Real analysis - norm of bounded limit points

Question

I've been struggling with a real analysis problem for 3 days, and I'd appreciate your help with it.

Let $||\cdot||$ be an arbitrary norm in $\mathbb{R}^n$ and let $(x_m)$ be a sequence in $\mathbb{R}^n$ such that $||x_m||\leq c$, $\forall m\in\mathbb{N}$. If $\lim x_m = a$, show that $||a||\leq c$.

I do understand what the exercise says (if the norm of a sequence is bounded for all $m$, then the norm of its limiting point should be bounded too), but I can't find a way to prove it right. So far I've tried the following:

\begin{eqnarray} |~||x_m||-c~| &=& |~||x_m||-c+||a||-||a||~| =|~(||x_m||-||a||+(||a||-c)~|\\ &\leq& |~||x_m||-||a||~|+|~||a||-c~|\tag{1} \end{eqnarray}

So the first $|\cdot|$ of the left hand of (1) should converge to zero (since $\lim x_m = a$ implies $\lim ||x_m|| = ||a||$, that was another exercise), but I can't get any further..

Thanks in advance

Answer 1

This can be shown directly using some properties of inf and sup on the sequence $x_m$.

Note that $||x_m|| \leq c \implies \sup(\{||x_m||\}) \leq c$. If $x_m \to a$, then $$\inf(\{||x_m||\}) \leq ||x_m|| \leq \sup(\{||x_m||\}) \implies \inf(\{||x_m||\}) \leq ||a|| \leq \sup(\{||x_m||\})$$ As $\sup(\{||x_m||\}) \leq c$, then $||a|| \leq c$.

Answer 2

You say that you know that $\lim x_m = a$ implies $\lim \|x_m\| = \|a\|$. From here (along with the fact $\|x_m\| \le c$ for all $m$) you can immediately arrive at $\|a\| \le c$. If this is not clear to you, consider the contrapositive: show that if $\|a\| > c$, then (since $\lim \|x_m\| = \|a\|$) there is some $m$ such that $\|x_m\| > c$ as well.

Answer 3

For an arbitrary $\epsilon > 0$, for sufficiently large $m$ we have $$| ||x_{m}|| - ||a|| | \leq \epsilon$$ thus $$||a|| \leq \epsilon + ||x_{m}|| \leq \epsilon + c$$ Given that $\epsilon$ is arbitrary: $$||a|| \leq c$$