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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.

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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.

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  • $\begingroup$ ok i'll work out the details. Thanks, but why the argument of the other proofs in the theorem doesn't work here? $\endgroup$ – Riccardo Jan 3 '14 at 23:12
  • $\begingroup$ Another doubt :) why infinite dimensional kernel entail finite dimensional range in your reasoning? Thanks in advance! $\endgroup$ – Riccardo Jan 3 '14 at 23:22
  • $\begingroup$ Infinite-dimensional kernel does not automatically imply finite-dimensional range. But since every weak neighbourhood contains the intersection of finitely many $\lambda^{-1}(\mathbb{D})$, the subspace is finite-codimensional. As to what exactly doesn't work for $(E,W) \to (F,S)$, it would help if I knew how Brezis proves the other cases. I'm trying to figure out the breakage in the proof I know, though. $\endgroup$ – Daniel Fischer♦ Jan 3 '14 at 23:26
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    $\begingroup$ The open unit disk of $\mathbb{C}$; in case of real scalars, the interval $(-1,1)$. $\endgroup$ – Daniel Fischer♦ Jan 3 '14 at 23:36
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    $\begingroup$ Oh, that direction, $(E,W) \to (F,S)$ continuity implies $(E,S)\to (F,S)$ continuity, is true. You can prove it by observing that a strengthening of the topology on the domain only introduces new continuous functions (if $f^{-1}(U)$ is open in the weaker topology, it also is open in the stronger), or, in this configuration, by the CGT. It's the other direction that fails, a $(E,S) \to (F,S)$-continuous map is generally not $(E,W)\to (F,S)$-continuous. $\endgroup$ – Daniel Fischer♦ Jan 4 '14 at 11:36

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