Let $p \in [1, \infty)$ and define the operator $T$, which maps a suitable $L^p(\Omega, \mathcal{F}, \mu,\mathbb{R})$-function onto its product with the $\mathbb{R}$-identity square, i.e. $$T: D(T) \rightarrow L^p, \quad (Tf)(x):= x^2 f(x), \quad f \in D(T), x \in \mathbb{R},$$ where $D(T) := \{f\in L^p \mid [x \mapsto x^2f(x)] \in L^p\}.$
My idea is to take an arbitrary convergent sequence in the graph of $T$ and show its limit lies in the graph. In symbols, let $(f_n)_n \subset D(T)$ such that $f_n \rightarrow f$ almost everywhere with $f \in L^p$ and $Tf_n= x^2f_n(x) \rightarrow g$ almost everywhere with $g \in L^p$. My goal is to show $f\in D(T)$ and $Tf = g$.
I'm new to operator theory and I do not know how to approach either claim stated. Basically, I have statements which are valid almost everywhere, but the spaces in questions are $L^p$. How can the claims be expressed in a way which relates to the $p$-norms? I haven't tried much, as I am not sure what exactly needs to be shown. All hints or pointers are appreciated.
You start with $L^p$-convergence, not pointwise almost everywhere convergence.
You take a sequence $(f_n)$ in $D(T)$ such that [for certain $f,g \in L^p$] you have a) $\lVert f_n - f\rVert_{L^p} \to 0$, and b) $\lVert T(f_n) - g\rVert_{L^p} \to 0$. You want to conclude $f \in D(T)$ and $g = T(f)$.
Now you take a subsequence such that $f_n \to f$ almost everywhere. It follows that $T(f_n)$ converges almost everywhere to $x \mapsto x^2f(x)$. Since by assumption $T(f_n) \to g$ in $L^p$ it follows that $g(x) = x^2 f(x)$ almost everywhere. This shows that $f \in D(T)$ and $g = T(f)$, and the proof that $T$ is closed is complete.