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Been sitting on this one for a few days and would really appreciate some help. I have included a definition and theorem that seemed useful. If anyone would be willing to critique or confirm my proof it would really make my day. Thanks in advance.

Suppose $X$ is a topological vector space which is of the second category in itself. Let $K$ be a closed, convex, absorbing subset of $X$. Prove that $K$ contains a neighborhood of $0$.

Suggestion: Show first that $H = K \cap (-K)$ is absorbing. By a catagory argument, $H$ has interior. Then use $$ 2H = H + H = H - H. $$ Show that the result is false without convexity of $K$, even if $X = \mathbb{R}^2$. Show that the result is false if $X$ is $L^2$ topologized by the $L^1$-norm

Definition: A convex set $K\subset X$ is called absorbing, if given $x\in X$ there exists $\lambda>0$ such that $\lambda x\in K$.

Theorem: Suppose $V$ is a neighborhood of $0$ in a topological vector space $X$. If $$ 0 < r_1 < r_2 < \dots < r_n \rightarrow \infty, then $$ $$ X = \bigcup_n r_n V. $$

Proof:

Let $X$ be a topoligical vector space which is of second category in itself. Let $K$ be a closed, convex absorbing subset of $X$. Since $K$ is absorbing, we observe that $X = \bigcup_n nK$. Consider $H = K \cap (-K)$. We observe that $H$ is also a closed convex set since the intersection of convex sets are convex, and finite intersections of closed sets are closed. Since $K\subset X$ is convex we note that $K$ is second category in its self since $K$ is everywhere dense in its self. We observe that $\overline{H}^\circ \not = \emptyset$ since the intersection of second category sets is second category. Hence H has non empty interior. Since $2H = H+H = H-H$; $H$ must be a neighborhood of $0$. By the theorem stated above, we may say that $H$ is absorbing since $H$ is a neighborhood of $0$. $H \subset K \Rightarrow K$ contains a neighborhood of $0$.

To emphasize the requirement of $K$ convex suppose that $X = \mathbb{R}^2$. Suppose $K$ is the set consisting of the whole space with the nonzero points of the parabola $y=x^2$. Then obviously $K$ is absorbing but does not contain a neighborhood of $0$.

If $X$ is $L^2$ topologized by the $L^1$ norm consider a set $K = \{f: \|f\|_2 \leq 1\}$ then we notice that $K$ is convex and absorbing, but there exists a sequence of functions $f_n \in X$ with $\| f_n\|_1 = 1/n \rightarrow 0$ and $\| f_n\|_{2} = 2$ for all $n$. So no ball around $0$ can be contained in $K$.

Ben
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1 Answers1

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Let $X$ be a topological vector space which is of second category in itself. Let $K$ be a closed , convex, absorbing subset of $X$.

Consider $H=K\cap (-K)$. Now for any $x\in X$ we have $s>0,t>0$ such that $sx\in K$ and $t(-x)=-tx\in K$. Also $K$ is convex in $X$, so that $\frac{t}{s+t}(sx)\ +\ \frac{s}{s+t}(-tx)=0\in K$. Now if $s>t>0$ then $tx=\frac{t}{s}(sx)\ +\ (1- \frac{t}{s})0\in K$ due to convexity of $K$ i.e. $s>t>0$ implies $tx\in K\cap (-K)=H$. Next if $t>s>0$ then $-sx=\frac{s}{t} (-tx)\ +\ (1-\frac{s}{t})0 \in K$ i.e. $sx\in K\cap (-K)=H$. So that $H$ is absorbing subset of X (the case $t=s>0$ is trivial) . Note also that $H$ is convex closed in $X$ being intersection of two convex closed subsets.

Now $X=\bigcup_{t>0}\ tH$ since $H$ is absorbing. Now due to convexity of $H$ we also have $X=\bigcup_{t>0,t\in \Bbb Q}\ tH$ i.e. $X$ is countable union of closed subsets of $X$ , but $X$ is second category , therefore $int(tH)\not=\phi$ for some $t>0,t\in \Bbb Q$. Since scalar multiplication is a homeomorphism we can say that interior of $H$ is also non-empty. So let $U$ be open in $X$ and $U\subseteq H$.

Now $$\{2h : h\in H\}= \{h_1+h_2: h_1,h_2\in H\}= \{h_1-h_2: h_1,h_2\in H\}$$ i.e. $2H=H+H=H-H$. The first equality due to convexity of $H$ and $0\in H$. The second equality due to $h\in K\cap (-K)$ implies that $-h\in K\cap (-K)$. Therefore $U-U\subseteq H-H=2H$ but $V:=U-U$ is an open nbd of $0$. Hence $\frac{1}{2} V$ is an open nbd of $0$ contained in $H=K\cap (-K)$. So $ K$ contains an open nbd of $0$.

Sumanta
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