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.
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).
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.
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.