Let $$M=\begin{bmatrix} 1 &\frac{1}{2}&\frac{1}{3}&\frac{1}{4} \dots\\ 0 &\frac{1}{2}&\frac{1}{3}&\frac{1}{4} \dots\\ 0 & 0 &\frac{1}{3} &\frac{1}{4} \dots\\ \vdots & \vdots &\dots \end{bmatrix}$$
I need to find out if this matrix defines a map from $l^2$ to $l^2$. For that I look at a typical $y(j)$ position, meaning if I multiply $M$ by $x=\left(x(1),x(2),..,x(j),...\right)$, I get $y=\left(y(1),y(2),..,y(j),..\right)$ and for $M$ to be a map each $y(j)$ should make sense and whole of $y$ should make sense.
For each $y(j)$ to make sense , $y(j)=\sum_{k=j}^{\infty} \dfrac{1}{k}x(k)$, should converge. So I look at $$\sum_{k=j}^{\infty}|\frac{1}{k} x(k)| \le \sqrt{\sum_{k=j}^{\infty}\frac{1}{k^2}} \times ||x||_2$$ So the series converges absolutely and hence it converges. No problem with this.
Now $y$ should be in $l^2$ for everything to fall through. So I look at $$||y||^{2}_2=\sum_{j=1}^{\infty} |\left(\sum_{k=j}^{\infty}\frac{1}{k}x(k)\right)|^2 \le \sum_{j=1}^{\infty}\left(\sum_{k=j}^{\infty}|\frac{1}{k}x(k)|\right)^2 $$ $$\le \sum_{j=1}^{\infty}\{\left(\sum_{k=j}^{\infty}\frac{1}{k^2}\right)\left(\sum_{k=j}^{\infty}|x(k)|^2\right)\} $$
This is where I am stuck . I can always pull out $||x||_2$ but that doesn't help. I have a double series here. To solve this problem it seems to me that i have to show that this is double series converges. Cauchy-Schwarz is the best possible approximation I can get. So I don't think I have to do something with that.
I tried to use the fact that the tail of the series goes to zero but that doesn't help either. (since there is a difference between going to zero and actually being zero)
Now suppose I multiply the matrix with $ x=(0,...0, j,0,0,0..) $ where $ j $ is at the $ k$ th place, then $ y=(1,1,1,...,1,0,0..0) $ where the last$1$ is at the $ k $ th place. Also $||y||_2=\sqrt {k}$.
This doesn't lead to anything. I think that finally it boils down to choosing $ x $ such that the series on the right diverges.
Suppose I multiply by an arbitrary $x=\left(x(1),x(2),..,x(j),..\dots \right)$, (of course $x \in l^2$)then $y=(y(1),y(2),..,y(j),..\dots )$, where $y(j)=\sum_{k=j}^{\infty} \dfrac{x(k)}{k}$.
I am unable to find an $x$. Thanks for the help!!
To prove this, it seems to me that one could follow the proof of the Schur test (e.g., Exercise 9 in II.1 of "A course on functional analysis" by Conway or Problem 45 in "A Hilbert space problem book" by Halmos).
I guess it could be summarized as follows (in loose terms). The sequence we are interested in is
$$ y_i=\sum_{j=i}^\infty\frac{1}{j}x_j $$
where $x$ has finite $l^2$-norm. One has
$$ \begin{align} \lvert\lvert y\rvert\rvert^2=\sum_{i=1}^\infty\,\lvert y_i\rvert^2 &= \sum_{i=1}^\infty\,\left\lvert\sum_{j=i}^\infty \frac{1}{j}x_j \right\rvert^2 \\ &= \sum_{i=1}^\infty\,\left\lvert\sum_{j=i}^\infty \frac{1}{j^{3/4}}\frac{x_j}{j^{1/4}} \right\rvert^2\leq\sum_{i=1}^\infty\,\left(\sum_{j=i}^\infty\frac{1}{j^{3/2}}\right)\left(\sum_{j=i}^\infty\frac{\lvert x_j\rvert^2}{j^{1/2}}\right)\\ &\leq\sum_{i=1}^\infty\,\frac{\alpha}{i^{1/2}}\left(\sum_{j=i}^\infty\frac{\lvert x_j\rvert^2}{j^{1/2}}\right) \end{align} $$
Here, the last inequality follows from the fact that the first sum over $j$ can be majorized by $\alpha/i^{1/2}$ for suitable $\alpha>0$. Now, we can interchange the order of summation in the $i$-$j$ plane to get
$$ \begin{align} \lvert\lvert y\rvert\rvert^2\leq\sum_{j=1}^\infty\,\sum_{i=1}^{j}\frac{\alpha}{i^{1/2}}\frac{\lvert x_j\rvert^2}{j^{1/2}} = \sum_{j=1}^\infty\,\frac{\lvert x_j\rvert^2}{j^{1/2}}\sum_{i=1}^{j}\frac{\alpha}{i^{1/2}} \end{align} $$
However, one can find $\beta>0$ such that $j^{-1/2}\sum_{i=1}^{j}i^{-1/2}\leq\beta$. In total, we thus obtain
$$ \lvert\lvert y\rvert\rvert^2\leq\alpha\beta\,\lvert\lvert x\rvert\rvert^2 $$
Note: Schur test (from Conway's book)
Let $\{\alpha_{i,j}\}_{i,j=1}^\infty$ be an infinite matrix such that $\alpha_{i,j}\geq0$ for all $i,j$ and such that there are scalars $p_i>0$ and $\beta,\gamma>0$ with
$$\sum_i \alpha_{i,j}p_i\leq \beta\,p_j $$ and $$\sum_j \alpha_{i,j}p_j\leq \gamma\,p_i $$
for all $i,j\geq1$. Then there is an operator $A$ on $l^2(\mathcal{N})$ with $\langle e_i\vert A e_j\rangle = \alpha_{i,j}$ and $\lvert\lvert A \rvert\rvert^2\leq \beta\gamma$.