I am having problems with the following proof and I need to fill in some details:

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  • I understand that continuity is being proven by the sequence definition but I do not get why (a) follows immediately after (2.1.10). Could I get the extra steps?

Also could I have a brief epsilon-delta proof for point (b) after having applied Cauchy-Swartz as I am unsure on how do to this by epsilon-delta?


The point is that we want to show that we can get $\|x_n\|$ as close to $\|x\|$ as we wish based on the assumption that we can get $\|x_n - x\|$ as close to $0$ as we wish. The bound $|\|x_n\| - \|x\|| \leq \|x_n -x\|$ guarantees this, because we can just take the limit on both sides: $$\lim_{n \to \infty} |\|x_n\| - \|x\|| \leq \lim_{n \to \infty} \|x_n - x\| = 0$$ where the last limit is equal to zero by assumption.

Here is the proof written out with epsilons. Assume that $\lim_{n \to \infty} \|x_n - x\| = 0$. We want to show that $\lim_{n \to \infty} \|x_n\| = \|x\|$.

Let $\epsilon > 0$. By assumption, we can find an $N$ such that $\|x_n - x\| < \epsilon$ when $n \geq N$. So when $n \geq N$, we get $$\left| \|x_n \| - \| x \| \right| \leq \|x_n - x\| < \epsilon$$ which proves the claim.

  • $\begingroup$ Thanks! could you give a brief epsilon-delta as you just did for the end of point (b)? I am having troubles dealing with $||x_n|| $ and $||y||$. $\endgroup$ – Monolite May 22 '15 at 22:25
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    $\begingroup$ Let $\epsilon > 0$. We want to find an $N$ such that $\|x_n\|\|y_n - y\| + \|y\|\|x_n - x\| < \epsilon$ when $n \geq N$. First of all, since $\|x_n\|$ converges, it is bounded, by $K$, say. Secondly, we can find $M$ and $M'$ such that $\|y_n - y\| < \epsilon/(2K)$ when $n \geq M$and $\|x_n - x\| < \epsilon/(2\|y\|)$ when $n \geq M'$. Now you can finish the argument by showing that $N = \max\{ M,M'\}$ works. $\endgroup$ – Ulrik May 23 '15 at 13:36
  • $\begingroup$ @Ulrik thanks for your comment! However, doesn't your argument fail if, by any chance, $||y||=0$? You are using $\frac{\epsilon}{2||y||}$ but if $||y||=0$ then this is undefined $\endgroup$ – Euler_Salter Oct 29 '18 at 19:26
  • $\begingroup$ @Ulrik I might have found a solution. We could use $\frac{\epsilon}{2(||y||+1)}$ and then use the fact that $||y||+1 \geq ||y||$ and therefore we have that $||x_n-x||||y|| \leq \frac{\epsilon}{2(||y||+1)} ||y|| \leq \frac{\epsilon}{2(||y||+1)} (||y|| + 1) = \frac{\epsilon}{2}$ $\endgroup$ – Euler_Salter Oct 29 '18 at 19:39

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