Uniform convergence - is this too easy?

Question

Let $f_n:\mathbb{R} \rightarrow \mathbb{R}$ converge uniformly to $f$ with $f$ is uniformly continous. Show that $f_n(x+a_n)$ converges uniformly to $f$ where $a_n \rightarrow 0$ for $n \rightarrow \infty$.

So whats the point here? I think the problem is solved just with $\lim_{n \rightarrow \infty} f_n(x+a_n) = f(x+\lim_{n \rightarrow \infty }a_n) = f(x)$ ??

This is actually a homework so I guess it cant be that easy. Is there something wrong?

Answer 1

Yes, there's something wrong. What you did would prove, at most, that $(f_n)_{n\in\mathbb N}$ converges pointwise to $f$.

Let $\varepsilon>0$. Take $\delta>0$ such that $|x-y|<\delta\implies\bigl|f(x)-f(y)\bigr|<\varepsilon$. Take $p\in\mathbb N$ such that $n\geqslant p\implies|a_n|<\delta$. Then $\bigl|(x+a_n)-x\bigr|<\delta$ (for each real $x$). So $\bigl|f(x+a_n)-f(x)\bigr|<\varepsilon$. In other words, $\bigl|f_n(x)-f(x)\bigr|<\varepsilon$.

Answer 2

No, your proof idea has problems. Where did you use uniform converge and uniform continuity?

Here's a hint for a proof:

$$|f_n(x+a_n)-f(x)| \le |f_n(x+a_n)-f(x+a_n)| + |f(x+a_n)-f(x)|.$$