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.

Answer 1

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 $\forall x \in E: ~ p(x) \le \frac{2}{r}\|x\|$.