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I'm trying to prove that if $X$ is infinite dimentional Banach space, $S_{X}$ (the unit sphere) is a dense $G_{\delta}$ set in $(B_X,\omega)$ (where $\omega$ is the weak topology). From here I'm trying to conclude that in an infinite dimensional space the norm is never $\omega$-continuous.

I showed that $S_X $ is $\omega$-dense in $B_X$ by showing that for every point $x_0\in B_X$ and every $\omega$-neighborhood $U$, $U\cap S_X\neq \varnothing$, using that the intersection $\cap_{i=1}^nf^{-1}(0)$ of linear functionals in infinite dimention space contains a nonzero element.

After that, to show that the norm is not $\omega$-continuous, I used that $\lVert \cdot\rVert^{-1}[(-1,1)]$ is bounded and therefore not $\omega$-open.

I'm having trouble proving that $S_X$ is a $G_{\delta}$ set. The obvious candidates would be $G_n=\{x\in B_X\big| \lVert x\rVert>1-\frac{1}{n}\}$ and $S_X=\cap_{n=1}^\infty G_n$, but I'm not sure that $G_n$ are $\omega$-open.

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  • $\begingroup$ What is $S_X{}$? $\endgroup$
    – tomasz
    Jun 12, 2018 at 13:10
  • $\begingroup$ And $\omega$, for that matter? $\endgroup$
    – tomasz
    Jun 12, 2018 at 13:12
  • $\begingroup$ The unit sphere in X, $S_X=\{x\in X \big| ||x||=1\}$, and $\omega$ is the weak topology, generated by all the continuous linear functionals on X $\endgroup$ Jun 12, 2018 at 13:12
  • $\begingroup$ I made some edits to the last two lines to fix some things that seemed like obvious typos. You might want to check my edits reflect what you meant. $\endgroup$ Jun 12, 2018 at 14:22
  • $\begingroup$ It is often a good strategy to show that a set is open (with respect to some topology) by showing that its complement is closed. (That works quite well here.) $\endgroup$ Jun 12, 2018 at 14:24

2 Answers 2

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Let $f$ be a norm $1$ functional. Then $G_{f,n}:=f^{-1}{\big(}(1-1/n,\infty){\big)}$ is weak open in $X$ and a subset of $\tilde G_n=\{x\in X\mid \|x\|>1-1/n\}$. For any $x\in \tilde G_n$ you can find via Hahn-Banach a norm $1$ functional $f_x$ with $f_x(x)=\|x\|$, so $x\in G_{f_x,n}$. For that reason $\tilde G_n=\bigcup_{f\in X^*,\|f\|=1} G_{f,n}$ and $\tilde G_n$ is open.

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Notice that $G_n = H_n \cap B_X$ where $$H_n = \{x \in X : \|x\| > 1 - \frac1n\}.$$ So it will suffice to show that $H_n$ is a weak-open subset of $X$. This follows from weak lower-semicontinuity of the norm. A function $f:X \to \mathbb{R}$ is lower-semicontinuous at every point in the space if and only if $\{x : f(x) > \alpha \}$ is open for every $\alpha \in \mathbb{R}$.

This question contains a few proofs that the norm is always weakly lower-semicontinuous.

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  • $\begingroup$ Still I'll need to use the Hahn-Banach theorem to show that there is a linear functional $f\in S_{X^*}$ such that $f(x)=\lVert x \rVert$ for every $x\in B_X$ $\endgroup$ Jun 12, 2018 at 18:07
  • $\begingroup$ @user3701033 Why? The function that you want to be weakly lower-semicontinuous is $x \mapsto \|x\|$. $\endgroup$ Jun 12, 2018 at 18:09
  • $\begingroup$ Also I don't see why you'd want to avoid using Hahn-Banach so I don't know why that would be an issue. The result you state is a standard corollary of Hahn-Banach. $\endgroup$ Jun 12, 2018 at 18:14
  • $\begingroup$ My bad, I misunderstood you. I need to use the norm, not the functional I stated. I wasn't trying to avoid using Hahn-Banach, I was just trying to understand what you were doing. I have never used LSC functions before. $\endgroup$ Jun 12, 2018 at 18:18
  • $\begingroup$ @user3701033 In that case, I wouldn't worry too much about them. My point was that any proof that the $G_n$ (or rather the $H_n$) are weakly open will boil down to proving that the norm is weakly lower-semicontinuous, by the definition pretty much. I just wanted to make you aware that you would be proving/using this more general result along the way. $\endgroup$ Jun 12, 2018 at 18:23

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