Hilbert-Schmidt Integral operators are usually defined from $H=L_2[a,b]$ into $H=L_2[a,b]$ as $$(Tf)(x) = \int_a^b K(x,y)f(y) dy,$$

provided that $K(x,y)$ is a Hilbert Schmidt kernel, namely
$$\int_a^b \int_a^b |K(x,y)|^2dx dy < \infty.$$

I was wondering if the following extension is used with the same name. Letting $H_1 = L_2[a,b]$, $H_2=L_2[c,d]$, and
$$\int_a^b \int_c^d |K(x,y)|^2dx dy < \infty,$$

consider the integral operators $T_1:H_2 \rightarrow H_1$ and $T_2:H_1 \rightarrow H_2$ $$(T_1 f)(x) = \int_c^d K(x,y)f(y) dy,$$ $$(T_2 f)(y) = \int_a^b K(x,y)g(x) dx.$$ Can one still call these operators and corresponding kernel of Hilbert-Schmidt type?


The abstract definition of Hilbert-Schmidt operators on a Hilbert space $H$ is that for a complete orthonormal set $A$ a bounded operator $$ T: H \to H $$ is a Hilbert-Schmidt operator if $$ \sum_{a \in A} \| T a \|^2 \lt \infty $$ We can, of course, without problems say that we actually have two different, but isomorphic, Hilbert spaces $H_1$ and $H_2$ and reformulate the definition for $$ T: H_1 \to H_2 $$ which would not change anything (definitions nor theorems) besides additional indexes. So, yes, I think one can and should talk about Hilbert-Schmidt operators in your example.

  • $\begingroup$ Are Hilbert Schmidt operators the same as Hilbert Schmidt Integral operators? $\endgroup$ – Tyler Hilton Oct 18 '13 at 22:28

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