Does $(f,Tg)_{L^2}$ define an inner product space?

Question

To my understanding inner product

$$(f,g)_{L^2(\mathcal{D})} = \int_\mathcal{D} f(\boldsymbol{x})g(\boldsymbol{x})\,\mathrm{d}\boldsymbol{x},~~\mathcal{D} \subset \mathbb{R}^N$$

defines an inner product space when completed by saying that elements of $L^2(\mathcal{D})$ are

.. equivalence classes of those functions, where $f$ is equivalent to $g$ if the Lebesgue integral of $|f-g|^2$ over $\mathcal{D}$ is zero.

-Kreyszig, Introductory Functional Analysis with Applications, p. 62

According to Kreyszig, this guarantees the validity of the norm axiom

$$ \| f \| = 0 \Leftrightarrow f = 0$$

and I can fully see where this comes from.

What I would like to ask is for what kind of operators $T:L^2 \mapsto L^2$ the statement

$$ (f,g)_T = (f,Tg)_{L^2(\mathcal{D})} $$

defines an inner product space? Is there any theorems or simple methods to proof that $(\cdot,\cdot)_T$ is an inner product?

In my particular case $T$ is a self-adjoint Hilbert-Schmidt integral operator whose kernel happens to be positive.

Answer 1

As Willie Wong remarked, the thing to check is positive definiteness. For a Hilbert-Schmidt integral operator, positivity (pointwise) of the kernel is not relevant here. The kernel can be expanded in a series $$A(x,y) = \sum_{i=1}^\infty \lambda_i \phi_i(x) \overline{\phi_i(y)}$$ converging in $L^2({ \cal D}^2)$, where $\phi_i$ are an orthonormal basis of $L^2$ and the $\lambda_i$ are the eigenvalues of $T$. You need all $\lambda_i > 0$.

Answer 2

Look at the definition of the inner product. Clearly you need $T$ to be bounded and self-adjoint for the putative inner product to take values in the reals (or complex numbers, depending if your functions are real or complex valued) and for the form to have conjugate symmetry. Sesquilinearity follows from that of the $L^2$ inner product (as long as $T$ is linear). The remaining thing to check is positive definiteness.