I'm working on a straightforward problem but my result is the opposite of what I want to show and I cannot find the mistake so any help would be much appreciated.
We have $K = \{ \sqrt{n} e_n | n \geq 1 \}$ where $\{e_n\}$ is an orthonormal basis of a Hilbert space $H$. We are asked to show that the function $f(x) = \begin{cases} 1 \textit{ if } x \in K \\ 0 \textit{ if } x \notin K\end{cases}$ is not weakly continuous in $0$ but that we have that $x_n$ converges weakly to $0$, we must have that $f(x_n)$ converges to $0$.
For the first part, the preimage of a small ball around $0$ is just the preimage of $0$ or $K^c$. We need to show that $\exists x \in K^c: \forall y_1,\dots,y_n \forall \epsilon >0:\exists z\in K: \forall i: |\langle y_i,z-x \rangle |< \epsilon$.
This I think can be done by setting $x=0$, since if $y_i=\sum_{n}a_{in}e_n$, we have $|\langle y_i, \sqrt{n}e_n\rangle|=\sqrt n |a_{in}| $ and this becomes arbitrary small since the coefficients $|a_{in}|$ have to converge faster to zero than $1/n$. So I think that is correct.
But now if $x_n= \sqrt n e_n$, we have now proven that $| \langle y,x_n \rangle | $ converges to zero for all $y$ so that $\langle x_n,y \rangle \rightarrow \langle 0,y \rangle$ and thus $x_n$ converges to $0$ weakly. But for this sequence $f(x_n) \rightarrow 1$. So I don't see why this approach seems to work for the first part and fail for the second part.
Any help?
thanks
Best Answer
With the choice $x_n=\sqrt ne_n$, the sequence $\left(\lVert x_n\rVert\right)_{n\geqslant 1}$ is not bounded in $H$ hence cannot converge weakly to zero.
Let $\left(x_n\right)_{n\geqslant 1}$ be a sequence converging weakly to $0$. We have to show that $f(x_n)\to 0$ and since $f$ takes only the values $0$ or $1$, we have to prove that there exists an $n_0$ such that $f(x_n)=0$ for $n\geqslant n_0$. If this $n_0$ does not exist, then we can find $n_k\uparrow \infty$ such that $x_{n_k}\in K$ for all $k$. Let $I =\{i\geqslant 1,x_{n_k}=\sqrt{i}e_i\mbox{ for some }k\}$. Then the set $I$ is finite (otherwise, this would contradict the boundedness of $(x_n)$. This contradict the weak convergence to zero.