Must the domain of a Hilbert space operator be limited to square-summable elements? If so, what type of space does not restrict, thus?

Question

Background: The operation of bandlimiting is a projection operator, $P$, into a separable Hilbert space. For example, with inner product, $\int_{-1/2}^{1/2}g(f)\overline{h(f)}df$, there is an orthonormal basis, $\{ e^{2i \pi n f}\}$, which in time domain is $\{ sinc(\pi(t-n)) \}$, for $n \in \mathbb{Z}$. Thus, for $g(f) \in L^2(-1/2,1/2)$, the expression of the operator, $Pg(f)$, is $\sum_{n \in \mathbb{Z}} c_n e^{2i \pi n f}$, with the coefficicents, $c_n = \int_{-1/2}^{1/2} g(f) e^{-2i \pi n f}df$.

Now, it is possible to apply bandlimiting to a Dirac delta, $\delta(f-f_0)$. For $|f_0|<1/2$, the result is $\delta(f-f_0)$, and zero for $|f_0|>1/2$. In time domain, the result is $e^{2i \pi f_0 t}, t \in \mathbb{R}$ or zero, (again depending upon $|f_0|$.)(For brevity I omit the case of $f_0=1/2$.)

For a bounded operator, $Q$, in a Hilbert space, $||Qx||< M|x|, M<\infty $. I know that the norm of the projection is one, $||P||=1$, so $P$ is bounded. I know that the this projection operator is not compact. I know that the sequence for the result upon the Dirac delta is not Cauchy, which roughly explains how the result can be $\delta(f-f_0)$.

What I am having trouble with is that the distribution, $\delta(f-f_0)$ is not in $L^2(-1/2,1/2)$, even though the operator, $P$ can handle it. Furthermore, $P$ being self-adjoint, must have its range the same as its domain, so the range is not limited to $L^2$.

Question: Is it a requirement for the domain of an operator, $A$, in a Hilbert space to be limited to square-summable elements, $\mathcal{D}(A):\{x|x\in L^2 \}$? If so, what is the type of space that correctly admits the Dirac, above? (If I have misunderstood something, what am I getting wrong?)

My motivation is that when describing this bandlimiting operation, I want to describe it correctly. I thought it to be a projection operator on a Hilbert space, with domain $\mathcal{D}(A):\{x|x\in L^2 \}$. When I started to consider it acting upon Diracs, I was not sure I had correctly described the operator. Maybe my question is best said, "How does one properly word the definition of that operation, to correctly include Dirac's in its domain?"

Answer 1

The so-called delta function, formally, is a tempered distribution where the tempered distributions (denoted $\mathcal S'$) are defined to be the dual of Schwartz space (denoted $\mathcal S$), i.e. $\mathcal S'$ is the collection of all continuous linear functionals $:\mathcal S\to\mathbb C$. In particular

• $\mathcal S$ is embedded naturally in $\mathcal S'$ by means of the (continuous) map $\iota:\mathcal S\to\mathcal S'$, $(\iota f)(g)=\int f(x)g(x)\,dx$ for all $f,g\in\mathcal S$.
• every continuous map $T:\mathcal S\to\mathcal S$ gives rise to a continuous adjoint $T':\mathcal S'\to\mathcal S'$ via $(T'\phi)(f):=\phi(Tf)$.

Thus given a bounded linear operator $T$ on $L^2$, making $\delta$ part of the domain means finding an extension of $T$ to the tempered distributions $\mathcal S'\ni\delta$. The standard way for such extensions is to check whether

1. $T(\mathcal S)\subseteq\mathcal S$ (i.e. $T$ maps Schwartz functions to Schwartz functions)
2. $T|_{\mathcal S}:\mathcal S\to\mathcal S$ is continuous (w.r.t. the topology of $\mathcal S$).

If this is the case then $\iota\circ T:\mathcal S\to\mathcal S'$ is continuous, hence there is at most one continuous extension $\hat T:\mathcal S'\to\mathcal S'$. Thus if one can find $T_0:\mathcal S\to\mathcal S$ continuous such that $$ T_0'\circ\iota=\iota\circ T $$ when $T_0'=:\hat T$ is the extension of $T$ we were looking for.

This is the rough roadmap for problems of this kind; for an introduction to this topic refer, e.g., to Chapter V.3 ff. in "Methods of Modern Mathematical Physics I: Functional Analysis" by Reed & Simon (1980).