# Problem of the Week # 268 - Jun 20, 2017

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#### Ackbach

##### Indicium Physicus
Staff member
Here is this week's POTW:

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Let $f(x)$ be a continuous function such that $f(2x^2-1)=2xf(x)$ for all $x$. Show that $f(x)=0$ for $-1\leq x\leq 1$.

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#### Ackbach

##### Indicium Physicus
Staff member
Re: Problem Of The Week # 268 - Jun 20, 2017

This was Problem B-4 in the 2000 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

For $t$ real and not a multiple of $\pi$, write $g(t) =\frac{f(\cos t)}{\sin t}$. Then $g(t+\pi) = g(t)$; furthermore, the given equation implies that
$g(2t) = \frac{f(2\cos^2 t - 1)}{\sin (2t)} =\frac{2(\cos t) f(\cos t)}{\sin(2t)} = g(t).$
In particular, for any integer $n$ and $k$, we have
$g(1+n\pi/2^k) = g(2^k + n\pi) = g(2^k) = g(1).$
Since $f$ is continuous, $g$ is continuous where it is defined; but the set $\{1+n\pi/2^k | n,k\in{\mathbb{Z}}\}$ is dense in the reals, and so $g$ must be constant on its domain. Since $g(-t) = -g(t)$ for all $t$, we must have $g(t) = 0$ when $t$ is not a multiple of $\pi$. Hence $f(x) = 0$ for $x \in (-1,1)$. Finally, setting $x=0$ and $x=1$ in the given equation yields $f(-1) = f(1) = 0$.

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