Define $f : L^2 \rightarrow \mathbb{R}$ by $f(x) = \sum_{n=1}^{\infty} \frac{x_n}{n}$. Is $f$ continuous? Problem I came across studying for qualify exams. I think I have it, if $x^k$ is a sequences of sequences in $L^2$ and it converges to $x$ in $L^2$, then $d(x^k_n, x_n) \rightarrow 0$ as $k \rightarrow \infty$ for all $n$. Then consider $d(f(x^k), f(x)) = \sum_{n=1}^{\infty} \frac{x^k_n - x_n}{n}$. But since $d(x^k_n, x_n) \rightarrow 0$ as $k \rightarrow \infty$, $d(f(x^k), f(x)) \rightarrow 0$ as $k \rightarrow \infty$. Is this right? Also, I was curious if it's even true that $f(x)$ is finite for all $x \in L^2$. I'm suspicious that it's not necessarily true but I couldn't figure out a proof or find a counterexample. 
 A: If I am understanding your question correctly, by L2 you mean the space of square summable sequences (which I am more familiar with the notation $\ell_2$ or $\ell^2$ or $\ell_2(\mathbb{N})$). 
In this case, the function $f:\ell_2\to \mathbb{R}$ given by $f((x_n)_{n=1}^\infty)=\sum_{n=1}^\infty \frac{x_n}{n}$ is continuous. This is because, if $a=(n^{-1})_{n=1}^\infty$, then $a\in \ell_2$ (since $\sum_{n=1}^\infty \frac{1}{n^2}<\infty$). Therefore $|f(x)|\leqslant \|x\|_2 \|a\|_2$ by the Cauchy-Schwarz inequality. 
Moreover, $|f(x^k)-f(x)|\leqslant \|x^k-x\|_2 \|a\|_2 \underset{k\to \infty}{\to} 0$, also by Cauchy-Schwarz. 
A: Hint:
$$|f(x) - f(y)| \le \frac{\pi}{\sqrt 6} \left( \sum_{k=1}^\infty (x_k - y_k)^2 \right)^{1/2}.$$
A: This is a map of vector spaces, so continuity of the map in general is equivalent to continuity at $0$, and furthermore, equivalent to boundedness of the map $f$. So then it is enough to prove or disprove that if $\sum_{n=1}^\infty x_n^2 = 1$, then $\sum_{n=1}^\infty \frac{x_n}{n} < \infty$. We will obtain this from the Cauchy-Schwarz inequality.
We have in general $$\bigg(\sum_{n=1}^\infty a_n b_n \bigg)^2 \leq \big(\sum_{n=1}^\infty a_n^2 \big) \big( \sum_{n=1}^\infty b_n^2 \big)$$
So taking $a_n = 1/n$ and $b_n = x_n$, we obtain 
$$ \sum \frac{ x_n}{n} \leq \sqrt{\big(\sum\frac{1}{n^2} \big)  \big(\sum x_n^2} \big)$$
But the first term is known to be $\pi^2/6 < \infty$ and the second term is $1$ by hypothesis.
So the map is continuous and we even computed it's norm to be $\pi/\sqrt{6}$, as the Cauchy-Schwarz inequality is sharp.
