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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?)

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    Is $T$ an operator on functions or values?2012-05-16
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    @copper.hat: $T$ is an orthogonal projection on functions from $H$. I think I know what you are thinking of: is the function $h$ belongs to $H$ or not!! So lets assume thats true.2012-05-16
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    Well, no. In the above equation, $T$ operates on elements of $H$. But inside the integral, it is operating on the range of $g$.2012-05-16
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    The idea is that any function $f\in H$ can be writen as $f(u)=\langle f(x),g(x,u)\rangle$, so I want to know if $T(f)=\langle f(x),T(g(x,u))\rangle$. (Maybe this is more clear!)2012-05-16
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    I see; the notation was confusing me. I would use the notation $T(\int f(x)g(x,\cdot)dx = \int f(x) T(g(x,\cdot)) dx$. This makes it clear that you are performing vector (as in $\in H$) valued integrations2012-05-16
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    Let $g_u\in H$ be the function defined by $g_u(x)=g(x,u)$. Where you wrote "$T(g(x,u))$", do you actually mean "$(T(g_u))(x)$? Also, where you wrote "$T(h(u))$", a pedant like me would write $(Th)(u)$, indicating that $T$ sends $h$ to the function $Th$, which is then evaluated at $u$. This may be what copper.hat was confused about, or maybe it's just me.2012-05-16
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    I think the issue here is how you define and assure the existence of the vector valued integral. Assuming this is not an issue, most vector integration schemes satisfy $\phi(\int f d\mu) = \int \phi(f) d\mu$, $\forall \phi \in H^*$. So, in your case, the interchange would be justified if $T$ is continuous.2012-05-16
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    Let me see if I understand the question. For each $f\in H$, let $h_f(u)=\langle f(\cdot),g(\cdot,u)\rangle$, and assume that $h_f$ is in $H$. Given an orthogonal projection $T$ on $H$, is it true that $(Th_f)(u) = \langle f(\cdot),Tg(\cdot,u)\rangle$? This would be equivalent to $Th_f = h_{T^*f}=h_{Tf}$ (with the last equality because $T^*=T$).2012-05-16
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    My model was if $h = \sum f_i g_i$, where $f_i$ are scalars, and $g_i \in H$, then $T(h) = \sum f_i T(g_i)$. So, I view '$x$' as an index as such. (Also, I don't think it is pedantic, I think much confusion would disappear with better choices.)2012-05-16
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    @Jonas Meyer: Yes, this is what I want! So is it true?2012-05-16
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    @Kim: Well, it would be true in the case of your second comment, in which $h_f=f$ for all $f\in H$, and therefore $Th_f=Tf=h_{Tf}$. (And this only uses the fact that $T=T^*$.) I don't see why it would be true in a more general setting, but haven't thought about it enough.2012-05-16
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    @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

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