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I encountered this problem on a graduate school entrance test :

Let $\{f_n\}$ be a sequence of real-valued continuous functions on $\mathbb{R}$ such that $$f_n\left(x + \frac{1}{n}\right) = f_n(x) \hspace{2mm} \forall \hspace{2mm} x \in \mathbb{R} \text{ and } n \in \mathbb{N}.$$ Suppose $f:\mathbb{R} \to \mathbb{R}$ is such that $\{f_n\}$ converges uniformly to $f$ on $\mathbb{R}$, then show that $f$ is a constant function.

My attempt :

Let $x,y \in \mathbb{R}$ and $\epsilon >0$ be arbitrary. It suffices to show that $|f(x) - f(y)| < \epsilon$. By triangle inequality, given any $n \in \mathbb{N}$ : $$|f(x) - f(y)| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)|$$ By uniform convergence, $\exists$ $N \in \mathbb{N}$ such that $|f(x) - f_n(x)| < \epsilon/3$ and $|f(y) - f_n(y)| < \epsilon/3 $ whenever $n > \mathbb{N}$.

So, now it suffices to show that $|f_n(x) - f_n(y)| \to 0$. Now comes the confusing part :

  • I fix an $n$.
  • Continuity of $f_n$ implies that existence of $\delta >0$ such that $|f_n(t)-f_n(y)| < \epsilon$ whenever $|t-y|<\delta$.
  • Find $n'$ such that $\frac{1}{n'} < \delta$. Then, $\exists$ $k \in \mathbb{N}$ such that $\left|\left(x+\frac{k}{n'}\right) - y\right| < \delta$.
  • But now I can't say that $|f_{n'}(x)-f_{n'}(y)| = |f_{n'}(x+\frac{k}{n'})-f_{n'}(y)| < \epsilon$ as $f_{n'}$ might require a smaller $\delta'$ than $f_n$.

I hope I have made my point clear. If not, feel free to ignore my attempt and post your own solution.

Any help/hints shall be highly appreciated.

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2 Answers 2

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Let $x \in \Bbb R$ be arbitrary and let $m = m(n) \in \Bbb Z$ be such that $y := x - \frac m n \in \left[0, \frac 1 n \right)$. Observe,

$$ f_n(x) = f_n \left(x - \frac 1 n \right) = f_n \left(x - \frac 2 n \right) = \dots = f_n \left(x - \frac m n \right) = f_n \left(y \right) $$

Since $f_n$ continuous, and $f_n \rightarrow f$ uniformly, by the Uniform Limit Theorem we have that $f$ is continuous.

Now, for any $\epsilon > 0$, we can find $N \in \Bbb{N}$ large enough, such that for all $n \ge N$ we have

$$ \left|f_n(x) - f(0)\right| =\\ \left|f_n(y) - f(0)\right| =\\ \left|f_n(y) - f(y) + f(y) -f(0)\right| \le\\ \left|f_n(y) - f(y)\right| + \left|f(y) - f(0)\right| \le\\ \dfrac \epsilon 2 + \dfrac \epsilon 2 =\\ \epsilon $$

where the bound on the first term is due to uniform convergence, and the bound on the second term is due to the continuity of $f$. Note that $\epsilon$ is a uniform bound.

Finally,

$$ |f(x) - f(0)| = \lim_{n \rightarrow \infty} |f_n(x) - f(0)| \le \epsilon $$

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    $\begingroup$ +1 I don't understand why this was not the accepted answer. It is so much clearer and simpler. $\endgroup$
    – Mark Viola
    Jun 16, 2022 at 13:15
  • $\begingroup$ Can’t cater to everyone :-) $\endgroup$
    – Joe Shmo
    Jun 16, 2022 at 13:30
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    $\begingroup$ +1. This method can easily adapt to prove the generalization that I stated in a comment to my A. $\endgroup$ Jun 16, 2022 at 16:24
  • $\begingroup$ yours was a great answer, too, @DanielWainfleet. $\endgroup$
    – Joe Shmo
    Jun 16, 2022 at 19:30
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For any $q= a/b\in \Bbb Q,$ with $a\in\Bbb Z$ and $b\in\Bbb N,$ for any $n\in \Bbb N$ we have $$\forall j\in\Bbb Z\,(\,f_{nb}(j/nb)=f_{nb}((j+1)/nb)\,)$$ so $f_{nb}(0)=f_{nb}(a/b)=f_{nb}(q).$ Therefore $$f(0)=\lim_{n\to\infty} f_{nb}(0)=\lim_{n\to\infty}f_{nb}(q)=f(q).$$ So $f$ is constant on $\Bbb Q.$ Now $f_m\to f$ uniformly, and each $f_m$ is continuous, so $f$ is continuous. So $f$ is continuous and $f$ is constant on $\Bbb Q$ so $f$ is constant.

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  • $\begingroup$ That is a very clever answer. Thanks a lot ! $\endgroup$ Jun 16, 2022 at 2:25
  • $\begingroup$ @Another_Ramanujan_Fan . I just thought of a slightly more difficult generalization: Suppose each $f_n$ is continuous and periodic with period $p_n.$ If $f_n\to f$ uniformly and $p_n\to 0$ then $f$ is constant. $\endgroup$ Jun 16, 2022 at 6:39
  • $\begingroup$ Yess ! Brilliant. $\endgroup$ Jun 16, 2022 at 10:10

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