1
$\begingroup$

I have a question which maybe looks very simple: Let $T$ be an orthogonal projection on a Hilbert space $H$. If $g(x,u)\in H$, for all $u\in \mathbb R$, and the inner product is defined by $\langle f(.), g(.,u)\rangle_{H}=\int_{\mathbb R}f(x)g(x,u)dx $ which is a function of $u$ (say $h(u)$), for all $f\in H$.

Now my question is that if we apply the projection to the resulting function $h(u)$, can we move the projection inside the integral, i.e.:

$T (h(u))= T\big( \int_{\mathbb R}f(x)g(x,u)dx \big)= \int_{\mathbb R}f(x)T(g(x,u))dx$

(If this is not always true what are the cases where we can do this?)

  • 0
    @Jonas Meyer: I just need it for the case where $h_{f}=f, f\in H$. But How to give an explicit proof for this?2012-05-16

1 Answers 1

1

This answer will first formulate a more precise statement of the special case that came out in the comments, and then show why it is true.

Suppose that $H$ is a (real) Hilbert space of real-valued functions on $\mathbb R$ with $L^2$ inner product. Suppose that $g:\mathbb R\times \mathbb R\to\mathbb R$ is a function such that for all $u\in \mathbb R$, $g(\cdot, u)$ is in $H$, and for all $f\in H$, $f(u)=\langle f,g(\cdot,u)\rangle$.

Claim: If $T$ is a symmetric linear operator on $H$, then for all $f\in H$ and $u\in \mathbb R$, $(Tf)(u)=\langle f,Tg(\cdot,u)\rangle$.

Proof: $(Tf)(u)=\langle Tf,g(\cdot,u)\rangle=\langle f,T^*g(\cdot,u)\rangle=\langle f,Tg(\cdot,u)\rangle,$ because $T=T^*$.