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I am not sure if I have seen integral transforms in the right way, but given a transform like Fourier transform - it's actually a basis transformation right ?

$$ F(y) = \int K(x,y) f(x) \text{d}x $$ where $K(x,y) = \text{e}^{-ixy}$ for the case Fourier transform. The functions $F(y)$ and $f(x)$ can be seen as $\left<y|F\right>$ and $\left<x|f\right>$ respectively. In such a case the above integral equation can be rewritten as -

$$ \left< y|F \right> = \left<y|\mathbb{\hat I}|F\right> = \sum_x \left<y |x\right> \left<x |f\right> $$

So is $\left<y |x\right>$ one way of looking at the integral kernel for all general cases ? If not, I wish to understand how one can precisely look at integral kernels.

EDIT 1: I also wish to know that can transforms like Laplace, Mellin etc. also be treated like that as Transformation matrix, also in which case it might not be unitary matrix in all cases, but rather just a map from one inner product space to another.

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  • $\begingroup$ Depending on the space over which you're integrating (I think you mean over all of $\Bbb R$, though) and the function $K$, you may "lose information" when going from $f$ to $F$. $\endgroup$ – Ben Grossmann Sep 3 '15 at 17:27
  • $\begingroup$ @Omnomnomnom : I would like to know how and why exactly would that happen ? Secondly when we say we will lose information, would it imply that it will affect the physical meaning attached to it ? $\endgroup$ – user35952 Sep 5 '15 at 16:19
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Intuitively, $K(x,y)$ is a "continuous" matrix $K$ acting on a "continuous" row vector vector $f$: $$ \sum_{x}f(x)K(x,y) = \int f(x)K(x,y)dx $$ The associated quadratic form would be $$ \sum_{x}f(x)K(x,y)g(y)^{\star} = \int\int K(x,y)f(x)\overline{g(y)}dxdy $$ This would be selfadjoint if $K(x,y)=\overline{K(y,x)}$, analogous to a Hermitian matrix, and would lead to an observable. A type of condition that allows everything to work really well is the Hilbert-Schmidt condition: $$ M^{2}= \iint |K(x,y)|^{2}dxdy < \infty. $$ That may be a little restrictive for what you're doing. I'm not sure, but it's very nice from a Mathematical point-of-view because it allows one to define a bounded linear operator $K : L^{2}(\mathbb{R})\rightarrow L^{2}(\mathbb{R})$: $$ Kf = \int K(x,y)f(x)dx. $$ This follows from Cauchy-Schwarz: $$ |Kf(y)|^{2} \le \int |K(x,y)|^{2}dx\int |f(x)|^{2}dx \\ \int |Kf(y)|^{2}dy \le \int\int |K(x,y)|^{2}dxdy\int|f(x)|^{2}dx \\ \|Kf\|^{2} \le \int\int|K(x,y)|^{2}dxdy \|f\|^{2} \\ \|Kf\| \le \left(\int\int|K(x,y)|^{2}dxdy\right)^{1/2}\|f\| \\ \|Kf\| \le M\|f\|. $$ Here $M$ is the Hilbert-Schmidt constant defined above. This describes how Hilbert originally started in generalizing matrices. Eventually von Neumann's operator theory replaced these ideas because (1) Matrices cannot distinguish between different linear operators and (2) linear operators allowed more general objects that were better suited for Quantum.

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  • $\begingroup$ The Hilbert Schmidt condition is probably too restrictive since it excludes the Fourier kernel. $\endgroup$ – Ben Grossmann Sep 3 '15 at 19:27
  • $\begingroup$ @Omnomnomnom : You may well require more general operators, but the ideas are basically the same, even though the Mathematical formalism to make everything precise becomes increasingly tricky. $\endgroup$ – DisintegratingByParts Sep 3 '15 at 19:38
  • $\begingroup$ @Omnomnomnom : In the case of the Fourier transform kernel, you can truncate in $x,y$ to $[-R,R]$ and let $R\rightarrow \infty$ to define the operators. That's a simplistic example of how one can deal with such things. $\endgroup$ – DisintegratingByParts Sep 3 '15 at 19:50
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    $\begingroup$ The Laplace transform maps $L^{2}[0,\infty)$ onto $H^{2}(\Pi_{+})$, where $H^{2}(\Pi_{+})$ consists of analytic functions on the right half plane $\Pi_{+}=\{ z \in \mathbb{C} : \Re z > 0 \}$ for which $\|f\|_{H^{2}(\Pi_+)}^{2}=\sup_{x > 0}\int_{-\infty}^{\infty}|f(x+iy)|^{2}dy$ is finite. Using these spaces $\mathscr{L} : L^{2}[0,\infty)\rightarrow H^{2}(\Pi_{+})$ is unitary (isometric and onto.) This is the Paley-Wiener Theorem. So this fits well, but just not with the same domain and range spaces. $\endgroup$ – DisintegratingByParts Sep 4 '15 at 13:29
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    $\begingroup$ Oh ! Thank you so much, that is one real insight to me about Laplace transform. I think the domain and range spaces will have to be different for different transforms. $\endgroup$ – user35952 Sep 4 '15 at 15:31

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