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Let $f : (−1, 1) \rightarrow R $ be a continuous function with the property that $f(x) = f(x^4)$ for all $x$ and $f(\frac{1}{2}) = a$. Show that the only function $f$ satisfying this property is the constant function $f(x) = a$.

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up vote 4 down vote accepted

We have $a=f(1/2)=f(1/2^4)=f(1/2^{16})$ and so on. So by continuity $f(0)=a$,

Suppose that $f(k)=b$ for some $k$ with $|k|\lt 1$. The same argument shows that $f(0)=b$.

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Note that if $x\in I=(-1,1)$, then $x^4 \in I$. Since $f(x) = f(x^4)$, we have $f(x) = f(x^{4^n})$ for $n \in \mathbb{N}$. Since $\lim_n x^{4^n} = 0$ and $f$ is continuous, we have $\lim_n f(x^{4^n}) = f(0)$. Hence $f(x) = f(0)$ for all $x \in I$. It follows that $f(x) = f(0) = f(\frac{1}{2})$.

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Since $f(x)=f(0) \forall x$ , we conclude $f(x)$ is free from x. – A.D Jan 27 '13 at 7:22

HINT: Let $A_0=\left[\frac1{2^4},\frac12\right]$, $A_1=\left[\frac1{2^{16}},\frac1{2^4}\right]$, and in general


then $f[A_k]=f[A_0]$ for all $k\in\Bbb N$. Let $u=\max_{x\in A_0}f(x)$; then for each $k\in\Bbb N$ there is an $x_k\in A_k$ such that $f(x_k)=u$. Now $$f\left(\frac1{2^{4^k}}\right)=a$$

for each $k\in\Bbb N$, and the points $1/2^{4^k}$ converge to $0$, so by continuity $f(0)=a$. On the other hand, the points $x_k$ also converge to $0$, so by continuity $f(0)=u$, and we conclude that $u=a$. Essentially the same argument shows that $\min_{x\in A_0}f(x)=a$, so $f$ is constant on $\left[0,\frac12\right]$. It’s not hard now to see that $f$ must in fact be constant on $[0,1]$, and since $(-x)^4=x^4$, that it’s constant on $(-1,1)$.

Added: This is not the most efficient argument; I’ll leave it partly because the technique may be useful in another setting and partly as an illustration of how it can pay to reexamine an argument to see whether it can be simplified. If one looks a little more closely at this argument, one may notice that it really just relies on the fact that if $x\in(-1,1)$, then the sequence $\left\langle x,x^4,\left(x^4\right)^4,\dots\right\rangle$ converges to $0$. At that point the simpler argument posted by André Nicolas and copper.hat quickly emerges.

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