Let $f \colon [0,1] \to \mathbb R$ be continuous and let $f_n$ be defined by $$ f_n(x):=f(x^n), \qquad x \in [0,1], \,\, n \in \mathbb N. $$ Suppose there exists a continuous function $g$ on $[0,1]$ such that $f_n \to g$ uniformly. Then $f$ is constant.

Let $\overline{x} \in [0,1)$ be fixed. Then we have that the sequence of real numbers $x_n:=\overline{x}^{n} \to 0$ as $n \to +\infty$.

On the one hand, by continuity, we have $$ f(x_n) \to f(0); $$ on the other hand, being $f(x_n) = f(\overline{x}^{n}) = f_n(\overline{x})$, by pointwise convergence, we deduce
$$ f(x_n) = f_n(\overline{x}) \to g(\overline{x}). $$ Hence for every $x \in [0,1)$ we have $g(x)= f(0)$. If $x=1$ then we have $f(1)=f_n(1) \to g(1)$. By continuity, $g$ must be constant and in particular we have $f(0)=f(1)$.

Now for any $\overline{x} \in (0,1)$ we have $$ f(\overline{x}) = f_n(\overline{x}^{1/n}) \to g(1) $$ because $f_n \to g$ uniformly and $\overline{x}^{1/n} \to 1$. Hence we must have $f(x)=g(1)$ for any $x \in (0,1)$.

Thus we deduce that $f$ is constant on $[0,1]$.

Is this correct? Thanks.

  • 1
    $\begingroup$ Yes, it's correct. $\endgroup$ – Hamou Jul 24 '14 at 21:31
  • $\begingroup$ Thanks for your reply. I wait if something else also replies. $\endgroup$ – Romeo Jul 25 '14 at 16:52
  • $\begingroup$ This is indeed correct; except that $f$ should be $g$ in your statement. $\endgroup$ – Etienne Jul 25 '14 at 16:55

Yes it is correct. Maybe the argument used to show that $f(\overline{x}) = f_n(\overline{x}^{1/n}) \to g(1)$ should be detailed a little bit more. We use the fact that if $f_n\to g$ uniformly and $x_n\to x$, then $f_n(x_n)\to g(x)$. By the way, it seems it is the only place where we use uniform convergence.


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