Let $H_1,H_2$ be two Hilbert Spaces and $T:H_1 \to H_2$ a finite rank linear operator, that is $dim[T(H_1)]=m<+\infty$.
I need to show that $T$ is a Hilbert-Schmidt operator, i.e., given a Hilbert basis $\{e_\alpha\}_{\alpha \in J}$
$$\|T\|^2_{HS} = \sum_{\alpha \in J}\|T(e_\alpha)\|^2<\infty.$$
Now, as $T$ has finite rank, there's an ortogonal basis $\{y_1,…,y_m\}\subset H_2$ of $T(H_1)$ such that
$$T(x) = \sum_{i=1}^m a_i(x)y_i$$
Using this we have
$$\|T\|^2_{HS} \leq \sum_{\alpha \in J} \sum_{i=1}^m |a_i(x)|^2 \|y_i\|^2 =\sum_{\alpha \in J} \sum_{i=1}^m |a_i(x)|^2 $$
But why would this be finite?
Apparently everyone seems to talk about Hilbert-Schmidt operators only in the context in which $T:H \to H$. In this case it is easy to show the claim as we can take an ortonormal basis $\{e_\alpha\}_{\alpha \in J}$ and as $T$ has finite rank, there is $I \subset J$ finite with $T(e_\alpha) = 0 \; \forall \alpha \in J\setminus I$ and the series of the Hilbert-Schmidt norm is actually a finite sum.
The problem is preciselly that we have different spaces $H_1$ and $H_2$. Can I somehow split $H_1 = A \bigoplus B$ so that $T(B) = 0$ and $dim(A)<\infty$? That would solve the problem.
Best Answer
The expression $$ \sum_{\alpha \in J} \sum_{i=1}^m |a_i(x)|^2 $$ is (usually) not finite, because you have an infinite sum over something which does not depend on $\alpha$. So if $|a_i(x)|\neq0$ for some $i$ and $x$, then the expression is infinite.
Hint for the general solution:
Decompose the operator $T$ into finitely many operators $T_i:H_1\to\Bbb C$. Then you can represent $T_i$ by some $y_i\in H_1$. Show that $T_i$ is Hilbert-Schmidt using $y_i$.
This should lead to an expression of the form $$ \sum_{\alpha\in J} |(y_i,e_\alpha)|^2, $$ (where $\{e_\alpha\}_{\alpha\in J}$ is a Hilbert basis) which can be shown to be finite.