Is the range of a weak convergent sequence also a weak convergent sequence?

Question

Under which condition the image of a weakly convergent sequence (xn) by an unbounded linear operator T is a weakly convergent sequence (Txn) ?

Answer 1

Since you didn't specify, I'll assume this is in a Banach space $X$, with $T: X \to Y$ (for another Banach space $Y$) a closed densely-defined operator, and all $x_n \in \mathscr D(T)$.

By the uniform boundedness principle, any weakly convergent sequence is bounded. So a necessary condition is that $T x_n$ is bounded.

If $Y$ is reflexive and the dual operator $T': Y' \to X'$ also densely defined, that is also sufficient: $f(T x_n) = (T' f)(x_n)$ converges for $f \in \mathscr D(T')$, and the map $f \mapsto \lim_n f(T x_n)$ extends by continuity to a member of $Y''$ which corresponds to a member of $Y$, and then we verify that this is the weak limit of $T x_n$.

For an example to show that the reflexivity of $Y$ is needed, consider $T: \ell^2 \to c_0$ with $(T x)_i = i x_i$, defined on $\mathcal D(T) = \{x \in \ell^2: \lim_{i \to \infty} i x_i = 0\}$, and $x_n = (1, 1/2, \ldots, 1/n, 0, 0, \ldots)$. In this case $x_n$ converges not just weakly but in norm, and $Tx_n = (1,\ldots, 1, 0, 0, \ldots)$ is bounded but does not converge weakly (it converges weak-$*$ in $\ell^\infty = (\ell^1)'$ to $(1,1,\ldots)$ which is not in $c_0$).