Suppose $T:\mathcal D(T)\rightarrow \mathcal Y$ is a closed operator from a Banach space $\mathcal X$ to a Banach space $\mathcal Y$. Is it true that $$ \|Tx\|\leq \liminf_{n\rightarrow\infty} \|T x_n\| $$ whenever $\|x-x_n\|\rightarrow 0$, i.e. is $x\mapsto \| Tx\|$ lower semi-continuous? (here it is implicit that $x_n,x\in \mathcal D(T)$).
When $T$ is a multiplication operator in some $L^ p$ space, the inequality reduces to Fatou's Lemma. More generally, I would like to know if $x\mapsto \| Tx+y\|$ is lower semi-continuous for all $y\in\mathcal Y$. Any help in either direction will be much appreciated!
The answer seems to be Yes if the space $Y$ is reflexive.
Extracting a subsequence if necessary, we may assume that the sequence $(\Vert Tx_n\Vert)$ has a limit in $[0,\infty]$.
If $\Vert Tx_n\Vert\to \infty$, the inequality is trivially satisfied. So assume that $\Vert Tx_n\Vert\to l<\infty$ and let us show that $\Vert x\Vert\leq l$.
Since the norm of $Y$ is weakly lower-semicontinuous, it is enough to show that $Tx_n\to Tx$ weakly; and since by our assumption the sequence $(Tx_n)$ is bounded, it is enough to show that $\langle Tx_n, y^*\rangle \to\langle Tx,y^*\rangle$ for every $y^*$ in some dense subspace of $Y^*$. Now, since $T$ is closed, the adjoint operator $T^*$ is densely defined. (It is here that one probably needs reflexivity: after a superficial look at Goldberg's "Unbounded linear operators", closedness of $T$ apparently implies that $\mathcal D(T^*)$ is $w^*$-dense in $Y^*$, hence norm-dense under the reflexivity assumption. This has to be checked carefully, though). And since $x_n\to x$ weakly, the definition of $T^*$ shows that $\langle Tx_n, y^*\rangle \langle Tx,y^*\rangle$ for every $y^*\in\mathcal D(T^*)$.
In general ($Y$ not reflexive), this proof does not seem to work. However, the inequality obviously holds provided that convergence of $x_n$ to $x$ implies convergence of $Tx_n$ to $Tx$ with respect to some topology $\tau$ on $Y$ such that the norm is lower-semicontinuous with respect to $\tau$. The very few unbounded operators that I know all satisfy this.