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I'm currently looking at Proposition 2 of Section VIII.3 of Reed and Simon's functional analysis text.

Proposition: Let $\langle M, \mu \rangle$ be a measurable space with $\mu$ a finite measure. Suppose that $f$ is a measurable, real valued function on $M$ which is finite a.e. $[\mu]$. Let $T_f : \varphi \mapsto f\varphi$ be the operator on $L^2(M, d\mu)$ with domain $$D(T_f) = \{ \varphi \ \vert \ f \varphi \in L^2(M, \mu) \}.$$ Suppose in addition that $f \in L^q(M, d\mu)$ for $2 < p < \infty$. Let $D$ be any dense set in $L^q(M, d\mu)$, where $q^{-1} + p^{-1} = 1/2$. Then $D$ is a core for $T_f$.

Proof: Let us first show that $L^q$ is a core for $T_f$. By Holder's inequality $\| g \|_2 \leq \| 1 \|_p \| g \|_q$ and $\| fg \|_2 \leq \| f \|_p \| g \|_q$, so $L^q \subset D(T_f)$. Moreover, if $f \in D(T_f)$, let $g_n$ be the function $$g_n = \begin{cases} g, & \left| g(m) \right| \leq n \\ 0, & \text{otherwise}. \end{cases}$$ By the dominated convergence theorem, $g_n \to g$ and $fg_n \to fg$ in $L^2$. Since each $g_n$ is in $L^q$, we conclude that $L^q$ is a core for $T_f$.

I understand the constituent components of the proof, I don't understand the conclusions. In particular

  • How does the Holder inequality argument lead us to conclude that $L^q \subset D(T_f)$?
  • How does the fact that each $g_n$ is in $L^q$ allow us to conclude that $L^q$ is a core for $T_f$?

I also don't have much of an intuition for what the core of an unbounded operator is, if someone could provide any intuition that would be great.

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    $\begingroup$ You mean $f\in L^p$, not $f\in L^q$. $\endgroup$ – user940 Apr 8 '17 at 23:33
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How does the Holder inequality argument lead us to conclude that $L^q \subset D(T_f)$?

Let $g \in L^q$. Since $\|g\|_2 \leq \|1\|_p \|g\|_q< \infty$ ($M$ has finite measure), you know that $L^q \subseteq L^2$. By $\|fg\|_2 \leq \|f\|_p \|g\|_q< \infty$, you know that $L^q$ is furthermore a subset of $D(T_f)$ (Note that $D(T_f)$ is by definition a subset of $L^2$, this is why the first application of Hölder was needed).

How does the fact that each $g_n$ is in $L^q$ allow us to conclude that $L^q$ is a core for $T_f$?

By the first part, we know that the restriction of $T_f$ to $L^q$ is well-defined. You should also convince yourself that $T_f$ is densely defined and closed. In this setting, the question of whether the closure of $T_f \upharpoonright L^q$ equals $T_f$ makes sense. What the proof shows is in fact $\overline{T_f \upharpoonright L^q} = T_f$, by showing that $\overline{\Gamma (T_f \upharpoonright L^q)} = \Gamma(T_f)$, where $\Gamma(A)$ denotes the graph of the function $A$.

Since $T_f$ is closed, we know that $\overline{\Gamma (T_f \upharpoonright L^q)} \subseteq \Gamma(T_f)$, so we (and accordingly Reed and Simon) only have to prove $\supseteq$. Therefore, let $(g, fg) \in \Gamma(T_f)$, i.e. $g \in D(T_f)$ and $fg \in L^2$. Then, define $g_n := \chi_{\{|g| > n\}} g$. Again, by Hölder you get $$ \|g_n\|_q \leq \| \chi_{\{|g| > n\}} \|_r \|g\|_2 < \infty, $$ with $\frac{1}{q} = \frac{1}{r} + \frac{1}{2}$, i.e. $r = \frac{2q}{2 - q} > 0$. Note that since $p > 2$ by assumption, we must have $q < 2$ so the expression for $r$ is well-defined. But now you have found a sequence $(g_n, f g_n) \in \Gamma(T_f \upharpoonright$ converging to $(g, fg)$ in the product topology.

I also don't have much of an intuition for what the core of an unbounded operator is, if someone could provide any intuition that would be great.

Here is my two cents. If it doesn't satisfy you, consider reposting this specific part of your original question:

The Proposition deals with two different types of domains: $D(T_f)$ on the one hand and $L^q$ and $D$ on the other hand. $D(T_f)$ sometimes is called "maximal domain of definition" by obvious reasons. $L^q$ and $D$ are "cores" for $T_f$. "Cores" to $D(T_f)$ correspond to the dense subspaces of $D(T_f)$. The philosophy behind considering cores: Work on a smaller "handy" set ($L^q$ is handier than just $D(T_f)$) without "loosing" your original operator - you can always go back to $T_f$ by considering the closure $\overline{T_f \upharpoonright D}$. Note that for the closure to be well-defined it is essential that $T_f$ is a closed operator.

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