# Prove f(x) is zero in range [-1,1]

jaus tail

## Homework Statement

A function f(x) is continuous in the range [-1,1]
f(x) = 2xf(x2 - 1)

Prove that f(x) = 0 everywhere in range [-1,1]

## Homework Equations

I don't know how to proceed.
By putting values as x = 0, I got f(0) = 2*0 = 0.
f(1) = 2*1*f(0) = 0.

And I also get f(negative values) = f(positive values). So it's like a cosine wave.

But how to prove that the values are equal to zero.

## The Attempt at a Solution

How to proceed? Continuous means value of f(x) when I approach from left side = value of f(x) when I approach from right side. Not sure how to imply it here.

## Answers and Replies

Homework Helper
And I also get f(negative values) = f(positive values)
How ? And: if you are right, and if you can also get the opposite, you have it !

• jaus tail
jaus tail
No, I think i wrote wrong.
Putting x = -1,
I get f(-1) = 2(-1)*f((-1)2 -1 )
which is f(-1) = -2f(0)
and f(1) = 2f(0)

f(0) = 0*f(-1) = 0.
So f(0) = 0
So f(-1) = 0 and so is f(1) from underlined parts.

So f(x) = 0 for x = -1, 0, 1. But how to prove for other values of x?

when x = 0.5. I get f(0.5) = 2*0.5 f(0.25 - 1) = f(0.75)
and when I put x = 0.75 I get
f(0.75) = 2*0.75 f(0.5625-1) = 1.5f(-0.4375)

I don't know how to prove that all these f(x) are equal to zero.

jaus tail
How ? And: if you are right, and if you can also get the opposite, you have it !
Although I goofed up there, how would I have it if I proved this. why would f(x) be zero if f(-x) = f(x).
A cosine wave is continuous in range -90 to + 90 but it's not always zero, and it satisfies condition f(-x) = f(x)

Homework Helper
Gold Member
2021 Award
Although I goofed up there, how would I have it if I proved this. why would f(x) be zero if f(-x) = f(x).
A cosine wave is continuous in range -90 to + 90 but it's not always zero, and it satisfies condition f(-x) = f(x)

This looks tricky. What happens if you start from any point in ##x_1 \in [-1, 1]## and generate a recursive sequence by ##x_{n+1} = x_n^2 - 1##?

• jaus tail
jaus tail
What's a recursive sequence?

I tried with mean value theorem.
a = -1, b = 1
there is a point c between a and b such that f'(c) = [ f(b) - f(a) ] / [b - a]
this is zero since f(-1) = f(1) = 0
So since f'(c) is zero, we get a constant value and since one constant value is zero we can say that f(x) = constant = 0.

Homework Helper
Gold Member
2021 Award
What's a recursive sequence?

I tried with mean value theorem.
a = -1, b = 1
there is a point c between a and b such that f'(c) = [ f(b) - f(a) ] / [b - a]
this is zero since f(-1) = f(1) = 0
So since f'(c) is zero, we get a constant value and since one constant value is zero we can say that f(x) = constant = 0.

It doesn't say that ##f## is differentiable. But, another idea is to use the fact that a continuous function attains its max and min on a closed interval.

Regarding my previous idea. If you start with a number. E.g. ##x_1 = 1/2##, then ##x_2 = x_1^2 - 1 = -3/4## and ##x_3 = x_2^2 - 1 = -7/16## and so on. This sequence may converge to a point ##x## where ##x^2 - 1 = x## (if there is such a point).

These are just ideas. I haven't spotted the solution yet, so they may not help!

Last edited:
Gigaz
Simple proof:

Go to the point where x²-1=x and call it P12. (Should be -0.6180339887498949 and 1.6180339887498949)
f(P12)=2P12 f(P12)
1/2=P12 except if f is zero. And because P12 is clearly not 1/2, f must be zero.

Edit: Wait im stupid. That says only that f(P12) must be zero...

• Delta2
Mentor
2021 Award
It doesn't say that ##f## is differentiable.
So maybe integration is an idea. What about ##F(x):= \int_{-1}^x f(y)dy\,##? With ##g(x)=x^2-1## we get easier conditions plus differentiability. However, I haven't found the trick, yet.

Homework Helper
Gold Member
2021 Award
So maybe integration is an idea. What about ##F(x):= \int_{-1}^x f(y)dy\,##? With ##g(x)=x^2-1## we get easier conditions plus differentiability. However, I haven't found the trick, yet.

Do you think it might be false?

The condition on the interval ##[0, 1]## is irrelevant. If you can satisfy the condition on ##[-1, 0]## that would do it.

Mentor
2021 Award
Do you think it might be false?
What could be false? I only thought continuity invites to be integrated to ##F'(x)=f(x)## with ##F(0)=F(-1)=F(1)=0## and ##F=F\circ g## if I didn't make a mistake. These are easier to handle.
The condition on the interval ##[0, 1]## is irrelevant. If you can satisfy the condition on ##[-1, 0]## that would do it.
Sure, but do we have it on one of them? I'm looking for an argument without iterations.

Homework Helper
Gold Member
2021 Award
What could be false?

That there might be a continuous function that meets the criteria?

Mentor
2021 Award
That there might be a continuous function that meets the criteria?
I don't get it, sorry. Since it is supposed to result in ##f(x)=0## it'll be both, continuous and differentiable. I think that this is the only solution, I just don't see the killer argument. I simply thought, that we can get differentiability for free by an integration.

Homework Helper
Dearly Missed
Although I goofed up there, how would I have it if I proved this. why would f(x) be zero if f(-x) = f(x).
A cosine wave is continuous in range -90 to + 90 but it's not always zero, and it satisfies condition f(-x) = f(x)

What could be false? I only thought continuity invites to be integrated to ##F'(x)=f(x)## with ##F(0)=F(-1)=F(1)=0## and ##F=F\circ g## if I didn't make a mistake. These are easier to handle.

Sure, but do we have it on one of them? I'm looking for an argument without iterations.

Iteration plus integration will do the job. Let
$$F(x) = \int_0^x f(t) \, dt,$$
so that we get
$$F(x) = F(x^2-1) - F(-1).$$
Since ##f## is an odd function, ##F## is even; thus ##F(-1) = F(1)##, so we have ##F(x) = F(x^2-1) - F(0)##. Using ##F(0)=0## we have ##F(1) = -F(1)##, so ##F(1) =0.## Therefore, we have
$$F(x) = F(x^2 - 1).$$

Now, for any ##x_0 \in (-1,1)## the iterative scheme ##x_1 = x_0^2-1, x_2 = x_1^2 - 1, \ldots## converges to the negative solution of ##x = x^2-1##, which is ##\tau \equiv (1-\sqrt{r})/2 \doteq -0.61803##. (We can see this by looking at the "cobweb diagram" of the iteration scheme ##x_{n+1} = x_n^2 - 1.##). Since ##F(x_{n+1}) = F(x_n)## for all ##n \geq 0## we have ##F(x_0) = \lim_{n \to \infty} F(x_n ) = F(\tau)## for any ##x_0##. In other words, ##F(x) = \text{constant}.##

• PeroK and fresh_42
Homework Helper
Gold Member
My idea is that ##f(x)=2xf(x^2-1)=2x2(x^2-1)f((x^2-1)^2-1)=...=2^np_n(x)f(q_n(x))## where ##p_n## and ##q_n## polynomials of degree ##2^n-1## and ##2^n##. If we can prove that ##\lim_{n->\infty}q_n(x)=0## then I think we got it.

Last edited:
Homework Helper
Gold Member
2021 Award
Now, for any ##x_0 \in (-1,1)## the iterative scheme ##x_1 = x_0^2-1, x_2 = x_1^2 - 1, \ldots## converges to the negative solution of ##x = x^2-1##, which is ##\tau \equiv (1-\sqrt{r})/2 \doteq -0.61803##.

I found that this sequence converges alternatingly to ##0## and ##-1##.

• fresh_42
Homework Helper
Gold Member
I found that this sequence converges alternatingly to ##0## and ##-1##.
That's only if you start with 0, 1 or -1. for any other x in (-1,1) it converges to -0.618...=##\frac{1-\sqrt{5}}{2}##.

Now I see clearly that the polynomial ##q_n(x)## in post #15 converges to -0.618.. for any x in (-1,0) or (0,1) and f(-0.618...)=0.

Homework Helper
Gold Member
2021 Award
That's only if you start with 0, 1 or -1. for any other x in (-1,1) it converges to -0.618...=##\frac{1-\sqrt{5}}{2}##

Can you prove that? Try starting at 0.5, for example, and see what happens.

Homework Helper
Dearly Missed
Can you prove that? Try starting at 0.5, for example, and see what happens.

Yes, you are right. However, the basic limiting argument still holds: ##F(x_0) = F(x_n) ## for all ##n##, and for very large ##n## we have that ##x_n## is either ##0## or ##-1## (alternating). But since ##F(0) = F(-1) = 0## we have ##F(x_0) = 0##. That is for any ##x_0 \in (-1,1)## different from ##(1-\sqrt{5})/2.##

Last edited:
Homework Helper
Gold Member
Can you prove that? Try starting at 0.5, for example, and see what happens.

Well I ve got to eat my hat, seems you are right. But still the sequence ##f(q_n(x))## converges alternatingly to f(0) or f(-1) and since they are equal to 0, converges to zero.

Homework Helper
Gold Member
I think he is right but still your proof is mostly correct because ##F(x_{2n})## converges to ##F(0) ## and ##F(x_{2n+1})## converges to ##F(-1)## but ##F(0)=F(-1)=0##.

Homework Helper
Dearly Missed
And, yet:

0.5, -0.75, -0.4375, -0.8086, -0.34618, -0.88016 etc.

I edited the message to remove those incorrect statements, but you responded before I made the changes. Apparently, the version you responded to remains unchanged. As the edited version makes plain, the basic argument still goes through.

• PeroK
Gold Member
Well, by repeatedly applying the IVT we have that f is continuous with infinitely-many zeros. Only non-zero function I know satisfies this is the distance function d(x,S) for ##S## a closed subspace ( of [-1,1] ). Can we exclude this ? (Yes) EDIT: Does anyone know of EDIT any other non-zero continuous function that can have infinitely-many zeros ( in a compact subset of Real line)? Obviously not a poly, but that does not narrow it too much. Maybe we can show the 0's are dense , then extending by continuity we can show the function is identically 0?

EDIT2: My bad, should have been obvious that a "contraction" of sinx or cosx would have produced an example:https://math.stackexchange.com/questions/48746/existence-of-non-constant-continuous-functions-with-infinitely-many-zeros
STILL, this gives us countably-many. Could we have uncountably-many zeros? This would imply the zero set would have a limit point in ##[-1,1]##, which excludes Complex-analytic functions ( if the function was analytic to start with ). Seems like a Baire-Cat -type argument could show it.

Last edited:
Homework Helper
Gold Member
Guys do you find my idea in post #15, further explained in post #20 as correct? My approach doesn't use integrals or derivatives or the fundamental theorem of calculus or IVT which I believe probably they aren't yet being taught to the OP...

Gold Member
My idea is that ##f(x)=2xf(x^2-1)=2x2(x^2-1)f((x^2-1)^2-1)=...=2^np_n(x)f(q_n(x))## where ##p_n## and ##q_n## polynomials of degree ##2^n-1## and ##2^n##. If we can prove that ##\lim_{n->\infty}q_n(x)=0## then I think we got it.
But, will the terms on polys remain within ##[-1,1]##?

Mentor
2021 Award
I still think that ##F = F \circ g## with ##g(x)=x^2-1\, , \,F'(x)=f(x)## and ##F(-1)=F(0)=F(-1)## is the key. To write it this way is too invitingly. Why should there be the factor two otherwise?

Homework Helper
Gold Member
I still think that ##F = F \circ g## with ##g(x)=x^2-1\, , \,F'(x)=f(x)## and ##F(-1)=F(0)=F(-1)## is the key. To write it this way is too invitingly. Why should there be the factor two otherwise?

yes you got a point here, the 2 factor "screams" that 2x is the derivative of ##x^2-1##.

Still I believe it can be done my way and without the use of integrals and derivatives. My main difficulty is proving that ##2^np_n(x)## is bounded but I believe it can be done.

It is ##p_n(x)=q_1(x)q_2(x)...q_{n-1}(x)##
For ##q_n(x)## we have that ##\lim_{n->\infty}{q_{2n}(x)}=0##, hence we can find a ##k(x)## such that ##q_{2n}(x)>-\frac{1}{4}## for ##n>k(x)##.

So it will be ##p_n(x)=q_1(x)...q_{2k(x)}(x)q_{2k(x)+1}(x)...q_n(x)=P_{2k(x)}(x)q_{2k(x)+1}(x)...q_n(x)##.

to prove that ##2^np_n(x)## is bounded we break it down to
##2^{2k(x)}P_{2k(x)}2^{n-2k(x)}q_{2k(x)+1}(x)...q_n(x)##

##2^{2k(x)}P_{2k(x)}## is bounded cause its finite. The rest part of product I believe it is also bounded for any n because ##q_{2i}(x)>\frac{-1}{4}## for any ##i>k(x)##.

Last edited:
Mentor
2021 Award
yes you got a point here, the 2 factor "screams" that 2x is the derivative of ##x^2-1##. Still I believe it can be done my way and without the use of integrals and derivatives. My main difficulty is proving that ##2^np_n(x)## is bounded but I believe it can be done.
But this still leaves the same convergence problem as with Ray's idea: ##q_n(x)## alternates between ##-1## and ##0## and I'm not sure that the boundaries won't cause problems. The compactness of ##[-1,1]## might play a role, but it is simultaneously an obstacle to get a "norm" below ##1##.

Homework Helper
Gold Member
But this still leaves the same convergence problem as with Ray's idea: ##q_n(x)## alternates between ##-1## and ##0## and I'm not sure that the boundaries won't cause problems. The compactness of ##[-1,1]## might play a role, but it is simultaneously an obstacle to get a "norm" below ##1##.
##q_n(x)## alternates but ##f(q_n(x))## converges because f(-1)=f(0)=0 .

Mentor
2021 Award
##q_n(x)## alternates but ##f(q_n(x))## converges because f(-1)=f(0)=0 .
I haven't checked, but what happens at ##\xi = \dfrac{1}{2}(1-\sqrt{5})## with ##q_n(\xi)## and ##p_n(\xi)\,##?

Homework Helper
Gold Member
##q_n(\xi)=\xi## , ##p_n(\xi)=\xi^n##. No problem here either because ##f(\xi)=0##

Mentor
2021 Award
Yes, but ##2^n\xi^n## explodes and as we don't know much about ##f##, it cannot be compensated by ##f(q_n(x))##, at least we don't know.

Homework Helper
Gold Member
you can prove another way ##f(\xi)=0## check post #8. We must exclude ##\xi## from this treatment and treat it as a special case.

Mentor
2021 Award
you can prove another way ##f(\xi)=0## check post #8. We must exclude ##\xi## from this treatment and treat it as a special case.
That's not my point. ##f(\xi)=0## directly follows from the defining equation. But the exclusion of ##\xi## might not be sufficient, because ##2^n (\xi \pm \varepsilon)^n## is also divergent and polynomials are continuous, so at first glance the construction looks unstable.