# For Hilbert-Schmidt operator, $\sum\lVert Te_n\rVert^2 = \lVert k \rVert^2$

Let $X = L^2(0,1)$ and $k \in L^2( (0,1) \times (0,1))$. Therefore we have a Hilbert-Schmidt integral operator:

$Tf(t)=\int_0^1 k(s,t)f(s)ds$

Let $\{e_n\}$ be an orthonormal basis of $X$, how to show $\sum_n \lVert Te_n \rVert^2=\lVert k \rVert^2_{L^2(0,1)\times(0,1)}$?

$\lVert Te_n \rVert^2=\sum_k |\langle Te_n,e_k\rangle|^2$ and $\sum_n\lVert Te_n \rVert^2=\sum_n \sum_k |\langle Te_n,e_k\rangle|^2$.

On the other hand, $\lVert k \rVert^2=\int_0^1 \int_0^1 |k(t,s)|^2\ dt \ ds$ and I cannot see any relationship between them.

• I think you made a typo, it should be $\lVert k \rVert^2=\int_0^1 \int_0^1 |k(t,s)|^2\ dt \ ds$. – TZakrevskiy May 18 '17 at 12:20
• @TZakrevskiy Thanks, I have fixed it. – userHasNoNumber May 18 '17 at 12:36

Hint: use that $f_{n,m}(s,t) = \overline{e_n(s)} e_m(t)$ is an orthonormal basis for $L^2\big((0,1)\times (0,1)\big)$.