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

1. Nov 29, 2017

### jaus tail

We've only been taught of limits and derivatives and integration.
I tried using derivatives.
f(x) = 2x(f(x2-a))
Let f(x^2 - 1) = g(x)
So
f(x) = 2x g(x)
f'(x) = 2xg'(x) + 2g(x)
f'(x) = 2xf'(x)(2x) + 2f(x^2 - 1)
f'(x) (1-4x^2) = 2f(x^2 - 1)
So we get f'(x) = 2f(x^2 - 1)/(1-4x^2)

Now we got that f(-1) = f(1) = 0.
So if f(x) is non zero at any point between 1 and -1, then f'(x) must be positive some place and negative some place.
But above we see that f'(x) is an even function.
So f'(x) is not satisfying condition of positive and negative. It's either positive or negative and if it's one of them then it can have only 1 zero.
And so we can say that f(x) is constant at zero.

Aren't all continuous functions derivative? I'm not sure what's the difference. Sine and consine are continous and differentiable. So i guess it should be true.

2. Nov 30, 2017

### PeroK

Not all continuous functions are differentiable. One example is the modulus $|x|$, which is not differentiable at $x = 0$.

There is, in fact, the Weierstrass function, which is continuous but nowhere differentiable!

https://en.wikipedia.org/wiki/Weierstrass_function

3. Nov 30, 2017

### jaus tail

So I guess we can take two conditions.
1) f(x) is differentiable then we can go with my solution.
and 2) f(x) is not differentiable. Any idea how to prove in this case?

4. Nov 30, 2017

### PeroK

First:

$g'(x) = 2xf'(x^2-1) \ne 2xf'(x)$

But also, I don't understand how you get that an even function can't take positive and negative values.

I wonder where you got this question, because it seems to be a little beyond your level at the moment.

5. Nov 30, 2017

### PeroK

See @Ray Vickson's solution in post #14.

6. Nov 30, 2017

### Delta²

If you are allowed to use integral and derivatives and all that, follow Ray Vickson solution at post #14. It is mostly correct, and it uses the theorem that $F(x)=\int_0^x{f(y)dy}$ is differentiable and it is $F'(x)=f(x)$ (that is the fundamental theorem of calculus). it doesn't need the assumption that f(x) is differentiable.

If you are not allowed to use integral and derivatives then you can follow my approach explained in post #15, #20 and #28 which I believe is also correct.

7. Nov 30, 2017

### Delta²

$2^np_n(x)$ is dominated by $p_n(x)$ for any $x\neq\xi$ its $\lim_{n->\infty}{p_n(x)}=0$ and the convergence of $p_n(x)$ to 0 is such that it overcomes $2^n$. I believe I prove this at post #28.

8. Nov 30, 2017

### PeroK

If you take an interval close enough to -1 then you can generate a sequence of function values of increasing magnitude without bound in that interval. Which contradicts continuity, unless the function is zero on the interval. After that your argument simplifies for other points.