On the proof of the continuity of the inner product.

Question

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

• 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?

Answer 1

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.