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.
Best Answer
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 $\varepsilon > 0$, we can find $N \in \Bbb{N}$ large enough, such that for all $n \ge N$ we have
$$ \left|f(x) - f(0)\right| =\\ \left|f(x) - f_n(x) + f_n(x) - f(y) + f(y) - f(0)\right| =\\ \left|f(x) - f_n(x) + f_n(y) - f(y) + f(y) - f(0)\right| \le\\ \left|f(x) - f_n(x)\right| + \left|f_n(y) - f(y)\right| + \left|f(y) - f(0)\right| \le\\ \dfrac \varepsilon 3 + \dfrac \varepsilon 3 + \dfrac \varepsilon 3 =\\ \varepsilon $$
where the bound on the first term is due to the convergence $f_n \to f$, the bound on the second term is due to the uniform convergence $f_n \rightrightarrows f$, and the bound on the third term is due to the continuity of $f$. Note that $\varepsilon$ is a uniform bound.