meaning of $f(H)$ for the operator $H$ 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!
 A: 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"!
A: 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.
