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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.

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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.

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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 $\lambda_i > 0$.

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  • $\begingroup$ To see that the positivity of the kernel does not help: consider what happens when $\mathcal{D} = [0,1]$ and $k(x,y) = 1$. $\endgroup$ Jul 20, 2012 at 17:34
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    $\begingroup$ @NateEldredge : that example is somewhat of a borderline case since $\lambda_i \ge 0$. You could also try $k(x,y) = 1 + (x-y)^2$ which has some negative eigenvalues, e.g. $Tf = - f/6$ for $f(x) = x - 1/2$. $\endgroup$ Jul 20, 2012 at 17:52
  • $\begingroup$ Indeed, that's much better. $\endgroup$ Jul 20, 2012 at 19:16
  • $\begingroup$ Thank you for a great answer Robert! I would still like to ask for an additional confirmation: What to do with the case $(f,f)_T = 0 \Rightarrow f = 0$ since to my understanding this does not simply follow without completing the space somehow? $\endgroup$
    – knl
    Jul 21, 2012 at 8:37
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    $\begingroup$ If $T$ is a positive operator (i.e. $(f,Tf) \ge 0$ for all $f$) then the only way you can have $(f,Tf) = 0$ is $Tf=0$. This is because Cauchy-Schwarz says $(g,Tf) = 0$ for all $g$, and that implies $Tf=0$. So, as I said, the question is whether all $\lambda_i > 0$ (not just $\ge 0$). $\endgroup$ Jul 22, 2012 at 6:16

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