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Here's a proof why $l^p(\mathbb N)$ is not locally convex, this is just for simplicity, it can be easily generalized. If it would be locally convex, then the unit ball $B_1(0)$ would contain a convex neighborhood U of $0$. Then there must be $\delta>0$ with $B_{2\delta}(0)\subset U$, hence also $\mathrm{conv}(B_{2\delta}(0))\subset U\subset B_1(0)$. Let ...


The closure of the domain in $L^2$ is simply $L^2$: Obviously it holds $C_0^\infty(0,1)\subset D(A_0)$. The set of smooth function is dense in $L^2(0,1)$, hence its closure is $L^2(0,1)$. This implies that the closure of $D(A_0)$ is $L^2(0,1)$ as well.


Yes, every continuous semi-norm $p$ on $M$ can be extended to a continuous semi-norm on $X$: Let $U=\lbrace x\in M: p(x)<1\rbrace$ be the unit ball of $p$. Since $U$ is open in $M$ and $0\in U$ there is a convex $0$-neighbourhood $V$ in $X$ such that $V\cap M\subseteq U$. Now, let $W$ be the convex hull of $U\cup V$. Since $U$ and $V$ are convex we have ...

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