# Thread: Prove that function is identically zero?

1. ## Prove that function is identically zero?

Let $f(x)$ be a continuous function in $[-1, 1]$ and satisfies $2f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]$. Prove that $f(x)$ is identically zero $\forall\ x\in [-1, 1]$.

2. Originally Posted by fardeen_gen
Let $f(x)$ be a continuous function in $[-1, 1]$ and satisfies $2f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]$. Prove that $f(x)$ is identically zero $\forall\ x\in [-1, 1]$.
$2f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]$

2 cancels out both sides, why it is there?

I think it may be $f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]$.

3. Well, I'm having a go at $f(2x^2-1)=2xf(x)$, which seems more sensible. To start with, putting $-x$ for $x$ shows that $f(-x)=-f(x)$, so we can concentrate on the interval $[0,1]$.

So put $x=\cos\theta$ for $\theta\in[0,\pi/2]$ to derive $f(\cos 2\theta)=2\cos\theta f(\cos\theta)$ (1).

Putting $x=\sin\theta$ gives $f(-\cos 2\theta)=2\sin\theta f(\sin\theta)$, or $f(\cos 2\theta)=-2\sin\theta f(\sin\theta)$ (2).

It is clear from (2) that $f(1)=0$.

Putting $\theta=\pi/4$ in (1) and (2) gives $f(0)=\surd 2f(1/\surd2)=-\surd2f(1/\surd2)$, so now $f(\sin 0)=f(\sin \frac\pi4)=f(\sin\frac\pi2)=0$.

Put $\theta=\pi/8$ in (1) and (2) now shows that $f(\sin\frac\pi8)=f(\sin\frac{3\pi}8)=0$.

Now direct substitutions, the identity $\sin(\pi/2-\theta)=\cos\theta$ and previous results show that $f(\sin\frac\pi{16})=f(\sin\frac{3\pi}{16})=f(\sin\ frac{5\pi}{16})=f(\sin\frac{7\pi}{16})=0$.

So far we know that

$f(\sin 0)=f(\sin\frac\pi{16})=f(\sin\frac{2\pi}{16})=f(\s in\frac{3\pi}{16})=f(\sin\frac{4\pi}{16})=f(\sin\f rac{5\pi}{16})=f(\sin\frac{6\pi}{16})$ $=f(\sin\frac{7\pi}{16})=f(\sin\frac{8\pi}{16})=0$.

You can see how to proceed. We can prove by induction that for all positive integers $n$, $f(\sin \theta)=0$ whenever $\theta\in S_n=\{\frac{k\pi}{2^{n+1}}:0\leq k\leq 2^n\}$.

The function $g(\theta)=f(\sin\theta)$ is continuous on $[0,\pi/2]$ and vanishes on $S=\bigcup_{n=1}^\infty S_n$. Moreover $S$ is dense in $[0,\pi/2]$ (anyone else want to prove this?)

It follows that $g(\theta)=0$ on $[0,\pi/2]$ and therefore $f(x)=0$ on $[0,1]$ and therefore on $[-1,1]$.