Note: this question has been asked before, here: How to find the norm of the operator $(Ax)_n = \frac{1}{n} \sum_{k=1}^n \frac{x_k}{\sqrt{k}}$?, but it wasn't definitively answered, so I'd like to post my thoughts on the problem, and ask for a solution if anyone has any ideas.
So, the question is:
For which $p \geq 1$ is $A: \ell^p \to \ell^p$, $$ A(x_{n})_{n \geq 1} = \big( \frac{1}{m} \sum_{n=1}^{m}\frac{x_{n}}{\sqrt{n}} \big)_{m \geq 1} $$ a bounded linear operator? For those $p$ for which $A$ is bounded, find $||A||$, where $||\cdot||$ is the operator norm.
Now, I had two ideas: use convexity of $x \mapsto x^p$, or use Hölder's inequality. However, each approach is far less sophisticated than Norbert's answer in the linked question, because I didn't know about Hardy's inequality before reading the question. The convexity approach got me nowhere, while the Hölder approach got me the following upper bound: $$||A(x_{n})_{n \geq 1}||_{p} \leq \sqrt[p]{\sum_{m=1}^{\infty} \frac{S_{m}}{m^p}} \cdot ||(x_{n})_{n \geq 1}||_{p},$$
where $S_{m} = (\sum_{n=1}^{m} \frac{1}{n^{\frac{q}{2}}})^{\frac{p}{q}}$. The series converges because $\sum_{n=1}^{m} \frac{1}{n^{q/2}} \sim m^{1-q/2}$, and so the summands behave like $m^{-p+\frac{p}{q}-\frac{p}{2}}$, and $\frac{p}{q}-\frac{p}{2}-p = p-1-\frac{p}{2}-p = -1-\frac{p}{2} <-1$.