Suppose that $k:[0,1]\times[0,1]\to\mathbb C$ is a Hilbert-Schmidt kernel, i.e. $$ \int_0^1\int_0^1|k(x,y)|^2\mathrm dx\mathrm dy<\infty. $$ The associated Hilbert-Schmidt integral operator $K:L^2([0,1];\mathbb C)\to L^2([0,1];\mathbb C)$ is given by $$ (Ku)(x)=\int_0^1k(x,y)u(y)\mathrm dy $$ for each $u\in L^2([0,1];\mathbb C)$. The Hilbert-Schmidt norm of the operator $K$ is given by $$ \|K\|_\mathrm{HS}^2=\|k\|_{L^2}^2=\int_0^1\int_0^1|k(x,y)|^2\mathrm dx\mathrm dy. $$ Suppose that we have another Hilbert-Schmidt operator $T:L^2([0,1];\mathbb C)\to L^2([0,1];\mathbb C)$ with the kernel $t$.

Is the Hilbert-Schmidt inner product $\langle K,T\rangle_\mathrm{HS}$ equal to $$ \int_0^1\int_0^1k(x,y)\overline{t(x,y)}\mathrm dx\mathrm dy, $$ where $\overline{x}$ is the complex conjugate of a complex number $x$?

It seems that this should be the case because $\langle K,K\rangle_{HS}=\|K\|_\mathrm{HS}^2$. However, I am not sure. I tried to find a textbook that gives the expression for the Hilbert-Schmidt inner product but I did not succeed.

Any help is much appreciated!


Well, we have that the bilinear functional $\langle K,T\rangle_\mathrm{HS}$ defined via $$ \langle K,T\rangle_\mathrm{HS}:=\int_0^1\int_0^1k(x,y)\overline{t(x,y)}\mathrm dx\mathrm dy, $$ is a scalar product on the space of Hilbert-Schmidt integral operators. It further induces a (the mentioned) norm by acknowledging that $z\cdot \overline z=|z|^2$ $$ \langle K,K\rangle_\mathrm{HS}=\|K\|_\mathrm{HS}^2=\|k\|_{L^2}^2=\int_0^1\int_0^1|k(x,y)|^2\mathrm dx\mathrm dy. $$ so we have $$ \sqrt{\langle K,K\rangle_\mathrm{HS}}:=\|K\|_\mathrm{HS}=\sqrt{\int_0^1\int_0^1|k(x,y)|^2\mathrm dx\mathrm dy} $$ Further we know that every norm - if induced by an inner product - is uniquely induced (by the parallelogram law and more specifically the polarization identity). Therefore, the above inner product is the inner product associated to the given norm.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.