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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!

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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.

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