Uniform convergence to a continuous function

Question

The problem goes as follows:

$$\text{Show that if } f_n \text{ uniformly converges to a continuous function } f \text{, then }\\ f_n(x+\frac{1}{n}) \text{ pointwise converges to} f(x) \text{ }\forall x \in \mathbb{R}$$

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I know that: $$ x + \frac{1}{n} \to x \text{ as } n \to \infty \Rightarrow f(x + \frac{1}{n}) \to f(x)\text{ as }n \to \infty $$

I also know that: $$\text{Since } f_n \text{ uniformly converges to } f, \forall \varepsilon >0 \text{ }\exists N\in\mathbb{N} \text{ such that } \forall x \in D_f, \forall n \ge N\\|f_n(x)-f(x)|<\varepsilon$$

Don't know how to proceed but I haven't used the fact that $ f $ continuous yet.

Thanks in advance!

Answer 1

Let $\epsilon>0$. By uniform convergence we can find $N$ such that $|f_{n}-f|<\epsilon/2$ for all $n>N$. By continuity we can find $\delta_{x}$ such that $|f(y)-f(x)|<\epsilon/2$ for all $y$ satisfying $|x-y|<\delta_x$. Let $M=\max(N,1/\delta_{x})$. Then, for all $n>M$, \begin{align*} \left|f_{n}(x+1/n)-f(x)\right| &=\left|f_{n}(x+1/n)-f(x+1/n)+f(x+1/n)-f(x)\right|\\ &\leq\left|f_{n}(x+1/n)-f(x+1/n)\right|+\left|f(x+1/n)-f(x)\right|\\ &<\epsilon/2 + \epsilon/2=\epsilon. \end{align*}

Answer 2

It can be proved that if $f_n $ is uniformly convergent to a continiuous function $f.$ Then for every sequence $x_n \to x$ we have $f_n (x_n ) \to f(x) .$ This follows from inequality

$$|f_n (x_n ) - f(x) |\leq |f_n (x_n) - f(x_n) | + |f(x_n ) - f(x) |$$