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.

 

[BvU]

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]

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)

 

[PeroK]

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]

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.

 

[PeroK]

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!

 

[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 I am stupid. That says only that f(P12) must be zero...

 

[fresh_42]

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.

 

[PeroK]

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.

 

[fresh_42]

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.

 

[PeroK]

What could be false?

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

 

[fresh_42]

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.

 

[Ray Vickson]

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, $F(x) = \text{constant}.$

 

[Delta2]

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.

 

[PeroK]

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 $-1$.

 

[Delta2]

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.

 

[PeroK]

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.

 

[Ray Vickson]

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 $(1-\sqrt{5})/2.$

 

[Delta2]

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.

 

[PeroK]

As I said, look at the "cobweb diagram"; see, eg., https://www.math.ubc.ca/~andrewr/620341/pdfs/ga_sum.pdf or http://mathworld.wolfram.com/WebDiagram.html .

And, yet:

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

 

[Delta2]

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 $F(0)=F(-1)=0$.

 

[Ray Vickson]

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.

 

[WWGD]

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.

 

[Delta2]

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...

 

[WWGD]

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 $[-1,1]$?

 

[fresh_42]

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?

 

[Delta2]

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 $i>k(x)$.

 

[fresh_42]

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 $1$.

 

[Delta2]

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 .