Let f be a continuous function defined on the interval [0, 1]. For each x in [0, 1], are there a and b in (0, 1) such that f(x)=af(a) +(1-b)f(b) ?

This is not possible. Counterexample is

f(x)=sin(2 pi x), which at both endpoints is 0.

Choose any x not equal 0, 0.5, or 1. The sine is nonzero there. You cannot express it as a sum of two zeros.

Árpád Bűrmen, the problem asks to show that given any x in the closed interval [0,1], we can find a and b in the open interval (0,1) such that af(a) +(1-b)f(b)=f(x), where f:[0,1]->R is continuous on [0,1]. If x is in the open interval (0,1), as in your example, then we can simply choose a=x and b=x, since then af(a) +(1-b)f(b)=xf(x) +(1-x)f(x)=f(x), and the statelent holds. Note that in this case; i.e., if x is not equal to either of the endpoints 0 or 1, the continuity of f is not needed. The difficult part is to show that there also exist such a and b for x=0 or 1.

We can show the statement for x=0 or 1 with the additional restriction that f(0)f(1)>0. In this case, setting b=1-a, we have af(a) +(1-b)f(b)=af(a) +af(1-a)=a(f(a)+f(1-a)). Note that for every a in (0,1), we have that 1-a is also in (0,1); hence, b is in (0,1), as needed. Now, considering the function F(a)=a(f(a)+f(1-a)) from [0,1] to R, we have that it is continuous on [0,1], and F(0)=0 and F(1)=f(1)+f(0). Since F is continuous on [0,1], we know that the image of F contains the closed interval [0,f(1)+f(0)] if f(0) and f(1) are both positive, or the closed interval [f(1)+f(0),0] if f(0) and f(1) are both negative. Further, we observe that both f(0) and f(1) are in the open interval (0,f(1)+f(0)) if f(0) and f(1) are both positive, or in the open interval (f(1)+f(0),0) if f(0) and f(1) are both negative. Hence, by the intermediate value theorem, in both cases, there exist a₁ and a₂ in (0,1) such that F(a₁)=f(0) and F(a₂)=f(1), which is what we wanted to prove.

Actually, the statement is false for every function f:[0,1]->R that is continuous and strictly increasing on [0,1], and vanishes at 0. For instance, if f:[0,1]->R such that f(x)=x, then there do not exist a and b in (0,1) such that f(0)=af(a)+(1-b)f(b), since the left side vanishes, while the right side is strictly positive.

It is a rather silly problem, isn't it? Unless it has been published with typos back in 1987 (it's all a blur in my mind, age is always a factor...)

Perhaps, this is not that much of a silly problem if it is slightly adjusted, i.e., if you tweak it a little - for example, by replacing the left-hand side with f(x)/2, and requiring that a-and-b be different from x, and finally requiring that f(x) be a positive function on [0,1].

I could be wrong, but it seems to me that the statement should be true in this case.

Technically,

f(x)/2 = area of the triangle with vertices at the points (0,0), (1,0) and (x, f(x))

then

f(x)/2 = x*f(x)/2 + (1-x)*f(x)/2 =

= area of the triangle with vertices at the points (0,0), (x,0) and (x, f(x))

+ plus +

area of the triangle with vertices at the points (x,0), (x, f(x)) and (1,0).

And, a*f(a) = area of rectangle

with vertices at the points (0,0), (a,0), (a, f(a)) and (0, f(a))

as well (1-b)*f(b)= also area of another rectangle

with vertices at the points (b,0), (1,0), (1, f(b)) and (b, f(b)).

Then we apply the Intermediate Value theorem for the function a*f(a) on the interval [0, x] where at the end points the values of the function are 0 and x*f(x).

Since x*f(x)/2 is between the two values then there exists a : such that

a*f(a) = x*f(x)/2.

Similar arguing is valid for the function (1-b)*f(b) on the interval [x,1].

Basically, the intermediate value theorem doesn't care about the positivity of the function f(x) - - it is useful only for a geometric interpretation, i.e., I guess it is enough to require x be such that f(x) be different from zero at a chosen point in (0,1).

Perhaps other adjustments might be needed here.

Thank you kindly, Steftcho ! You saved this problem from oblivion with a beautiful argument.

If I may, let me recap for our colleagues and readers, that Steftcho proved the following statement:

One remark. If f(x) = 0 we can take a_α = 0 and b_α = 1 (for any α > 1). If f(x) ≠ 0, then 0 < a_α < x < b_α < 1 (the inequalities are strict).

Another remark. If we want to introduce x = 0 and x = 1 into the problem, then:

• If f(0) = 0 or f(1) = 0 we can take a_α = 0 and b_α = 1 for any α > 1;

• If f(0) ≠ 0 take a_α = 0;

• If f(1) ≠ 0 take b_α = 1.

Great summary and generalization, thanks , in my case α = 2 was just making good geometric visualization/ interpretation.

It may be useful to note that since for the proof of the previous, modified version of the initial statement only the intermediate value property of f is needed, it is not necessary that f be continuous; it suffices to be a Darboux function on [0,1]; i.e., a function that has the intermediate value property on [0,1].

You are very welcome, Steftcho, and thank you again.

You are correct, Spiros, Darboux functions are enough for this problem.

Related problems:

1. Let f be a continuous real function defined on the interval [0, 1]. Are there a,b in [0,1] such that af(a)+(1-b)f(1-b)=0 ?

2. Let f be a continuous real function defined on the interval [0, 1]. Are there a,b in [0,1] such that af(a)-(1-b)f(1-b)=0 ?

3. Let f be a continuous real function defined on the interval [0, 1]. For each x in [0, 1], are there a and b in (0, 1) such that f(x)=af(1-a) +(1-b)f(b) ?

4. Let f be a continuous real function defined on the interval [0, 1]. For each x in [0, 1], are there a and b in (0, 1) such that f(x)=af(a) +(1-b)f(1-b) ?

Very good questions, Dinu, thank you kindly. With the understanding that we consider the non-trivial cases a ≠ b, a + b ≠ 1 (as the case may be for each one of the questions).

Yes George, of course !

Regarding problems 1-4, we refer, of course, to non-constant functions.

Since for every real number b in [0,1], the real number 1-b is also in [0,1], and conversely, the equations to be shown in problems 1,2, and 4 could be rewritten more conveniently as af(a)+bf(b)=0, af(a)-bf(b)=0, and f(x)=af(a)+bf(b), respectively, where a=/b. Note also that the problem 3 is the initial question of the thread, with the additional restriction that a=/b (i suppose this applies) or a+b=/1, or both.

What actually asks the problem 2 is to say whether the function xf(x) is injective or not if f is continuous on [0,1]. One should make extra assumptions to decide, and the same applies to problem 1.

Comme que la fonction f(x) est définit sur l’intervalle; 0 ≤ f(x) ≤ 1

On définit a et b appartenant à l’intervalle (0,1); 0 ≤ a ≤ 1 ; 0 ≤ b ≤ 1

Cela donne 0 ≤ f(a) ≤ 1 et 0 ≤ f(b) ≤ 1

Donc 0 ≤ a f(a) ≤ a 0 ≤ a f(x) ≤ 1 et 0 ≤ b f(b) ≤ b 0 ≤ b f(b) ≤ 1

cela done encore : 0 ≤ a f(a)+ f(b) ≤ 2

0 ≤ a f(a)+ f(b) -bf(b) ≤ 2-1 ≤ 1

Donc 0 ≤ a f(a)+ (1-b)f(b) ≤ 1

Et comme que f(x) est définit tel que 0 ≤ f(x) ≤ 1 donc on peut toujours trouver a et b appartenant à l’intervalle qui permettent d’avoir f(x) = a f(a)+ (1-b)f(b).

Ce serait correct, Hademine, mais la fonction f est seulement définie sur [0, 1], elle ne prend pas de valeurs dans [0, 1], i.e., 0 ≤ f(x) ≤ 1 est faux.

If f:[0,1]->R is a (continuous) function, then every expression of the form xf(y) ( x,y in (0,1] ) is in fact the area of a rectangle, a positive area if f(y)>0 and a negative area if f(y)<0.

As we easy can see, George's problem and 1-4 problems have connections with such areas. In this context, problem(question) 1 is the simplest, because if f>0 or f<0 on [0,1], then the answer about problem 1 is NO.

Suggest this fact we need the assumption f(c)=0 for at least one point in (0,1] ?

Anyway, George's problem and problems 3 and 4 are clearly more difficult, because, among others, they require a quantity representing a lenght is equal to a quantity representing the sum of two areas of bounded objects from the euclidean plane !

The answer to the question in Dinu’s last post is yes. Actually, that is a sufficient condition, too; i.e., if f changes sign on [0,1], then the answer to problems 1 and 2 is yes.

Proof

Assuming that f changes sign on [0,1], we have that there exists x₀ on (0,1] such that f(x₀)=0. Indeed, suppose to the contrary that for every x₀ on (0,1] we have f(x₀)=/0. Since f is continuous on (0,1], we have that f has constant sign on (0,1]. Further, since f is continuous on (0,1], we have that f(0)=0 or f(0) has the same sign as f(x) for every x in (0,1], and in both cases, the function f does not change sign on [0,1], which is a contradiction.

Since f(x₀)=0 and x₀=/0, choosing a=0 and b=x₀, we have af(a)+bf(b)=0 and af(a)-bf(b)=0, where a=/b, and the conditions of problems 1 and 2 are satisfied.

Spiros Konstantogiannis Yes Spiros, it's correct what you say! However, a=0 means we are in a trivial case. We should have an a=/0 ! If this is possible!

Regarding problem 2, we can consider the function

g:[0,1]->R , g(t)=tf(t)-(1-t)f(1-t).

g(0)=-f(1), g(1)=f(1), then g(0)g(1)=-[f(1)]², so g has at least one zero in [0,1]. If f(1) is not zero, then g has at least one zero in (0,1), therefore it exists c in (0,1) such that g(c)=cf(c)-(1-c)f(1-c)=0 and now, considering the initial statement of problem 2, we take a=b=c.

But somehow we are also in a trivial case, because a=b !

Regarding problems 3, 4 and George's problem, I think a solution about must use the fact that f, being continuous on a compact set, is bounded, i.e. m<=f(x)<=M for all x in [0,1], and moreover, it exist u,v in [0,1] such that f(u)=m, f(v)=M. m is the minimum of f on [0,1], M the maximum.

Referring to problems 1 and 2 given by Dinu, I would like to add the following statement.

Let f:[0,1]->R be continuous. If the zero set of f on (0,1); i.e., the set containing the points in (0,1) where f vanishes, is nonempty and it is not dense in itself, then the set {(x,y): x,y in (0,1) and x=/y and x|f(x)|-y|f(y)|=0} is uncountable.

Note that the equation x|f(x)|-y|f(y)|=0 is equivalent to the equation xf(x)-yf(y)=0 or xf(x)+yf(y)=0, which are the equations in problems 2 and 1, respectively.

Proof

Since the zero set, say S, of f on (0,1) is nonempty and it is not dense in itself, there exists a point x₀ of S such that x₀ is not a limit point of S. As a result, there exists a positive δ₁ such that the open interval (x₀-δ₁,x₀+δ₁) does not contain points of S other than x₀. Since x₀ belongs to S and S is a subset of (0,1), we know that x₀ belongs to (0,1), as well, and thus it is an interior point of (0,1). Hence, there exists a positive δ₂ such that (x₀-δ₂,x₀+δ₂) is contained in (0,1). Choosing δ=min{δ₁,δ₂}, we have that δ is positive, and (x₀-δ,x₀+δ) is contained in (0,1) and does not contain points of S other than x₀. As a result, f does not vanish in (x₀-δ,x₀+δ)\{x₀}.

Next, considering the function F:[0,1]->R defined by F(x)=x|f(x)|, we have that F is continuous on [0,1], the zero set of F on (0,1) is equal to S, and F is positive on (x₀-δ,x₀) and on (x₀,x₀+ δ). By the intermediate value theorem for F on [x₀-δ/2,x₀] and on [x₀,x₀+δ/2], the images of the previous two closed and bounded intervals are also closed and bounded intervals, and the intersection of the two images contains the interval (0,m], where m is the minimum of F(x₀-δ/2) and F(x₀+δ/2), and thus it is positive; i.e., m>0; hence, (0,m] is nonempty, and thus uncountable. As a result, every z in (0,m] has a preimage x in [x₀-δ/2,x₀) and a preimage y in (x₀,x₀+δ/2], under F, and thus F(x)=z=F(y), whence F(x) =F(y), and thus x|f(x)|-y|f(y)|=0. Since F is a function, distinct points in (0,m] have distinct preimages in [x₀-δ/2,x₀), and distinct preimages in (x₀,x₀+δ/2], respectively, and since (0,m] is uncountable, it follows that the equation x|f(x)|-y|f(y)|=0 is satisfied for uncountably infinitely many pairs (x,y) in [x₀-δ/2,x₀)x(x₀,x₀+δ/2], which is a subset of (0,1)x(0,1), and this proves the statement.

Some remarks

1. If the zero set S is dense in itself, then, since it is also nonempty, it is clearly infinite, and thus F(x)=0=F(y) for infinitely many x and y in S, with x=/y. Hence, x|f(x)|-y|f(y)|=0 for infinitely many pairs (x,y) in (0,1)x(0,1) with x=/y.

2. If the zero set is dense in (0,1), then, by the continuity of f, we derive that f vanishes everywhere in (0,1), and by the continuity of f again, f vanishes in [0,1], and thus f is constant, which is a trivial case.

3. All previous considerations hold for F(x)=g(x)|f(x)|, where g:[0,1]->R is continuous on [0,1] and positive on (0,1).

Spiros Konstantogiannis , Very nice your solution, very nice your approach, dear Spiros!

But I think condition "the set S is not dense in itself" must be reformulated. Because, if f is continuous, then f^-1({0}) ( f^-1(A) denotes the preimage of A) is closed, so S is closed, therefore cl(S)=S, where cl(S) is the closure of S.

So the condition can be " S is a rare set" in the sense of Baire, that is int(cl(S)) is the empty set, and in our case this signifies int(S) is the empty set.

Dear Dinu, thank you. From what I know, if a subset of a topological space has empty interior, this means that the subset does not contain open sets in the space. However, this does not necessarily mean that the said subset has isolated points. For instance, the Cantor set, which is also closed, has empty interior, and yet it does not contain isolated points; all its points are limit points. My aim was to find a condition that would ensure that the zero set has at least one isolated point in (0,1).

Dear Dinu and All, consider the function f:[0,1]->R defined by f(x)=(x-1/4)(x-1/2) if x is in [0,1/2), f(x)=0 if x is in [1/2,3/4] and f(x)=x-3/4 if x is in (3/4,1]. Clearly, f is continuous on [0,1]\{1/2,3/4}. Further, it is easily seen that both the left and right hand limits of f at 1/2 and at 3/4 exist and are equal to 0, whence f is continuous on {1/2,3/4}, as well. Consequently, f is continuous on [0,1]. The zero set of f is the union of the one-element set {1/4} and the closed interval [1/2,3/4], and its interior is the open interval (1/2,3/4); thus nonempty. However, choosing the isolated point 1/4 of the zero set, we can apply the proof given in my previous post and derive that the set {(x,y): x,y in (0,1) and x=/y and x|f(x)|-y|f(y)|=0} is uncountable.

I just add for the readers that a rare set is also called a nowhere dense set.

See also https://en.wikipedia.org/wiki/Nowhere_dense_set.

Spiros Konstantogiannis Yes Spiros, I understand all what you say! But:

The set S from your solution is closed due to the continuity of f and, being closed, we have S is dense in itself. Then S cannot satisfy the condition from your statement, i.e. "S is not dense in itself". That's why I said the condition must be reformulated, must be replaced with something else, with something appropiate.

In fact you need S to contain at least one point which is not a limit point of S. So that the condition be not too restrictive, we must find a more general condition that guarantees the existence of at least one point that is not a limit point. Therefore I proposed int(S) is empty, but your example about Cantor set show us I was wrong and we must find something else!

If we find it's ok! If we don't....it's ok and condition remains "S contains at least one isolated point in (0,1)".

Dear Dinu, I agree that we can use the condition “S contains at least one isolated point in (0,1)”. I think that we mean different things by saying “S is dense in itself” and for this reason, I did not understand your point at first. For me, “S is dense in itself” means that S has no isolated point in the topological space I consider that set; i.e., in (0,1) with the subspace topology. For you, it means that “S is dense in S”; i.e., that the closure of S is S. I should have clarified what I meant, but there are authors who use this notion as I did. Please, read the following extract from wikipedia (https://en.wikipedia.org/wiki/Dense-in-itself):

, Spiros Konstantogiannis An idea possible to be used in some way to solve George's problem:

Let m be the minimum of f on [0,1] and M the maximum. The function f from [0,1] to [m,M] is surjective, f having the intermediate value property.

J=(integral from 0 to 1)(f(t))=(integral from 0 to 1)(f(t)-tf(t)+tf(t))=

=(integral from 0 to 1)(tf(t))+(integral from 0 to 1)((1-t)f(t))=

=af(a)+(1-b)f(b) for some a,b in [0,1], due to mean value theorem for integrals.

J is in [m,M], then af(a)+(1-b)f(b) is in [m,M].

Now let x be a point in [0,1] => f(x) is in [m,M] =>......

( Maybe the convexity of [m,M] can be used somehow !!?)

Apparently, what I above wrote does not lead anywhere....but maybe it will generate an idea for someone else!

Another fact which can be useful:

(integral from 0 to s)(f(t)) is in [m,M] for every s in (0,1)

Dinu Teodorescu, I guess the last fact is slightly disputable at the left end of the interval [m,M]. That is, perhaps the fact is okay on the interval [0,M].

I am talking about this fact (statement) that you made:

"(integral from 0 to s)(f(t)) is in [m,M] for every s in (0,1)"

Technically, if we chose f(x) = 1 + (x-0.5)^2 then in your notations we have m=1.

However, integral from 0 to 0.5 (for exmple, or to any value less than 0.5) is 13/24 which is less than m=1.

Or, maybe I'm missing something here.

Steftcho P. Dokov Many thanks dear Steftcho ! Obviously you are right! Of course the statement "(integral from 0 to s)(f(t)) is in [m,M] for every s in (0,1)" is not generally true. If m=0, then yes, it is true!

Dear Dinu, if you are referring to the initial problem of the discussion, given by George, then if the function f vanishes at one or at both of the endpoints of [0,1], and does not vanish in (0,1), then, by continuity, f has constant sign in (0,1), and since a and 1-b are positive for all a,b in (0,1), the expression af(a)+(1-b)f(b) has the sign of f for all a,b in (0,1), and thus the equation f(x)=af(a)+(1-b)f(b) is not satisfied at (at least) one of the endpoints of [0,1], where f vanishes. I think that your idea to use the mean value theorem must be combined with some extra assumption, or some question that will lead to some extra assumption, as in the case of problems 1 and 2 given by you.

PS: I made an awful mistake in my last post. I wrote that according to my interpretation of the notion “dense in itself”, “S is dense in itself” means that S has at least one isolated point; actually, this is what “S is NOT dense in itself” means. I corrected it. Apologies.

Actually, by the mean value theorem, the integral from 0 to s of f is in [ms,Ms] for every s in (0,1), where m and M are the minimum and maximum value of f in [0,1], respectively.

This is an attempt for a counterexample for last problem

4. Let f be a continuous real function defined on the interval [0, 1]. For each x in [0, 1], are there a and b in (0, 1) such that f(x)=af(a) +(1-b)f(1-b) ?

Let consider f(x) = ln(2+x) which is continuous on [0, 1].

Then, we want to see if there are

a and b in (0, 1) such that f(x)=af(a) +(1-b)f(1-b).

If we assume that such pair (a, b) exists then

ln(2+x) = a*ln(a) + (1-b)*ln(1-b)

i.e.

ln(2+x) = ln(a^a) + ln( (1-b)^(1-b) )

ln(2+x) = ln( (a^a)*( (1-b)^(1-b) ) )

thus

2+x = (a^a)*( (1-b)^(1-b) )

and

x = (a^a)*( (1-b)^(1-b) ) -2

As a contradiction, we want to show that for each x in [0, 1]

there are not a-and-b in (0, 1) which satisfy the last equality.

Technically,

we will show that the RHS, (a^a)*( (1-b)^(1-b) ) -2, is negative for a-and-b in (0, 1).

This negativity can be seen e.g. with a google plot

or at

Hopefully, there isn't some silly mistake / typo.

Steftcho P. Dokov All clear now!

Indeed

(a^a)*( (1-b)^(1-b) ) -2 <= -1 <0 <= x for all a,b,x in [0,1].

So your counterexample invalidates problem 4.

Moreover ,your counterexample, dear Steftcho, invalidates too problem 3 and George's problem, because

((1-a)^a)*( b^(1-b) ) -2 <= -1 <0 <= x for all a,b,x in [0,1]

and

(a^a)*( b^(1-b) ) -2 <= -1 <0 <= x for all a,b,x in [0,1].

Therefore George's problem has a flawed statement, probably due to a typo mistake in Gazeta Matematica.

But, thanks to the discussions in this thread, we still have some other nice problems:

P1) Let f:[0,1]->R be continuous. If the zero set of f on (0,1); i.e., the set containing the points in (0,1) where f vanishes, is nonempty and has at least one isolated point in (0,1), then the set {(x,y): x,y in (0,1) and x=/y and x|f(x)|-y|f(y)|=0} is uncountable.

P2) Let f:[0,1]->R be continuous. Then there are a and b in [0, 1] such that (integral from 0 to 1)(f)=af(a) +(1-b)f(b).

Dear Steftcho P. Dokov, if f(x)=ln(2+x) for x in [0,1], the expression af(a)+(1-b)f(1-b) reads aln(2+a)+(1-b)ln(3-b), for a and b in (0,1), and the last expression reads ln[(2+a)^a*(3-b)^(1-b)]. Hence, the equation f(x)=af(a)+(1-b)f(1-b) reads ln(2+x)=ln[(2+a)^a*(3-b)^(1-b)], and since the function ln is injective, the last equation impiles that x=(2+a)^a*(3-b)^(1-b)-2. The right side of the last equation can be positive; for instance, if a=b=1/2, we have (2+a)^a=(3-b)^(1-b)=sqrt(2.5); hence, (2+a)^a*(3-b)^(1-b)-2=0.5>0.

Thanks a lot Spiros Konstantogiannis!

I was suspecting some silly mistake.

Sorry for the long unusable writing.

I would like to add that the statement of the initial problem is "for every x in [0,1], there exist a and b in (0,1) such that f(x)=af(a)+(1-b)f(1-b)". The negation of this statement is the statement "there exists x in [0,1] such that for every a and b in (0,1), we have f(x)=/af(a)+(1-b)f(1-b)". A counterexample to the initial statement is a function f such that the last statement is true. For this, it suffices to find only one point x in [0,1] such that af(a)+(1-b)f(1-b) cannot be equal to f(x) for all a and b in (0,1). If f is strictly positive or strictly negative in (0,1), then af(a)+(1-b)f(1-b) is likewise strictly positive or strictly negative for all a and b in (0,1). Hence, if f vanishes at (at least) one of the endpoints 0 or 1 of the interval [0,1], the point x can be chosen to be the point(s) where f vanishes. This seems to be a somewhat trivial counterexample, but it is a valid one. For instance, the function f:[0,1]->R defined by f(x)=x(x-1) provides a counterexample, since f(0)=/af(a)+(1-b)f(1-b) and f(1)=/af(a)+(1-b)f(1-b) for all a and b in (0,1).

Let me try to modify slightly the counterexample for problem 4.

Let f(x)=ln(0.1+x)

then af(a)+(1-b)ln(1-b)=a*ln(0.1+a)+(1-b)*ln(1.1-b)

and, according to Spiros important remark about negation of problem 4, we have to show that:

"there exists x in [0,1] such that for every a and b in (0,1), we have f(x)=/af(a)+(1-b)f(1-b)".

We select x=0.1, and we will show that

ln(0.1+0.1)=ln(0.2) < is less than < a*ln(0.1+a)+(1-b)*ln(1.1-b)= RHS

for every pair (a, b) in (0, 1).

We have ln(0.2)=-1.6094 < less than < -0.56,

which less than, nearly equal to, the minimum of the RHS = a*ln(0.1+a)+(1-b)*ln(1.1-b)

seen next

or

Hopefully, this one is error free.

Your comments are highly appreciated.

Sorry for another long writing.

Clearly, the way is written, the problem is false. A first thing that I am working on is to add conditions or to employ methods to ensure that a ≠ b. Maybe building not one, but two auxiliary functions, to which we apply various mean value theorem(s). We'll keep you posted.

I apologize for all wrong statements I posted!

Don't worry, Dinu, I started all this with the wrong problem. I now have a sort of replacement, this time with a valid proof. Not exactly what I wanted, but... I'll let you enjoy it and, if not solved, later I will post the proof as well. Here goes:

Let f : [0, 1] → R be a continuous zero mean (∫₀¹ f(t) dt = 0) function, and g : [0, 1] → R continuous such that g(0) · g(1) < 0. Then there exist 0 < a < c < b < 1 such that [f(a) - f(b)] · g(c) = f(a) · f(b).

Dear George, and All, if the there exist 0<a<c<b<1 such that [f(a)-f(b)]g(c)=f(a)f(b), then we have [f(a)-f(b)]g(c)=f(a)f(b)=f(b)f(a)=[f(b)-f(a)]g(c), whence 2[f(a)-f(b)]g(c)=0, and thus [f(a)-f(b)]g(c)=0 (1). By means of (1), the equation [f(a)-f(b)]g(c)=f(a)f(b) reads f(a)f(b)=0, whence f(a)=0 or f(b)=0. Hence, (1) reads f(b)g(c)=0 (2) or f(a)g(c)=0 (3), respectively.

Now, if we choose as f any function from [0,1] to R that is continuous on [0,1], its integral vanishes on [0,1], and f vanishes on (0,1) only once, and as g any function from [0,1] to R that is continuous on [0,1] and vanishes only once on [0,1] and at the same point as f, we have g(0)g(1)<0. Further, by (2), we have g(c)=0, since f(b) must be nonzero, which is a contradiction, since c=/a. Similarly, by (3), we have g(c)=0, since f(a) must be nonzero, which is also a contradiction, since c=/b.

Dear Spiros,

[f(a)-f(b)]g(c) = f(a)f(b) does not imply that f(b)f(a) = [f(b)-f(a)]g(c).

The above implication would be true if

[f(a)-f(b)]g(c) = f(a)f(b) holds for all a, b ∈ (0, 1),

which is not the case.

Dear George, you are right. Since a<b, the points a and b are strictly ordered; hence, they cannot be interchanged.

I would like to make the following remarks on George's last problem. Since f is continuous and its integral vanishes on [0,1], we know that f vanishes at least once in (0,1). Similarly, since g is continuous on [0,1] and g(0)g(1)<0, we know that g also vanishes at least once in (0,1).

If f vanishes more than once in (0,1), then there exist 0<a<b<1 such that f(a)=f(b)=0, and choosing any c between a and b, we have (f(a)-f(b))g(c)=0=f(a)f(b), whence (f(a)-f(b))g(c) =f(a)f(b).

If f vanishes only once in (0,1) and g does not vanish at the point where f vanishes, then there exist 0<x₀,x₁<1 such that x₀=/x₁ and such that f(x₀)=g(x₁)=0. As a result, x₀<x₁ or x₀>x₁. In the first case, we choose a=x₀, c=x₁, and b any point in (x₁,1), and we have 0<a<c<b<1 and f(a)=0=g(c)=0, and thus (f(a)-f(b))g(c)=0=f(a)f(b), whence (f(a)-f(b))g(c) =f(a)f(b). In the second case, we choose as a any point in (0,x₁), c=x₁ and b=x₀, and we have 0<a<c<b<1 and f(b)=0=g(c), and thus (f(a)-f(b))g(c)=0=f(a)f(b) , whence (f(a)-f(b))g(c) =f(a)f(b).

The previous remarks show that the only non-trivial case, non-trivial in the sense that the two sides of the equation (f(a)-f(b))g(c) =f(a)f(b) must be nonzero, is the case where f vanishes only once in (0,1) and g also vanishes at the point where f vanishes. Further, if g vanishes more than once in (0,1), we can repeat the previous reasoning choosing as x₁ a point where g vanishes and f does not vanish. Such a point exists if f vanishes once only and g vanishes more than once. Hence, we are left with the case where both f and g vanish only once in (0,1) and they vanish at the same point.

Regarding the last problem given by George, I have the following solution for the remaining, and non-trivial, case for the functions f and g.

As explained, both f and g vanish at least once in (0,1), and if f vanishes more than once or if there exists a point in (0,1) where only one of f and g vanish, then there exist 0<a<c<b<1 such that [f(a)-f(b)]g(c)=0=f(a)f(b), whence [f(a)-f(b)]g(c) =f(a)f(b). Hence, we are left with the case where f and g vanish only once and at the same point, say x₀, in (0,1).

As explained, too, f changes sign in (0,1), as a result of its continuity and its integral on [0,1] being equal to zero, and so does g, as a result of its continuity, too, and the condition g(0)g(1)<0.

Consequently, both f and g vanish at x₀ only and change sign at x₀.

By the additivity of the integral, we have that for every x in (0,1), the sum of the integrals of f from 0 to x and from x to 1 equals the integral of f on [0,1], which is equal to zero. Hence the sum of the two integrals is equal to zero. Applying the mean value theorem to the two integrals, we have that there exist y in (0,x) and z in (x,1) such that the first integral is equal to f(y)x and the second integral is equal to f(z)(1-x). As a result, we have f(y)x+ f(z)(1-x)=0 (1), where 0<y<x<z<1 (2). We emphasize that both points y and z depend on the point x. Since x and 1-x are positive for every x in (0,1), we have, by (1), that f(y)=f(z)=0 or f(y) and f(z) have different sign.

If f(y)=f(z)=0, then y=z=x₀, since y and z are in (0,1) and f vanishes in (0,1) only at x₀. Hence, (2) yields x₀<x<x₀, which is a contradiction. Consequently, f(y) and f(z) have different signs. As a result, f(y)=/f(z), and thus f(y)-f(z)=/0 for every x in (0,1).

Next, by (1), we have f(y)-f(z)=-f(z)/x and f(y)f(z)=(1-1/x)f²(z), whence f(y)f(z)/[f(y)-f(z)]=(1-x)f(z). Hence, for every x in (0,1), the expression f(y)f(z)/[f(y)-f(z)] is equal to the integral of f from x to 1, which, by (1), is equal to the opposite of the integral of f from 0 to x.

Next, considering the function F:[0,1]->R defined by F(x)=-indefinite integral of f with basepoint 0, we have, by the fundamental theorem of calculus, that F is continuous on [0,1]. Also, F(0)=0, and since the integral of f on [0,1] vanishes, we have F(1)=0, too.

Note that, for every x in (0,1), we have F(x)= f(y)f(z)/[f(y)-f(z)].

Next, we consider the function G:[0,1]->R defined by G(x)=g(x)-F(x). Clearly G is continuous on [0,1]. Also, since F vanishes at 0 and at 1, we have G(0)G(1)=g(0)g(1)<0. Consequently, there exists c in (0,1) such that G(c)=0, whence g(c)=F(c), and since F(x)= f(y)f(z)/[f(y)-f(z)] in (0,1), we have g(c)=f(a)f(b)/[f(a)-f(b)], whence [f(a)-f(b)]g(c)=f(a)f(b), where a=y(c) and b=z(c), and by (2), we have 0<a<c<b<1.

Along the same lines as Spiros, please see attached.