Lipschitz and Continuity. Proof assistance request.

Question

I was hoping to get some assistance with my proof. The question asks:

Define $g: \ell_2 \rightarrow \mathbb{R}$ by $g(x)=\sum_{n=1}^{\infty} \frac{x_n}n$. Is g continuous?

I have a feeling that if I can show that this is Lipschitz then the function is continuous. Here is what I have so far:

Let $x\in \ell_2$ then $g(x)=\sum_{n=1}^{\infty} \frac{x_n}n$. But because $x\in \ell_2$ I know that $[ \sum_{k=1}^{\infty}(x_k)^2]^\frac{1}2<\infty$ . And so $\sum\frac{x_n}n<C[ \sum_{k=1}^{\infty}(x_k)^2]^\frac{1}2<\infty$. Take $C=n^2$ then we have that $\sum\frac{x_n}n<n^2[ \sum_{k=1}^{\infty}(x_k)^2]^\frac{1}2$ proving that g(x) is continuous.

I don't know about my choice of c. It seemed to make sense because because of how g(x) was defined. So $x_1+\frac{x_2}2 + ...<n^2(x_1^2+x_2^2+...)^\frac{1}2$. The problem I'm having is that I don't know what n actually is!! The sum goes to infinity!

Answer 1

Let $y$ be the sequence $(1/n).$ Then $g(x) = \langle x,y \rangle, x \in l^2.$ Any map of this form is a continuous linear functional on $l^2.$

Answer 2

$g$ is linear so it suffices to consider $x_0=0$. Then, by Cauchy-Schwarz,

$\vert g(x)\vert=\vert \sum_{n=1}^{\infty} \frac{x_n}n\vert \le \left \| x \right \|_{2}\sqrt{\sum_{n=1}^{\infty}\frac{1}{n^{2}}}\to 0$ as $\left \| x \right \|_{2}\to 0$.