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.

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.

Great summary and generalization, thanks G. Stoica

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].

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) ?

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).

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)

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

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 , G. Stoica

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):

In general topology, a subset A of a topological space is said to be dense in itself or crowded if A has no isolated point. Equivalently, A is dense in itself if every point of A is a limit point of A.

The notion of dense set is distinct from dense in itself. This can sometimes be confusing, as “X is dense in X” (always true) is not the same as “X is dense in itself” (no isolated point).

G. Stoica

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.

https://www.wolframalpha.com/input?i=integrate+1%2B%28x-1%2F2%29%5E2+dx+from+x%3D0+to+1%2F2

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

https://www.google.com/search?q=plot+z%3D%28x%5Ex%29*%28%281-y%29%5E%281-y%29%29-2

or at

https://www.wolframalpha.com/input?i=maximize+%28x%5Ex%29*%28%281-y%29%5E%281-y%29%29-2+on+0%3Cx%3C1%2C+0%3Cy%3C1

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

Steftcho P. Dokov All clear now!

Indeed

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

https://www.wolframalpha.com/input?i=minimize+a*ln%280.1%2Ba%29%2B%281-b%29*ln%281.1-b%29+on+0%3C%3Da%3C%3D1%2C+0%3C%3Db%3C%3D1

or

https://www.google.com/search?q=plot+x*ln%280.1%2Bx%29%2B%281-y%29*ln%281.1-y%29

Hopefully, this one is error free.

Your comments are highly appreciated.

Sorry for another long writing.

I apologize for all wrong statements I posted!

Dear George, and All, if the there exist 0

Dear George, you are right. Since a

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)

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

Yes, for a general continuous function f:[0,1]→Rf: [0,1] \to \mathbb{R}, it is not always guaranteed that for each x∈[0,1]x \in [0,1], there exist a,b∈(0,1)a, b \in (0,1) such that:

f(x)=af(a)+(1−b)f(b)f(x) = a f(a) + (1 - b) f(b)

Let's critically analyze this.

1. Given Equation:

We are asked if for all x∈[0,1]x \in [0,1], there exist a,b∈(0,1)a, b \in (0,1) such that:

f(x)=af(a)+(1−b)f(b)f(x) = a f(a) + (1 - b) f(b)

This is a nonlinear condition, and whether it holds for all ff and every xx depends on the nature of ff.

2. Counterexample:

Let’s take a simple function:

f(x)=xf(x) = x

Then the equation becomes:

x=a⋅a+(1−b)⋅b=a2+b−b2x = a \cdot a + (1 - b) \cdot b = a^2 + b - b^2

We are asking: for any x∈[0,1]x \in [0,1], are there a,b∈(0,1)a, b \in (0,1) such that:

x=a2+b−b2x = a^2 + b - b^2

This is a constraint on the image of a specific nonlinear function in two variables, and it's not obvious that the range of this expression fills the entire interval [0,1][0,1]. In fact, we can check:

• At a=b=0.5a = b = 0.5: x=0.25+0.5−0.25=0.5x = 0.25 + 0.5 - 0.25 = 0.5
• But can we get x=0x = 0? That would require:a2+b−b2=0⇒a2=b2−ba^2 + b - b^2 = 0 \Rightarrow a^2 = b^2 - bBut b2−b≤0b^2 - b \le 0 for b∈(0,1)b \in (0,1), so a2≤0⇒a=0a^2 \le 0 \Rightarrow a = 0, contradicting a∈(0,1)a \in (0,1)

Thus, for x=0x = 0, no such a,b∈(0,1)a, b \in (0,1) exist. Hence, the condition fails for some xx.

✅ Conclusion:

The equation

f(x)=af(a)+(1−b)f(b)f(x) = a f(a) + (1 - b) f(b)

does not hold for all x∈[0,1]x \in [0,1] for an arbitrary continuous function ff. Therefore, the answer is:

No, there does not necessarily exist such a,b∈(0,1)a, b \in (0,1) for every x∈[0,1]x \in [0,1], even if ff is continuous.