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I am just wondering if anyone can help me out with the following question:

assuming $H$ is some operator, say Laplacian $\Delta$. What is the meaning of $f(H)$ for some function $f$? In case of Laplacian, I think it can be described by means of Fourier transform, but is there another interpretation of $f(H)$ without using the Fourier space. Also, it kind of makes sense, if $f$ is polynomial. But what if $f$ is some cut-off function, for example?

I would be very glad to get any reference! Thank you!

Kerr
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    The formal treatment of this question is referred to as the Borel functional calculus – Ben Grossmann Oct 21 '20 at 22:03
  • Notably: if $f$ is the characteristic function $\chi_A$ of a Borel set $A \subset \Bbb R$, then $f(\Delta)$ is the projection onto the span of all eiegnvectors of $\Delta$ corresponding to some $\lambda \in A$. – Ben Grossmann Oct 21 '20 at 22:08
  • The case of $\Delta$ is easy to deal with in that it has a compact "inverse" $\Delta^{-1}$, and $f(\Delta)$ can be defined to be $f(\Delta) = f(i(\Delta^{-1})$, where $i(x) = 1/x$. – Ben Grossmann Oct 21 '20 at 22:10
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    @BenGrossmann thank you very much! Borel functional calculus is exactly what I was looking for. I found pretty nice chapter on it in the first volume of Reed and Simon. – Kerr Oct 21 '20 at 22:34

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If you are mostly interested in the "meaning", perhaps an intuitive answer would be satisfactory. As you mentioned, when $f$ is a polynomial, then the meaning of $f(T)$ is clear: just substitute $T$ for the polynomial variable.

If $T$ is bounded and self-adjoint, and $f$ is a continuous function on the spectrum of $T$, then one has by Stone-Weiestrass that $$ f=\lim_n p_n, $$ where the $p_n$ are polynomials. One may then prove that $\lim_n p_n(T)$ exists, as an operator, so it makes a lot of sense to call the limit $f(T)$.

If $T$ is just bounded, and $f(\lambda)=(z-\lambda)^{-1}$, for $z$ not in the spectrum of $T$, then it also makes sense to set $$ f(T)=(z-T )^{-1}. $$

More generally, if $f$ is holomorphic on some open set $U$ containing the spectrum of $T$, and if $\gamma $ is a closed curve in $U$ winding arround every point of $\sigma (T)$ counter-clockwise once, then Cauchy's integral formula says that $$ f(\lambda ) = {1\over 2\pi i}\int_\gamma {f(z)\over z-\lambda }\,dz, $$ for every $\lambda $ in $\sigma (T)$. Again it makes sense to define $$ f(T) = {1\over 2\pi i}\int_\gamma f(z)(z-T)^{-1}\,dz. $$

The list goes on and it is possible to give meaning to $f(T)$ in many other situations, such as when $f$ is (possibly unbounded) and self-adjoint and $f$ is Borel measurable. If you are looking for references I suggest you search for the terms "functional calculus"!

Ruy
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  • thank you! from the comments above, I found a nice chapter regarding this question in Reed and Simon, volume 1. – Kerr Oct 21 '20 at 22:36
  • I am glad I could help. Indeed the various forms of functional calculus are often the main goal of many basic functional analysis books! – Ruy Oct 21 '20 at 22:39
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    Good explanation. This question & answer contains the very important example of the unbounded operator $\frac 1 i \frac{d}{dx}$, whose functional calculus essentially yields the theory of the Fourier transform. – Giuseppe Negro Oct 21 '20 at 23:08
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You can view the more general case as a limiting case where you integrate around the real axis, which is where the spectrum of $A$ is found. The limit exists as a strong (vector) limit, but not necessarily as an operator limit: $$ f_{a,b}(A)x = \lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_a^bf(t)\left((t-i\epsilon-A)^{-1}x-(t+i\epsilon-A)^{-1}x\right) dt $$ Then the following strong limits exists and give you the functional calculus for $A$: $$ f(A)x = \lim_{r\rightarrow-\infty \\ s\rightarrow\infty}f_{r,s}(A)x $$ There are quite a few details to work out, but the arguments are intuitive based on Complex Analysis and taking a limiting integral around the real axis, where the spectrum is found.

Disintegrating By Parts
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