Show that a $\sigma$-additive semi-norm is continuous.

Question

Suppose that $E$ is a Banach space and $p$ is a $\sigma$-subadditive semi-norm, that is, if $\sum_{n=1}^{\infty}x_n$ converges in $E$, then \begin{equation} p(\sum_{n=1}^{\infty}x_n) \leq \sum_{i=1}^{\infty}p(x_n) \leq \infty. \end{equation} Then we claim that $p$ is continuous: \begin{equation} \exists M>0 \quad \text{s.t.} \quad p(x)\leq M\Vert x \Vert, \quad \forall x \in E. \end{equation} I wanted to solve it by contradiction but was fruitless. This proposition can be used to prove three traditional theorems in functional analysis: The Banach-Steinhaus theorem, the open mapping theorem and the closed graph theorem, if we choose proper semi-norm $p$.

Any help will be appreciated. Thanks.

score 4 · Accepted Answer · answered Dec 11 '24 at 14:19

Let $B :=\{x\in E : p(x) < 1\}$ be the open unit ball for $p$. For $t > 0$ we have $$ tB = \{x\in E : p(x) < t\} $$ (by the semi-norm property $p(\lambda x) = |\lambda| p(x)$). Hence, if $x \in E$ we can choose $t > p(x)$ and then we have $x \in tB$. This means $$ E= \bigcup_{t>0} tB, $$ that is $B$ is absorbing. Moreover $B$ is convex, since it is a ball with respect to a semi-norm. Now also its closure $\overline{B}$ is convex and absorbing. It is known that a closed, convex and absorbing subset of a Banach space has $0$ as an inner point, that is there is a closed norm-ball $K=K(0,r)$ with $K \subseteq \overline{B}$.

Now it is sufficient to find a number $s>0$ such that $p(x) \le s$ if $\|x\| \le r$, since then $$ \forall x \in E: ~ p(x) \le \frac{s}{r}\|x\|. $$

We fix $x \in K$. Since then $x \in \overline{B}$, we find some $x_1$ in $B$ such that $\|x − x_1\| \le r/2$. Now $$ x − x_1 \in \frac{1}{2} K ⊆ \frac{1}{2}\overline{B}= \overline{2^{-1}B}, $$ and we find some $x_2$ in $2^{-1}B$ such that $\|x − x_1 − x_2\| \le r/2^2$. Next, we find $x_3$ in $2^{-2}B$ such that $\|x − x_1 − x_2 − x_3\| \le r/2^3$, and by induction there is a sequence $(x_n)$ such that $$ x_n \in 2^{−n+1}B, \quad \|x − \sum_{j=1}^n x_j \| \le \frac{r}{2^n}, $$ for each $n \in \mathbb{N}$. It follows that $p(x_n) < 2^{-n+1}$ for each $n$ and that $$ x = \sum_{n=1}^\infty x_n. $$ The $\sigma$-subadditivity of $p$ yields $$ p(x) = p(\sum_{n=1}^\infty x_n) \le \sum_{n=1}^\infty p(x_n) < \sum_{n=1}^\infty 2^{-n+1} =2. $$ Thus we can choose $s=2$ and have $\forall x \in E: ~ p(x) \le \frac{2}{r}\|x\|$.

That's true... Also note that this is false without completeness — Evangelopoulos Foivos, Dec 11 '24 at 14:26
In the meantime I learned that this is a result of Zabreiko. A different proof using the open mapping theorem is here: https://math.stackexchange.com/questions/842870/zabreikos-lemma — Gerd, Dec 11 '24 at 15:08
(+) Your proof is similar to the proof of the open mapping theorem. The aim in OP is derive the open mapping theorem from the problem on seminorms so the asker didn't want to use that theorem. — Ryszard Szwarc, Dec 11 '24 at 15:34
@RyszardSzwarc Yes you are right. Since the open mapping theorem and Zabreiko's Lemma are quite equivalent it's not obvious how to find a proof without common building blocks. — Gerd, Dec 11 '24 at 15:45
Thank you all for answering my question. Your idea is similar to that in the open mapping theorem and now I know how powerful it is. — Zhi-Chu Luo, Dec 12 '24 at 04:10

Show that a $\sigma$-additive semi-norm is continuous.

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