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I would like to prove the following:

Lemma. Let $f:X \rightarrow \mathbb{R}^n$ a continuous function. Let $x_n \in X$ such that $\lim_n x_n = a \in X$ and $\Vert f(x_n)\Vert < c$ for all $x_n \in X$. Then $\Vert f(a)\Vert \leq c$.

My attempt

We know that $\lim_n x_n = a$. Since $f$ is continuous, $\lim_n f(x_n) = f(a)$. Let $\epsilon >0$. Then, there exists $n_0$ such that $\Vert f(x_n) - f(a)\Vert < \epsilon$ whenever $n > n_0$. Then, $$\Vert f(a)\Vert - \Vert f(x_n)\Vert \leq \Vert f(x_n) - f(a)\Vert < \epsilon.$$ Using $\Vert f(x_n)\Vert < c$, it follows that $\Vert f(a)\Vert < \epsilon + c$. Therefore, taking $\epsilon \downarrow 0$ yields the desired result.

Is that correct?

parsiad
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Joãonani
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    The strict inequality will not survive the limiting procedure. Consider $X=\mathbb{R}$, $n=1$ and $f(x)=x$. Then take $x_n =1-1/n$, $a=1$ and $c=1$ to see that it will not be true. – Severin Schraven Aug 24 '19 at 18:44
  • can you give me a hint? – Joãonani Aug 24 '19 at 18:49
  • What do you mean? It is not true in this form and I gave you a counterexample. It becomes true if you write $\Vert f(a) \Vert \leq c$. Then your proof works. – Severin Schraven Aug 24 '19 at 18:56
  • I believe your answer is correct, but as Severin said you can't take strict inequality at the end, just less than or equal to – MathTrain Aug 24 '19 at 19:01
  • oh, I just made a typo mistake, now is it correct? – Joãonani Aug 24 '19 at 19:05
  • Typo still exists at the top of the post. You also don't mention that $f$ is continuous outside of the title. You can edit your post to fix these issues. – parsiad Aug 24 '19 at 19:28
  • ok, thank you, it's done. – Joãonani Aug 24 '19 at 19:31
  • As a side note, you don't have to resort to proving it from first principles. Left you a short proof below. – parsiad Aug 24 '19 at 19:34
  • @user638057: Some $\LaTeX$ tips. Don't use ||, use \Vert. Also, don't use \displaystyle (unless you absolutely have to) in inline math. The result is hard to read as it messes with the line spacing. I edited your post; you can look at the source. – parsiad Aug 24 '19 at 19:44

1 Answers1

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Let $g\equiv\Vert f\Vert$ and note that $g$ is continuous since it is the composition of continuous functions. Take limits on both sides of $g(x_{n})\leq c$ to get $\lim_{n}g(x_{n})\leq c$. Since continuous functions and limits commute, it follows that $\Vert f(a)\Vert=g(a)=g(\lim_{n}x_{n})=\lim_{n}g(x_{n})\leq c$, as desired.

parsiad
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