Exercise on convergence of sequences of functions Let $f: \mathbb{R} \to \mathbb{R}$ be a function. For every $n \in \mathbb{N}_{0}$, define the function $f_{n}: \mathbb{R} \to \mathbb{R}: x \mapsto f(x+\frac{1}{n})$.
Now, I have to examine if the following statements are true or false:
a) If $f$ is continuous, then $(f_{n})_{n}$ converges pointwise on $\mathbb{R}$ to $f$.
b) If  $f$ is uniformly continuous, then $(f_{n})_{n}$ converges uniformly on $\mathbb{R}$ to $f$.
I think that the first statement is true and I also think that I got a proof for it. I also think that the second statement is true but I have some doubts about it, is there someone that can help me?
 A: a) Let $x \in \mathbb R$ and define $x_n: =x+\frac{1}{n}$. Then $x_n \to x$ and therefore, since $f$ is continuous: $f(x_n) \to f(x)$. But this means: $f_n(x) \to f(x)$.
b) Let $\epsilon>0$. There is $ \delta >0$ such that $|f(x)-f(y)|< \epsilon$ if $|x-y| < \delta$.
Now let $N \in \mathbb N$ such that $1/N < \delta$. For $x \in \mathbb R$ and $n>N$ we then have
$|f_n(x)-f(x)|< \epsilon.$
A: *

*Observe that, for any fixed $x$, you can define the sequence $(x_n)_n$ by $x_n\stackrel{\rm def}{=} x+\frac{1}{n}$. Then, clearly, $\lim_{n\to\infty} x_n=x$, and furthermore by definition
$$f_n(x) = f(x_n)$$
Now, if $f$ is continuous, what can you say about convergence of $f(x_n)$? (This is the sequential characterization of continuity.)

*For the second... pick any $\varepsilon >0$, and let $\delta>0$ be a corresponding delta of uniform continuity: that is, such that
$$
\forall x,y,\text{ if }\lvert x-y\rvert \leq \delta \text{ then } \lvert f(x)-f(y)\rvert \leq \varepsilon \tag{$\dagger$}
$$
By definition, for any real $x$ and integer $n\geq1$, we have
$$
\lvert f_n(x)-f(x) \rvert= \left\lvert f\left(x+\frac{1}{n}\right)-f(x) \right\rvert
$$
Fix $N\stackrel{\rm def}{=}\left\lceil 1/\delta\right\rceil$ (note that it does not depend on $x$). For any $n\geq N$ we have $\lvert x+\frac{1}{n}-x \rvert \leq \delta$ for every real $x$, and therefore by $(\dagger)$
$$
\lvert f_n(x)-f(x) \rvert \leq \varepsilon.
$$
Since this holds for every $x$, we have
$$
\forall n\geq N,\, \sup_{x\in\mathbb{R}} \lvert f_n(x)-f(x) \rvert \leq \varepsilon
$$
showing uniform convergence.
