Linearity of infinite dimensional matrix

Question

The operator $T: \ell^2 \to \ell^2$ is given by the infinite-dimensional matrix with matrix elements

\begin{align*} t_{kl} = \frac{\vert k - l \vert}{k^2l^2} \end{align*}

for all $k, l \in \mathbb{N}$. We are using the complete orthonormal system $e_n := ((\delta_{kn})_k)_n$.

The question is proving its linearity (in general I am struggling to do this for similar operators). Take for example the condition $T(x+y) = T(x) + T(y)$. I know that I can write \begin{align*} x = \sum_k \langle x \vert e_k \rangle e_k, \ \ y = \sum_k \langle y \vert e_k \rangle e_k. \end{align*} It is now easy to prove that for the first $n$ terms in this sum we have linearity for all $n \in \mathbb{N}$. I guess the linearity of $T$ should then maybe follow from the fact that it is bounded and/or that we can prove that the "tail" of our sum goes to $0$, but I just don't get what is and what isn't necessary to show our result.

Could anyone help me pointing out what steps need to be taken to formally prove such a question?

PS. I saw the somewhat similar question Proving linearity of an operator using boundedness., but I am unsure why it follows from the continuity that the follows can be swapped around and whether it is even applicable for the operator I proposed or for a general 'infinite matrix'.

Answer 1

We have $$\sum_{k,l}t_{kl}^2 \le \sum_{k,l}{2(k^2+l^2)\over k^4l^4}=4\sum_k{1\over k^2}\sum_l{1\over l^4}<\infty $$ Hence $T$ a Hilbert-Schmidt operator. In particular $T$ is bounded and linear on $\ell^2.$

The linearity follows from the fact that the operator is given by $$(Tx)_k=\sum_{l=1}^\infty t_{kl}x_l$$ and the series is absolutely convergent, because $$\sum_{l=1}^\infty t_{kl}|x_l|\le \left ( \sum_{l=1}^\infty t_{kl}^2\right )^{1/2}\left ( \sum_{l=1}^\infty |x_l|^2\right )^{1/2}$$ Hence $$(T(x+y))_k=\sum_{l=1}^\infty t_{kl}[x_l+y_l]= \sum_{l=1}^\infty t_{kl}x_l+\sum_{l=1}^\infty t_{kl}y_l=(Tx)_k+(Ty)_l$$ Similarly $$(T(\lambda x))_k=\sum_{l=1}^\infty t_{kl}[\lambda x_l]=\lambda \sum_{l=1}^\infty t_{kl}x_l=\lambda (Tx)_k$$