# Uniform convergence of $\{f_n\}$ satisfying $f_n\left(x + \frac{1}{n}\right) = f_n(x)$ implies that the limit is a constant function

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.

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,

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

• +1 I don't understand why this was not the accepted answer. It is so much clearer and simpler. Jun 16, 2022 at 13:15
• Can’t cater to everyone :-) Jun 16, 2022 at 13:30
• +1. This method can easily adapt to prove the generalization that I stated in a comment to my A. Jun 16, 2022 at 16:24
• yours was a great answer, too, @DanielWainfleet. Jun 16, 2022 at 19:30

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.

• That is a very clever answer. Thanks a lot ! Jun 16, 2022 at 2:25
• @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. Jun 16, 2022 at 6:39
• Yess ! Brilliant. Jun 16, 2022 at 10:10