are well defined and they converge quadratically
to zero:
δ
k+1
≤
3
2
³
δ
k
1−δ
k
´
2
≤
8
3
δ
2
k
≤
2
3
δ
k
, k ≥ 0.
(3.6)
2. Minimal eigenvalues of all Hessians f
00
(x
k
) lie within the following bounds:
e
−1
λ
n
(f
00
(x
0
)) ≤ λ
n
(f
00
(x
k
)) ≤ e
3/4
λ
n
(f
00
(x
0
)).
(3.7)
3. The whole sequence {x
i
} converges quadratically to a point x
∗
, which is a
non-degenerate local minimum of function f(x). In particular, for any k ≥ 1 we
have
kf
0
(x
k
)k ≤ λ
2
n
(f
00
(x
0
))
9e
3/2
16L
¡
1
2
¢
2
k
.
(3.8)

Cubic regularization of Newton method and its global performance 11
Proof. Assume that for some k ≥ 0 we have f
00
(x
k
) Â 0. Then the corresponding
δ
k
is well defined. Assume that δ
k
≤
1
4
. From equation (2.5) we have
r
M
k
(x
k
) = kT
M
k
(x
k
) −x
k
k = k(f
00
(x
k
) + r
M
k
(x
k
)
M
k
2
I)
−1
f
0
(x
k
)k ≤
kf
0
(x
k
)k
λ
n
(f
00
(x
k
))
.
(3.9)
Note also that f
00
(x
k+1
) º f
00
(x
k
) −r
M
k
(x
k
)LI. Therefore
λ
n
(f
00
(x
k+1
) ≥ λ
n
(f
00
(x
k
)) −r
M
k
(x
k
)L
≥ λ
n
(f
00
(x
k
)) −
Lkf
0
(x
k
)k
λ
n
(f
00
(x
k
))
= (1 − δ
k
)λ
n
(f
00
(x
k
)).
Thus, f
00
(x
k+1
) is also positive definite. Moreover, using inequality (2.9) and the
upper bound for M
k
we obtain
δ
k+1
=
Lkf
0
(x
k+1
)k
λ
2
n
(f
00
(x
k+1
))
≤
3L
2
r
2
M
k
(x
k
)
2λ
2
n
(f
00
(x
k+1
))
≤
3L
2
kf
0
(x
k
)k
2
2λ
4
n
(f
00
(x
k
))(1−δ
k
)
2
=
3
2
³
δ
k
1−δ
k
´
2
≤
8
3
δ
2
k
.
Thus, δ
k+1
≤
1
4
and we prove (3.6) by induction. Note that we also get δ
k+1
≤
2
3
δ
k
.
Further, as we have already seen,
ln
λ
n
(f
00
(x
k
))
λ
n
(f
00
(x
0
))
≥
∞
P
i=0
ln(1 −δ
i
) ≥ −
∞
P
i=0
δ
i
1−δ
i
≥ −
1
1−δ
0
∞
P
i=0
δ
i
≥ −1.
In order to get an upper bound, note that f
00
(x
k+1
) ¹ f
00
(x
k
) + r
M
k
(x
k
)LI.
Hence,
λ
n
(f
00
(x
k+1
) ≤ λ
n
(f
00
(x
k
)) + r
M
k
(x
k
)L ≤ (1 + δ
k
)λ
n
(f
00
(x
k
)).
Therefore
ln
λ
n
(f
00
(x
k
))
λ
n
(f
00
(x
0
))
≤
∞
P
i=0
ln(1 + δ
i
) ≤
∞
P
i=0
δ
i
≤
3
4
.
It remains to prove Item 3 of the theorem. In view of inequalities (3.9) and
(3.7), we have
r
M
k
(x
k
) ≤
1
L
λ
n
(f
00
(x
k
))δ
k
≤
e
3/4
L
λ
n
(f
00
(x
0
))δ
k
.
Thus, {x
i
} is a Cauchy sequence, which has a unique limit point x
∗
. Since the
eigenvalues of f
00
(x) are continuous functions of x, from (3.7) we conclude that
f
00
(x
∗
) > 0.
Further, from inequality (3.6) we get
δ
k+1
≤
δ
2
k
(1−δ
0
)
2
≤
16
9
δ
2
k
.
Denoting
ˆ
δ
k
=
16
9
δ
k
, we get
ˆ
δ
k+1
≤
ˆ
δ
2
k
. Thus, for any k ≥ 1 we have
δ
k
=
9
16
ˆ
δ
k
≤
9
16
ˆ
δ
2
k
0
<
9
16
¡
1
2
¢
2
k
.
Using the upper bound in (3.7), we get (3.8). ut

12 Yu. Nesterov, B. Polyak
4. Global efficiency on specific problem classes
In the previous section, we have already seen that the modified Newton scheme
can be supported by a global efficiency estimate (3.4) on a general class of non-
convex problems. The main goal of this section is to show that on more specific
classes of non-convex problems the global performance of the scheme (3.3) is
much better. To the best of our knowledge, the results of this section are the
first global complexity results on a Newton-type scheme. The nice feature of
the scheme (3.3) consists in its ability to adjust the performance to a specific
problem class automatically.
4.1. Star-convex functions
Let us start from a definition.
Definition 1. We call a function f(x) star-convex if its set of global minimums
X
∗
is not empty and for any x
∗
∈ X
∗
and any x ∈ R
n
we have
f(αx
∗
+ (1 − α)x) ≤ αf(x
∗
) + (1 − α)f(x) ∀x ∈ F, ∀α ∈ [0, 1]. (4.1)
A particular example of a star-convex function is a usual convex function. How-
ever, in general, a star-convex function does not need to be convex, even for
scalar case. For instance, f(x) = |x|(1 − e
−|x|
), x ∈ R, is star-convex, but not
convex. Star-convex functions arise quite often in optimization problems related
to sum of squares, e.g. the function f (x, y) = x
2
y
2
+x
2
+y
2
belongs to this class.
Theorem 4. Assume that the objective function in (3.1) is star-convex and the
set F is bounded: diam F = D < ∞. Let sequence {x
k
} be generated by method
(3.3).
1. If f(x
0
) −f
∗
≥
3
2
LD
3
, then f(x
1
) −f
∗
≤
1
2
LD
3
.
2. If f(x
0
) − f
∗
≤
3
2
LD
3
, then the rate of convergence of process (3.3) is as
follows:
f(x
k
) −f(x
∗
) ≤
3LD
3
2(1+
1
3
k)
2
, k ≥ 0.
(4.2)
Proof. Indeed, in view of inequality (2.10), upper bound on the parameters M
k
and definition (4.1), for any k ≥ 0 we have:
f(x
k+1
) −f (x
∗
)
≤ min
y
£
f(y) − f(x
∗
) +
L
2
ky − x
k
k
3
: y = αx
∗
+ (1 −α)x
k
, α ∈ [0, 1]
¤
≤ min
α∈[0,1]
£
f(x
k
) −f(x
∗
) −α(f(x
k
) −f (x
∗
)) +
L
2
α
3
kx
∗
− x
k
k
3
¤
≤ min
α∈[0,1]
£
f(x
k
) −f(x
∗
) −α(f(x
k
) −f (x
∗
)) +
L
2
α
3
D
3
¤
.

Cubic regularization of Newton method and its global performance 13
The minimum of the objective function in the last minimization problem in
α ≥ 0 is achieved for
α
k
=
q
2(f(x
k
)−f(x
∗
))
3LD
3
.
If α
k
≥ 1, then the actual optimal value corresponds to α = 1. In this case
f(x
k+1
) −f(x
∗
) ≤
1
2
LD
3
.
Since the process (3.3) is monotone, this can happen only at the first iteration
of the method.
Assume that α
k
≤ 1. Then
f(x
k+1
) −f(x
∗
) ≤ f(x
k
) −f (x
∗
) −
£
2
3
(f(x
k
) −f(x
∗
))
¤
3/2
1
√
LD
3
.
Or, in a more convenient notation, that is α
2
k+1
≤ α
2
k
−
2
3
α
3
k
< α
2
k
. Therefore
1
α
k+1
−
1
α
k
=
α
k
−α
k+1
α
k
α
k+1
=
α
2
k
−α
2
k+1
α
k
α
k+1
(α
k
+α
k+1
)
≥
α
2
k
−α
2
k+1
2α
3
k
≥
1
3
.
Thus,
1
α
k
≥
1
α
0
+
k
3
≥ 1 +
k
3
, and (4.2) follows. ut
Let us introduce now a generalization of the notion of non-degenerate global
minimum.
Definition 2. We say that the optimal set X
∗
of function f(x) is globally non-
degenerate if there exists a constant γ > 0 such that for any x ∈ F we have
f(x) −f
∗
≥
γ
2
ρ
2
(x, X
∗
),
(4.3)
where f
∗
is the global minimal value of function f(x), and ρ(x, X
∗
) is the Eu-
clidean distance from x to X
∗
.
Of course, this property holds for strongly convex functions (in this case
X
∗
is a singleton), however it can also hold for some non-convex functions. As
an example, consider f (x ) = (kxk
2
− 1)
2
, X
∗
= {x : kxk = 1}. Moreover, if
the set X
∗
has a connected non-trivial component, the Hessians of the objective
function at these points cannot be non-degenerate. However, as we will see, in this
situation the modified Newton scheme ensures a super-linear rate of convergence.
Denote
¯ω =
1
L
2
¡
γ
2
¢
3
.
Theorem 5. Let function f (x) be star-convex. Assume that it has also a globally
non-degenerate optimal set. Then the performance of the scheme (3.3) on this
problem is as follows.
1. If f (x
0
)−f(x
∗
) ≥
4
9
¯ω, then at the first phase of the process we get the following
rate of convergence:
f(x
k
) −f(x
∗
) ≤
h
(f(x
0
) −f(x
∗
))
1/4
−
k
6
q
2
3
¯ω
1/4
i
4
.
(4.4)
This phase is terminated as soon as f(x
k
0
) −f (x
∗
) ≤
4
9
¯ω for some k
0
≥ 0.
2. For k ≥ k
0
the sequence converges superlinearly:
f(x
k+1
) −f(x
∗
) ≤
1
2
(f(x
k
) −f(x
∗
))
q
f(x
k
)−f(x
∗
)
¯ω
.
(4.5)

14 Yu. Nesterov, B. Polyak
Proof. Denote by x
∗
k
the projection of the point x
k
onto the optimal set X
∗
. In
view of inequality (2.10), upper bound on the parameters M
k
and definitions
(4.1), (4.3), for any k ≥ 0 we have:
f(x
k+1
) −f(x
∗
)
≤ min
α∈[0,1]
£
f(x
k
) −f (x
∗
) −α(f(x
k
) −f(x
∗
)) +
L
2
α
3
kx
∗
k
− x
k
k
3
¤
≤ min
α∈[0,1]
·
f(x
k
) −f(x
∗
) −α(f(x
k
) −f (x
∗
)) +
L
2
α
3
³
2
γ
(f(x
k
) −f(x
∗
))
´
3/2
¸
.
Denoting ∆
k
= (f (x
k
) −f(x
∗
))/¯ω, we get inequality
∆
k+1
≤ min
α∈[0,1]
h
∆
k
− α∆
k
+
1
2
α
3
∆
3/2
k
i
. (4.6)
Note that the first order optimality condition for α ≥ 0 in this problem is
α
k
=
q
2
3
∆
−1/2
k
.
Therefore, if ∆
k
≥
4
9
, we get
∆
k+1
≤ ∆
k
−
¡
2
3
¢
3/2
∆
3/4
k
.
Denoting u
k
=
9
4
∆
k
we get a simpler relation:
u
k+1
≤ u
k
−
2
3
u
3/4
k
,
which is applicable if u
k
≥ 1. Since the right-hand side of this inequality is
increasing for u
k
≥
1
16
, let us prove by induction that
u
k
≤
h
u
1/4
0
−
k
6
i
4
.
Indeed, inequality
h
u
1/4
0
−
k+1
6
i
4
≥
h
u
1/4
0
−
k
6
i
4
−
2
3
h
u
1/4
0
−
k
6
i
3
clearly is equivalent to
2
3
h
u
1/4
0
−
k
6
i
3
≥
h
u
1/4
0
−
k
6
i
4
−
h
u
1/4
0
−
k+1
6
i
4
=
1
6
[
h
u
1/4
0
−
k
6
i
3
+
h
u
1/4
0
−
k
6
i
2
h
u
1/4
0
−
k+1
6
i
+
h
u
1/4
0
−
k
6
ih
u
1/4
0
−
k+1
6
i
2
+
h
u
1/4
0
−
k+1
6
i
3
],
which is obviously true.
Finally, if u
k
≤ 1, then the optimal value for α in (4.6) is one and we get
(4.5). ut

Cubic regularization of Newton method and its global performance 15
4.2. Gradient-dominated functions
Let us study now another interesting class of problems.
Definition 3. A function f(x) is called gradient dominated of degree p ∈ [1, 2]
if it attains a global minimum at some point x
∗
and for any x ∈ F we have
f(x) −f (x
∗
) ≤ τ
f
kf
0
(x)k
p
, (4.7)
where τ
f
is a positive constant. The parameter p is called the degree of domina-
tion.
Note that we do not assume that the global minimum of function f is unique.
For p = 2, this class of functions has been introduced in [13].
Let us give several examples of gradient dominated functions.
Example 1. Convex functions. Let f be convex on R
n
. Assume it achieves its
minimum at point x
∗
. Then, for any x ∈ R
n
, kx − x
∗
k ≤ R, we have
f(x) −f (x
∗
) ≤ hf
0
(x), x − x
∗
i ≤ kf
0
(x)k · R.
Thus, on the set F = {x : kx − x
∗
k ≤ R}, function f is a gradient dominated
function of degree one with τ
f
= R . ut
Example 2. Strongly convex functions. Let f be differentiable and strongly con-
vex on R
n
. This means that there exists a constant γ > 0 such that
f(y) ≥ f (x) + hf
0
(x), y − xi +
1
2
γky − xk
2
, (4.8)
for all x, y ∈ R
n
. Then, (see, for example, [10], inequality (2.1.19)),
f(x) −f (x
∗
) ≤
1
2γ
kf
0
(x)k
2
∀x ∈ R
n
.
Thus, on the set F = R
n
, function f is a gradient dominated function of degree
two with τ
f
=
1
2γ
. ut
Example 3. Sum of squares. Consider a system of non-linear equations:
g(x) = 0 (4.9)
where g(x) = (g
1
(x), . . . , g
m
(x))
T
: R
n
→ R
m
is a differentiable function. We
assume that m ≤ n and that there exists a solution x
∗
to (4.9). Let us assume
in addition that the Jacobian
J(x) = (g
0
1
(x), . . . , g
0
m
(x))
is uniformly non-degenerate on a certain convex set F containing x
∗
. This means
that the value
σ ≡ inf
x∈F
λ
n
¡
J
T
(x)J(x)
¢

16 Yu. Nesterov, B. Polyak
is positive. Consider the function
f(x) =
1
2
m
X
i=1
g
2
i
(x).
Clearly, f(x
∗
) = 0. Note that f
0
(x) = J(x)g(x). Therefore
kf
0
(x)k
2
= h
¡
J
T
(x)J(x)
¢
g(x), g(x)i ≥ σkg(x)k
2
= 2σ(f(x) − f(x
∗
)).
Thus, f is a gradient dominated function on F of degree two with τ
f
=
1
2σ
. Note
that, for m < n, the set of solutions to (4.9) is not a singleton and therefore the
Hessians of function f are necessarily degenerate at the solutions. ut
In order to study the complexity of minimization of the gradient dominated
functions, we need one auxiliary result.
Lemma 7. At each step of method (3.3) we can guarantee the following decrease
of the objective function:
f(x
k
) −f(x
k+1
) ≥
L
0
·kf
0
(x
k+1
)k
3/2
3
√
2·(L+L
0
)
3/2
, k ≥ 0.
(4.10)
Proof. In view of inequalities (2.11) and (2.9) we get
f(x
k
) −f (x
k+1
) ≥
M
k
12
r
3
M
k
(x
k
) ≥
M
k
12
³
2kf
0
(x
k+1
)k
L+M
k
´
3/2
=
M
k
kf
0
(x
k+1
)k
3/2
3
√
2·(L+M
k
)
3/2
.
It remains to note that the right-hand side of this inequality is increasing in
M
k
≤ 2L. Thus, we can replace M
k
by its lower bound L
0
. ut
Let us start from the analysis of the gradient dominated functions of degree
one. The following theorem states that the process can be partitioned into two
phases. The initial phase (with large values of the objective function) terminates
fast enough, while at the second phase we have O(1/k
2
) rate of convergence.
Theorem 6. Let us apply method (3.3) to minimization of a gradient dominated
function f(x) of degree p = 1.
1. If the initial value of the objective function is large enough:
f(x
0
) −f(x
∗
) ≥ ˆω ≡
18
L
2
0
τ
3
f
· (L + L
0
)
3
,
then the process converges to the region L(ˆω) superlinearly:
ln
¡
1
ˆω
(f(x
k
) −f(x
∗
)
¢
≤
¡
2
3
¢
k
ln
¡
1
ˆω
(f(x
0
) −f(x
∗
)
¢
.
(4.11)
2. If f(x
0
) − f(x
∗
) ≤ γ
2
ˆω for some γ > 1, then we have the following estimate
for the rate of convergence:
f(x
k
) −f(x
∗
) ≤ ˆω ·
γ
2
(
2+
3
2
γ
)
2
(
2+
(
k+
3
2
)
·γ
)
2
, k ≥ 0. (4.12)

Cubic regularization of Newton method and its global performance 17
Proof. Using inequalities (4.10) and (4.7) with p = 1, we get
f(x
k
) −f(x
k+1
) ≥
L
0
·(f(x
k+1
)−f(x
∗
))
3/2
3
√
2·(L+L
0
)
3/2
·τ
3/2
f
= ˆω
−1/2
(f(x
k+1
) −f(x
∗
))
3/2
.
Denoting δ
k
= (f (x
k
) −f (x
∗
))/ˆω, we obtain
δ
k
− δ
k+1
≥ δ
3/2
k+1
. (4.13)
Hence, as far as δ
k
≥ 1, we get
ln δ
k
≤
¡
2
3
¢
k
ln δ
0
,
and that is (4.11).
Let us prove now inequality (4.12). Using inequality (4.13), we have
1
√
δ
k+1
−
1
√
δ
k
≥
1
√
δ
k+1
−
1
p
δ
k+1
+δ
3/2
k+1
=
p
δ
k+1
+δ
3/2
k+1
−
√
δ
k+1
√
δ
k+1
p
δ
k+1
+δ
3/2
k+1
=
1
p
1+
√
δ
k+1
·
³
1+
p
1+
√
δ
k+1
´
=
1
1+
√
δ
k+1
+
p
1+
√
δ
k+1
≥
1
2+
3
2
√
δ
k+1
≥
1
2+
3
2
√
δ
0
.
Thus,
1
δ
k
≥
1
γ
+
k
2+
3
2
γ
, and this is (4.12). ut
The reader should not be confused by the superlinear rate of convergence
established by (4.11). It is valid only for the first stage of the process and de-
scribes a convergence to the set L(ˆω). For example, the first stage of the process
discussed in Theorem 4 is even shorter: it takes a single iteration.
Let us look now at the gradient dominated functions of degree two. Here two
phases of the process can be indicated as well.
Theorem 7. Let us apply method (3.3) to minimization of a gradient dominated
function f(x) of degree p = 2.
1. If the initial value of the objective function is large enough:
f(x
0
) −f(x
∗
) ≥ ˜ω ≡
L
4
0
324(L+L
0
)
6
τ
3
f
,
(4.14)
then at its first phase the process converges as follows:
f(x
k
) −f (x
∗
) ≤ (f(x
0
) −f (x
∗
)) ·e
−k·σ
, (4.15)
where σ =
˜ω
1/4
˜ω
1/4
+(f(x
0
)−f(x
∗
))
1/4
. This phase ends on the first iteration k
0
, for
which (4.14) does not hold.
2. For k ≥ k
0
the rate of convergence is super-linear:
f(x
k+1
) −f (x
∗
) ≤ ˜ω ·
³
f(x
k
)−f(x
∗
)
˜ω
´
4/3
.
(4.16)

18 Yu. Nesterov, B. Polyak
Proof. Using inequalities (4.10) and (4.7) with p = 2, we get
f(x
k
) −f(x
k+1
) ≥
L
0
·(f(x
k+1
)−f(x
∗
))
3/4
3
√
2·(L+L
0
)
3/2
·τ
3/4
f
= ˜ω
1
/
4
(f(x
k+1
) −f (x
∗
))
3
/
4
.
Denoting δ
k
= (f (x
k
) −f (x
∗
))/˜ω, we obtain
δ
k
≥ δ
k+1
+ δ
3/4
k+1
. (4.17)
Hence,
δ
k
δ
k+1
≥ 1 + δ
−1/4
k
≥ 1 + δ
−1/4
0
=
1
1−σ
≥ e
σ
,
and we get (4.15). Finally, from (4.17) we have δ
k+1
≤ δ
4/3
k
, and that is (4.16).
ut
Comparing the statement of Theorem 7 with other theorems of this section we
see a significant difference: this is the first time when the initial gap f (x
0
)−f (x
∗
)
enters the complexity estimate of the first phase of the process in a polynomial
way; in all other cases the dependence on this gap is much weaker.
Note that it is possible to embed the gradient dominated functions of degree
one into the class of gradient dominated functions of degree two. However, the
reader can check that this only spoils the efficiency estimates established by
Theorem 7.
4.3. Nonlinear transformations of convex functions
Let u(x) : R
n
→ R
n
be a non-degenerate operator. Denote by v(u) its inverse:
v(u) : R
n
→ R
n
, v(u(x)) ≡ x.
Consider the following function:
f(x) = φ(u(x)),
where φ(u) is a convex function with bounded level sets. Such classes are typical
for minimization problems with composite objective functions. Denote by x
∗
≡
v(u
∗
) its minimum. Let us fix some x
0
∈ R
n
. Denote
σ = max
u
{kv
0
(u)k : φ(u) ≤ f(x
0
)},
D = max
u
{ku −u
∗
k : φ(u) ≤ f(x
0
)}.
The following result is straightforward.
Lemma 8. For any x, y ∈ L(f(x
0
)) we have
kx −yk ≤ σku(x) − u(y)k. (4.18)

Cubic regularization of Newton method and its global performance 19
Proof. Indeed, for x, y ∈ L(f(x
0
)), we have φ(u(x)) ≤ f(x
0
) and φ(u(y)) ≤
f(x
0
). Consider the trajectory x(t) = v(tu(y) + (1 − t)u(x)), t ∈ [0, 1]. Then
y − x =
1
R
0
x
0
(t)dt =
µ
1
R
0
v
0
(tu(y) + (1 − t)u(x))dt
¶
· (u(y) −u(x)),
and (4.18) follows. ut
The following result is very similar to Theorem 4.
Theorem 8. Assume that function f has Lipschitz continuous Hessian on F ⊇
L(f(x
0
)) with Lipschitz constant L. And let the sequence {x
k
} be generated by
method (3.3).
1. If f(x
0
) −f
∗
≥
3
2
L(σD)
3
, then f(x
1
) −f
∗
≤
1
2
L(σD)
3
.
2. If f (x
0
) −f
∗
≤
3
2
L(σD)
3
, then the rate of convergence of the process (3.3) is
as follows:
f(x
k
) −f(x
∗
) ≤
3L(σ D)
3
2(1+
1
3
k)
2
, k ≥ 0.
(4.19)
Proof. Indeed, in view of inequality (2.10), upper bound on the parameters M
k
and definition (4.1), for any k ≥ 0 we have:
f(x
k+1
) −f (x
∗
) ≤ min
y
[ f(y) − f(x
∗
) +
L
2
ky − x
k
k
3
:
y = v(αu
∗
+ (1 −α)u(x
k
)), α ∈ [0, 1] ].
By definition of points y in the above minimization problem and (4.18), we have
f(y) − f(x
∗
) = φ(αu
∗
+ (1 − α)u(x
k
)) −φ(u
∗
) ≤ (1 − α)(f(x
k
) −f (x
∗
)),
ky − x
k
k ≤ ασku(x
k
) −u
∗
k ≤ ασD.
This means that the reasoning of Theorem 4 goes through with replacement D
by σD. ut
Let us prove a statement on strongly convex φ. Denote ˇω =
1
L
2
¡
γ
2σ
2
¢
3
.
Theorem 9. Let function φ be strongly convex with convexity parameter γ > 0.
Then, under assumptions of Theorem 8, the performance of the scheme (3.3) is
as follows.
1. If f (x
0
)−f(x
∗
) ≥
4
9
ˇω, then at the first phase of the process we get the following
rate of convergence:
f(x
k
) −f(x
∗
) ≤
h
(f(x
0
) −f(x
∗
))
1/4
−
k
6
q
2
3
ˇω
1/4
i
4
.
(4.20)
This phase is terminated as soon as f(x
k
0
) −f (x
∗
) ≤
4
9
ˇω for some k
0
≥ 0.
2. For k ≥ k
0
the sequence converges superlinearly:
f(x
k+1
) −f(x
∗
) ≤
1
2
(f(x
k
) −f(x
∗
))
q
f(x
k
)−f(x
∗
)
ˇω
.
(4.21)

20 Yu. Nesterov, B. Polyak
Proof. Indeed, in view of inequality (2.10), upper bound on the parameters M
k
and definition (4.1), for any k ≥ 0 we have:
f(x
k+1
) −f (x
∗
) ≤ min
y
[ f(y) − f(x
∗
) +
L
2
ky − x
k
k
3
:
y = v(αu
∗
+ (1 −α)u(x
k
)), α ∈ [0, 1] ].
By definition of points y in the above minimization problem and (4.18), we have
f(y) − f(x
∗
) = φ(αu
∗
+ (1 − α)u(x
k
)) −φ(u
∗
) ≤ (1 − α)(f(x
k
) −f (x
∗
)),
ky − x
k
k ≤ ασku(x
k
) −u
∗
k ≤ ασ
q
2
γ
(f(x
0
) −f(x
∗
)).
This means that the reasoning of Theorem 5 goes through with replacement L
by σ
3
L. ut
Note that the functions discussed in this section are often used as test func-
tions for non-convex optimization algorithms.
5. Implementation issues
5.1. Solving the cubic regularization
Note that the auxiliary minimization problem (2.4), which we need to solve in
order to compute the mapping T
M
(x), namely,
min
h∈R
n
£
hg, hi +
1
2
hHh, hi +
M
6
khk
3
¤
,
(5.1)
is substantially nonconvex. It can have isolated strict local minima, while we
need to find a global one. Nevertheless, as we will show in this section, this
problem is equivalent to a convex one-dimensional optimization problem.
Before we present an “algorithmic” proof of this fact, let us provide it with
a general explanation. Introduce the following objects:
ξ
1
(h) = hg, hi +
1
2
hHh, hi, ξ
2
(h) = khk
2
,
Q =
©
ξ = (ξ
(1)
, ξ
(2)
)
T
: ξ
(1)
= ξ
1
(h), ξ
(2)
= ξ
2
(h), h ∈ R
n
ª
⊂ R
2
,
ϕ(ξ) = ξ
(1)
+
M
6
¡
ξ
(2)
¢
3/2
+
.
where (a)
+
= max{a, 0}. Then
min
h∈R
n
£
hg, hi +
1
2
hHh, hi +
M
6
khk
3
¤
≡ min
h∈R
n
h
ξ
1
(h) +
M
6
ξ
3/2
2
(h)
i
= min
ξ∈Q
ϕ(ξ).

Cubic regularization of Newton method and its global performance 21
Theorem 2.2 in [14] guarantees that for n ≥ 2 the set Q is convex and closed.
Thus, we have reduced the initial nonconvex minimization problem in R
n
to
a convex constrained minimization problem in R
2
. Up to this moment, this
reduction is not constructive, because Q is given in implicit form. However, the
next statement shows that the description of this set is quite simple.
Denote
v
u
(h) = hg, hi +
1
2
hHh, hi +
M
6
khk
3
, h ∈ R
n
,
and
v
l
(r) = −
1
2
h
¡
H +
Mr
2
I
¢
−1
g, gi −
M
12
r
3
.
For the first function sometimes we use the notation v
u
(g; h). Denote
D = {r ∈ R : H +
M
2
rI Â 0, r ≥ 0}.
Theorem 10. For any M > 0 we have the following relation:
min
h∈R
n
v
u
(h) = sup
r∈D
v
l
(r). (5.2)
For any r ∈ D, direction h(r) = −
¡
H +
Mr
2
I
¢
−1
g satisfies equation
0 ≤ v
u
(h(r)) −v
l
(r) =
M
12
(r + 2kh(r)k)(kh(r)k − r)
2
=
4
3M
·
r+2kh(r)k
(r+kh(r)k)
2
· v
0
l
(r)
2
.
(5.3)
Proof. Denote the left-hand side of relation (5.2) by v
∗
u
, and its right-hand side
by v
∗
l
. Let us show that v
∗
u
≥ v
∗
l
. Indeed,
v
∗
u
= min
h∈R
n
£
hg, hi +
1
2
hHh, hi +
M
6
khk
3
¤
= min
h∈R
n
,
τ=khk
2
h
hg, hi +
1
2
hHh, hi +
M
6
(τ)
3/2
+
i
= min
h∈R
n
,
τ∈R
sup
r∈R
h
hg, hi +
1
2
hHh, hi +
M
6
(τ)
3/2
+
+
M
4
r
¡
khk
2
− τ
¢
i
≥ sup
r∈D
min
h∈R
n
,
τ∈R
h
hg, hi +
1
2
hHh, hi +
M
6
(τ)
3/2
+
+
M
4
r
¡
khk
2
− τ
¢
i
≡ v
∗
l
.
Consider now an arbitrary r ∈ D . Then
g = −Hh(r) −
M
2
rh(r).
Therefore
v
u
(h(r)) = hg, h(r)i +
1
2
hHh(r), h(r)i +
M
6
kh(r)k
3
= −
1
2
hHh(r), h(r)i −
M
2
rkh(r)k
2
+
M
6
kh(r)k
3
= −
1
2
h
¡
H +
Mr
2
I
¢
h(r), h(r)i −
M
4
rkh(r)k
2
+
M
6
kh(r)k
3
= v
l
(r) +
M
12
r
3
−
M
4
rkh(r)k
2
+
M
6
kh(r)k
3
= v
l
(r) +
M
12
(r + 2kh(r)k) · (kh(r)k − r)
2
.

22 Yu. Nesterov, B. Polyak
Thus, relation (5.3) is proved.
Note that
v
0
l
(r) =
M
4
(kh(r)k
2
− r
2
).
Therefore, if the optimal value v
∗
l
is attained at some r
∗
> 0 from D, then
v
0
l
(r
∗
) = 0 and by (5.3) we conclude that v
∗
r
= v
∗
l
. If r
∗
=
2
M
(−λ
n
(H))
+
, then
equality (5.2) can be justified by continuity arguments (since v
∗
u
≡ v
∗
u
(g) is a
concave function in g ∈ R
n
; see also the discussion below). ut
Note that Proposition 1 follows from the definition of set D.
Theorem 10 demonstrates that in non-degenerate situation the solution of
problem (5.2) can be found from one-dimensional equation
r = k
¡
H +
Mr
2
I
¢
−1
gk, r ≥
2
M
(−λ
n
(H))
+
.
(5.4)
A technique for solving such equations is very well developed for the needs of
trust region methods (see [2], Chapter 7, for exhaustive expositions of the dif-
ferent approaches). As compared with (5.4), the equation arising in trust region
schemes has a constant left-hand side. But of course, all possible difficulties in
this equation are due to the non-linear (convex) right-hand side.
For completeness of presentation, let us briefly discuss the structure of equa-
tion (5.4). In the basis of eigenvectors of matrix H this equation can be written
as
r
2
=
n
P
i=1
˜g
2
i
(λ
i
+
M
2
r)
2
, r ≥
2
M
(−λ
n
)
+
,
(5.5)
where λ
i
are eigenvalues of matrix H and ˜g
i
are coordinates of vector g in the
new basis.
If ˜g
n
6= 0, then the solution r
∗
of equation (5.5) is in the interior of the
domain:
r >
2
M
(−λ
n
)
+
,
and we can compute the displacement h(r
∗
) by the explicit expression:
h(g; r
∗
) = −
³
H +
Mr
∗
2
I
´
−1
g.
If ˜g
n
= 0 then this formula does not work and we have to consider different
cases. In order to avoid all these complications, let us mention the following
simple result.
Lemma 9. Let ˜g
n
= 0. Define g(²) = ˜g + ²e
n
, where e
n
is the nth coordinate
vector. Denote by r
∗
(²) the solution of equation (5.5) with ˜g = g(²). Then any
limit point of the trajectory
h(g(²); r
∗
(²)), ² → 0,
is a global minimum in h of function v
u
(g; h).

Cubic regularization of Newton method and its global performance 23
Proof. Indeed, function v
∗
u
(g) is concave for g ∈ R
n
. Therefore it is continuous.
Hence,
v
∗
u
(g) = lim
²→0
v
∗
u
(g(²)) = lim
²→0
v
u
(g(²); h(g(²); r
∗
(²))).
It remains to note that the function v
u
(g; h) is continuous in both arguments.
ut
In order to illustrate the difficulties arising in the dual problem, let us look
at an example.
Example 4. Let n = 2 and
˜g =
µ
−1
0
¶
, λ
1
= 0, λ
2
= −1, M = 1.
Thus, our primal problem is as follows:
min
h∈R
2
(
ψ(h) ≡ −h
(1)
−
1
2
¡
h
(2)
¢
2
+
1
6
·
q
¡
h
(1)
¢
2
+
¡
h
(2)
¢
2
¸
3
)
.
Following to (2.5), we have to solve the following system of non-linear equations:
h
(1)
2
q
¡
h
(1)
¢
2
+
¡
h
(2)
¢
2
= 1,
h
(2)
2
q
¡
h
(1)
¢
2
+
¡
h
(2)
¢
2
= h
(2)
.
Thus, we have three candidate solutions:
h
∗
1
=
µ
√
2
0
¶
, h
∗
2
=
µ
1
√
3
¶
, h
∗
3
=
µ
1
−
√
3
¶
.
By direct substitution we can see that
ψ(h
∗
1
) = −
2
√
2
3
> −
7
6
= ψ(h
∗
2
) = ψ(h
∗
3
).
Thus, both h
∗
2
and h
∗
3
are our global solutions.
Let us look at the dual problem:
sup
r
·
φ(r) ≡ −
r
3
12
−
1
2
·
1
0+
1
2
r
= −
r
3
12
−
1
r
: −1 +
1
2
r > 0
¸
.
Note that φ
0
(r) = −
r
2
4
+
1
r
2
. Thus, φ
0
(2) = −
3
4
< 0 and we conclude that
r
∗
= 2, φ
∗
= −
7
6
.
However, r
∗
does not satisfy the equation φ
0
(r) = 0 and the object h(r
∗
) is not
defined. ut

24 Yu. Nesterov, B. Polyak
Let us conclude this section with a precise description of the solution of primal
problem in (5.2) in terms of the eigenvalues of matrix H. Denote by {s
i
}
n
i=1
an
orthonormal basis of eigenvectors of H, and let
ˆ
k satisfy the conditions
˜g
(i)
6= 0 for i <
ˆ
k,
˜g
(i)
= 0 for i ≥
ˆ
k.
Assume that r
∗
is the solution to the dual problem in (5.2). Then the solution
of the primal problem is given by the vector
h
∗
= −
ˆ
k−1
X
i=1
˜g
(i)
s
i
λ
i
+
M
2
r
∗
+ σs
n
,
where σ is chosen in accordance to the condition kh
∗
k = r
∗
. Note that this rule
works also for
ˆ
k = 1 or
ˆ
k = n + 1.
We leave justification of above rule as an exercise for the reader. As far as we
know, a technique for finding h
∗
without computation of the basis of eigenvalues
is not known yet.
5.2. Line search strategies
Let us discuss the possible computational cost of Step 1 in the method (3.3),
which consists of finding M
k
∈ [L
0
, 2L] satisfying the equation:
f(T
M
k
(x)) ≤
¯
f
M
k
(x
k
).
Note that for M
k
≥ L this inequality holds. Consider now the strategy
while f(T
M
k
(x)) > f(x
k
) do M
k
:= 2M
k
; M
k+1
:= M
k
.
(5.6)
It is clear that if we start the process (3.3) with any M
0
∈ [L
0
, 2L], then the
above procedure, as applied at each iteration of the method, has the following
advantages:
– M
k
≤ 2L.
– The total amount of additional computations of the mappings T
M
k
(x) during
the whole process (3.3) is bounded by
log
2
2L
L
0
.
This amount does not depend on the number of iterations in the main process.
However, it may be that the rule (5.6) is too conservative. Indeed, we can
only increase our estimate for the constant L and never come back. This may
force the method to take only the short steps. A more reasonable strategy looks
as follows:
while f(T
M
k
(x)) > f(x
k
) do M
k
:= 2M
k
;
x
k+1
:= T
M
k
(x
k
); M
k+1
:= max{
1
2
M
k
, L
0
}.
(5.7)

Cubic regularization of Newton method and its global performance 25
Then it is easy to prove by induction that N
k
, the total number of computations
of mappings T
M
(x) made by (5.7) during the first k iterations, is bounded as
follows:
N
k
≤ 2k + log
2
M
k
L
0
.
Thus, if N is the number of iterations in this process, then we compute at most
2N + log
2
2L
L
0
mappings T
M
(x). That seems to be a reasonable price to pay for the possibility
to go by long steps.
6. Discussion
Let us compare the results presented in this paper with some known facts on
global efficiency of other minimization schemes. Since there is almost no such
results for non-convex case, let us look at a simple class of convex problems.
Assume that function f(x) is strongly convex on R
n
with convexity param-
eter γ > 0 (see (4.8)). In this case there exists its unique minimum x
∗
and
condition (4.3) holds for all x ∈ R
n
(see, for example, [10], Section 2.1.3). As-
sume also that Hessian of f(x) is Lipschitz continuous:
kf
00
(x) −f
00
(y)k ≤ Lkx − yk, ∀x, y ∈ R
n
.
For such functions, let us obtain the complexity bounds of method (3.3) using
the results of Theorems 4 and 5.
Let us fix some x
0
∈ R
n
. Denote by D the radius of its level set:
D = max
x
{kx −x
∗
k : f(x) ≤ f(x
0
)}.
From (4.3) we get
D ≤
h
2
γ
(f(x
0
) −f (x
∗
))
i
1/2
.
We will see that it is natural to measure the quality of starting point x
0
by the
following characteristic:
κ ≡ κ(x
0
) =
LD
γ
.
Let us introduce three switching values
ω
0
=
γ
3
18L
2
≡
4
9
¯ω, ω
1
=
3
2
γD
2
, ω
2
=
3
2
LD
3
.
In view of Theorem 4, we can reach the level f(x
0
) − f(x
∗
) ≤
1
2
LD
3
in one
additional iteration. Therefore without loss of generality we assume that
f(x
1
) −f(x
∗
) ≤ ω
2
.
Assume also that we are interested in a very high accuracy of the solution. Note
that the case κ ≤ 1 is very easy since the first iteration of method (3.3) comes
very close to the region of super-linear convergence (see Item 2 of Theorem 5).

26 Yu. Nesterov, B. Polyak
Consider the case κ ≥ 1. Then ω
0
≤ ω
1
≤ ω
2
. Let us estimate the duration
of the following phases:
Phase 1: ω
1
≤ f(x
i
) ≤ ω
2
,
Phase 2: ω
0
≤ f(x
i
) ≤ ω
1
,
Phase 3: ² ≤ f(x
i
) ≤ ω
0
.
In view of Theorem 4, the duration k
1
of the first phase is bounded as follows:
ω
1
≤
3LD
3
2(1+
1
3
k
1
)
2
.
Thus, k
1
≤ 3
√
κ. Further, in view of Item 1 of Theorem 5, we can bound the
duration k
2
of the second phase:
ω
1/4
0
≤ (f (x
k
1
+1
) −f (x
∗
))
1/4
−
k
2
6
ω
1/4
0
≤ (
1
2
γD
2
)
1/4
−
k
2
6
ω
1/4
0
.
This gives the following bound: k
2
≤ 3
3/4
2
1/2
√
κ ≤ 3.25
√
κ.
Finally, denote δ
k
=
1
4ω
0
(f(x
k
) −f (x
∗
)). In view of inequality (4.5) we have:
δ
k+1
≤ δ
3/2
k
, k ≥
¯
k ≡ k
1
+ k
2
+ 1.
At the same time f(x
¯
k
) − f(x
∗
) ≤ ω
0
. Thus, δ
¯
k
≤
1
4
, and the bound on the
duration k
3
of the last phase can be found from inequality
4
(
3
2
)
k
3
≤
4ω
0
²
.
That is k
3
≤ log
3
2
log
4
2γ
3
9²L
2
. Putting all bounds together, we obtain that the
total number of steps N in (3.3) is bounded as follows:
N ≤ 6.25
q
LD
γ
+ log
3
2
³
log
4
1
²
+ log
4
2γ
3
9L
2
´
.
(6.1)
It is interesting that in estimate (6.1) the parameters of our problem interact
with accuracy in an additive way. Recall that usually such an interaction is
multiplicative. Let us estimate, for example, the complexity of our problem for so
called “optimal first-order method” for strongly convex functions with Lipschitz
continuous gradient (see [10], Section 2.2.1). Denote by
ˆ
L the largest eigenvalue
of matrix f
00
(x
∗
). Then can guarantee that
γI ≤ f
00
(x) ≤ (
ˆ
L + LD)I ∀x, kx − x
∗
k ≤ D.
Thus, the complexity bound for the optimal method is of the order
O
µ
q
ˆ
L+LD
γ
ln
(
ˆ
L+LD)D
2
²
¶
iterations. For gradient method it is much worse:
O
³
ˆ
L+LD
γ
ln
(
ˆ
L+LD)D
2
²
´
.

Cubic regularization of Newton method and its global performance 27
Thus, we conclude that the global complexity estimates of the modified New-
ton scheme (3.3) are incomparably better than the estimates of the gradient
schemes. At the same time, we should remember, of course, about the difference
in the computational cost of each iteration.
Note that the similar bounds can be obtained for other classes of non-convex
problems. For example, for nonlinear transformations of convex functions (see
Section 4.3), the complexity bound is as follows:
N ≤ 6.25
q
σ
γ
LD + log
3
2
³
log
4
1
²
+ log
4
2γ
3
9σ
6
L
2
´
.
(6.2)
To conclude, note that in scheme (3.3) it is possible to find elements of
Levenberg-Marquardt approach (see relation (2.7)), or a trust-region approach
(see Theorem 10 and related discussion), or a line-search technique (see the rule
of Step 1 in (3.3)). However, all these facts are consequences of the main idea of
the scheme, that is the choice of the next test point as a global minimizer of the
upper second-order approximation of objective function.
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