TITLE: clarification about theorem $3.10 $ of Brezis functional analysis book.
QUESTION [5 upvotes]: I'm referring to the theorem at page $61$. It shows that for a linear operator $T $ between $E $ and $ F$ Banach space are equivalent
(notation: $S$ means strong topology, the norm ome, $W $ weak topology)
$T$ is continuous from $ E, S $ to $F, S$
$T$ is continuous from $E, W$ to $F, W$
$T$ is continuous from $E, S$ to $F, W$
Where by $E, S$ i mean $E$ equipped with the strong topology.
My question is why the same argument is not valid for $T$ between $E, W $ to $F, S$?
I don't know what doesn't work.

REPLY [3 votes]: Take $E = F$ and $T = \operatorname{id}$. It's continuous for the other combinations of topologies, but not for $(E,W) \to (F,S)$.
That shows that $(E,W) \to (F,S)$ continuity is a stronger condition than the others.
Since a weak neighbourhood of $0$ contains a neighbourhood of the form
$$\{ e : \lvert \lambda_i(e)\rvert < 1, \; 1 \leqslant i \leqslant n\}$$
for finitely many $\lambda_1,\dotsc,\lambda_n \in E'$, it contains
$$K := \bigcap_{i=1}^n \ker \lambda_i,$$
a subspace of finite codimension. The image of $K$ under $T$ is a subspace that is contained in a ball of finite radius, hence $K \subset \ker T$, and therefore, the range of $T$ is finite-dimensional. That is a very restrictive condition.