Elementary question on pointwise convergence and norm continuity

Question

Let $\;a_n \in \mathbb R\;$ be a bounded sequence, then by Bolzano-Weierstrass Theorem it follows there exists a subsequence $\;a_{n_k}\subset a_n\;$ such that $\;a_{n_k} \to a \in \mathbb R\;$. If $\;f:\mathbb R \to \mathbb R^m\;$ a continuous function, then $\;f(x-a_{n_k}) \to f(x-a)\;$ pointwise.

In addition, $\;\forall g \in C(\mathbb R;\mathbb R^m)\;$ it holds: $\;{\vert g(x)-f(x-a_{n_k}) \vert}^2 \to {\vert g(x)-f(x-a) \vert}^2\;$ also pointwise.

NOTE: $\; \vert \cdot \vert \;$ stands above for the Euclidean norm

As I was studying, the above part of my notes confused me a little bit.

QUESTIONS:

1. Why is $\;f(x-a_{n_k}) \to f(x-a)\;$ pointwise? I know that uniform convergence is much stronger than pointwise but why is that the case here?
2. Why does the second convergence hold? Is it because $\;\vert \cdot \vert\;$ is also continuous?

Any help would be valuable. Thanks in advance!

Answer 1

For fixed $x$, then $\{x-a_{n_{k}}\}_{k}$ is a sequence such that $x-a_{n_{k}}\rightarrow x-a$. Then by sequential characterisation of the continuity of $f$, we have $f(x-a_{n_{k}})\rightarrow f(x-a)$.

For the second one, appeal to the composition with the continuous map $|\cdot|^{2}$.

Answer 2

1. Since $\lim_{k\to\infty}a_{n_k}=a$ and since $f$ is continuous, $\lim_{k\to\infty}f(x-a_{n_k})=f(x-a)$.
2. Since $\lim_{k\to\infty}a_{n_k}=a$ and since $f$ and $g$ are continuous, $\lim_{k\to\infty}\bigl|g(x)-f(x-a_{n_k})\bigr|^2=\bigl|g(x)-f(x-a)\bigr|^2$.

I didn't understand your reference to uniform continuity. It is not relevant here.

Answer 3

Consider $y_{n_k}=x-a_{n_k}$ and $y =x-a $ for any fixed $x\in \mathbb{R}$, then $y_{n_k} \rightarrow y $ in $\mathbb{R}$. Now as both $f$ and $g$ is continuous in $\mathbb{R},$ you can have both the convergence. Also you have to use the fact $||$ is continuous and composition of two continuous maps is continuous to get the last result.