I am trying to prove that the closure of $S=\{x\in X : ||x||=1\}$ in weak topology is the closure of $B_1(0)$ .
I have a doubt about what i am doing is correct and for that i need to know whether the intersection of kernels of finite linear functionals is non-trivial . I think here comes the main point about $X$ being infinite dimensional .
And it looks like the closure seems to be the whole space , the neighbourhood of a point $x_0$ contains the lines passing through $x_0$, so i am finding it a bit tricky to get an idea about the closure .
Best Answer
The closure is not the whole space. The fact that neighbourhoods contain lines is irrelevant, since the closure contains no neighbourhood.
Now let me show below that the intersection of the kernels of a finite set of functionals has to be non-trivial if $X$ is infinite-dimensional.
If you have functionals $f_1,\ldots,f_n$, consider the bounded linear map $\tilde f:X\to\mathbb C^n$ given by $\tilde f(x)=(f_1(x),\ldots,f_n(x))$. Then $\ker\tilde f=\bigcap_k\ker f_k$. So, if $\bigcap_k\ker f_k=\{0\}$, we would have $\tilde f$ linear and injective; it would then be an injection into a finite-dimensional space, which would imply $X$ is finite-dimensional. In conclusion, $\bigcap_k\ker f_k\ne\{0\}$