Proving this is not a weakly l.s.c. function

Question

In the following, l.s.c. stands for lower semicontinuous, and s.l.s.c. for sequentially l.s.c. So, l.s.c. can mean one of the following three:

1. The preimage of any $(-\infty,a]$ with $a\in\mathbb{R}$ is closed;
2. The preimage of any $(a,+\infty]$ (yeah well, the image of the function may include infinity, but not minus infinity) is open;
3. For all $x_0$ in the domain and $\epsilon>0$ there is an open neighborhood $V\ni x_0$ such that the image of $V$ is contained in the half-line between the image of $x_0$ minus $\epsilon$ and $+\infty$.

s.l.s.c., instead, means that $x_n\to x$ implies:

$$\phi(x)\leq\liminf\phi(x_n),$$

where $\phi$ is the s.l.s.c. function.

I was given the following assignment:

Let $E=\{\{x_n\}_{n=1}^\infty\subset\mathbb{R}:\sum n^2x_n^2<\infty\}$ with the norm $\|\{x_n\}\|=\sqrt{\sum n^2x_n^2}$.

1. Prove $(E,\|\cdot\|)$ is a Banach space;
2. Consider $\phi(\{x_n\})=-\sum x_n^2$; show it is sequentially l.s.c. w.r.t. the weak topology;
3. Show that $\phi$ is not weakly l.s.c.;
4. Deduce, by virtue of the last exercise, that the weak topology is not metrizable.

"The last exercise" was to prove that in a first countable space s.l.s.c. implies l.s.c..

For 1), I viewed $(E,\|\cdot\|)$ as $L^2(\mathbb{N},C')$, where $C'(\{n\}):=n^2$ and the measure is then extended by $\sigma$-additivity. Thus the space is not only Banach, but Hilbert.

For 2), if we view $x$ as a function on $\mathbb{N}$, it is clear that $\phi(x)=-\int_{\mathbb{N}}x(n)^2dC(n)$, where $C$ is the counting measure which if $A\subseteq\mathbb{N}$ yields $C(A)=|A|$. So we have a problem, since Fatou's lemma would imply s.l.s.c. for $-\phi$, as weak convergence implies pointwise convergence, seen as the projections are all linear and continuous. Am I missing something?

For 3), see below.

For 4), it is obvious that this space cannot be first countable, but a metric space always is, hence the topology here is not metrizable.

I do not really know how to do 3). I can try:

1. Proving the first definition of l.s.c. does not hold, i.e. there is a preimage of a closed low-unbounded half-line that isn't closed;
2. Proving the second one doesn't hold, i.e. find a half-line unbounded from above whose preimage is not open;
3. Constructing a weakly convergent net for which the images do not satisfy the l.s.c. condition, i.e. proving $\phi$ is not l.s.c. under nets;
4. Disproving the third definition of l.s.c.;
5. Using the fact a Hilbert basis of the dual are the projections in order to characterized weak open sets;
6. Finding a Hamel basis of the dual and using it to describe weak open sets.

5 and 6 would anyway require to use one of the above strategies. So what can I do here? And am I right in claiming there is a wrong minus sign in that $\phi$?

Update concerning 2)

The teacher suggested to try using Dominated Convergence to show $\phi(x)$ is precisely the limit of $\phi(x_n)$ as $x_n\to x$. I tried thinking of how to find a dominating function, but the only thing I could come up with is to try proving that if a series $\sum x_n$ with positive terms converges then it must eventually be strictly less than $\frac1n$, thus less than or equal to $\frac{1}{n^{1+\epsilon}}$ for some $\epsilon$, applying this to the sums of $x_n(m)^2$ to conclude they are all eventually less than $\frac{1}{m^{1+\epsilon}}$ and thus $x_n(m)^2\leq\frac{1}{m^{\frac12+\frac12\epsilon}}$ and thus I find this function which is integrable and dominates the $x_n$ and therefore I can use dominated convergence to infer $L^2$ convergence from pointwise convergence which in turn comes from the weak convergence in $E$. Is there any smarter way to do this? And is the statement about converging serieses true?

Update 2 concerning 2)

No need for that statement: since the sequence is bounded in the space seen above, one has $|x_n(m)|\leq\frac{C}{m}$ for all $m$, grow the constant and this extends to $x$, and $\frac{C}{m}$ is in $\ell^{(2)}$, so we have the integrable domination and using d.c. we conclude sequential continuity for both $\phi$ and $-\phi$.

Update concerning 3)

Asked the teacher for a suggestion, she said try proving there is a non-weakly-open strict superlevel set, i.e. a n-w-o $\phi^{-1}((\alpha,+\infty))$, by showing there is a bounded such set. I however seem to have proven all superlevel sets are unbounded. Indeed, $x_n=\frac{c}{n+\epsilon n^2}$, for all $c,\epsilon$ with $\epsilon>0$, belongs to my space, since its square goes to zero like $\frac{1}{n^4}$ and so the squares times $n^2$ have a convergent series. However, their norms explode, since their limit has infinite $L^2$ norm with that measure $C'$ defined above, and the integral of the limit is by Fatou at most the limit of the integral, which is therefore infinity. However, the images of those are all bounded in absolute value by $\sum\frac{c^2}{n^2}$, which for appropriate $c$s is any desired value, so for any $\alpha$, if $\phi^{-1}(\alpha,\infty)\neq\varnothing$ (i.e. $\alpha<0$), I can find an unbounded sequence contained in the level set. So Houston, we have a problem :).

Answer 1

The thing here is that there is a whole lot more needed for weak openness than being unbounded.

If $V$ is weakly open in a Banach space $E$ and nonempty and $x_0\in V$, then a whole line through $x_0$ is contained in $V$, so $x_0+tv_0\in V$ for all $t\in\mathbb{R}$ for some $v_0$.

That is because a basis for the weak topology in $x_0$ is made of the sets of the form $\bigcap_1^kf_i^{-1}(\omega_i)$ where $k\in\mathbb{N}$, $\omega_i$ is open in $\mathbb{R}$ and $f_i\in E'$. So there is a $k$ and $f_1,\dotsc,f_k$ and $\epsilon>0$ such that:

$$\bigcap_{i=1}^kf_i^{-1}(f_i(x_0)-\epsilon,f_i(x_0)+\epsilon)\subseteq V.$$

Take $T(x)=(f_1(x),\dotsc,f_k(x))$ from our Banach space to $\mathbb{R}^k$. If $E$ is infinite-dimensional, $T$ cannot be injective or, being linear, it would be an isomorphism onto its image, but the domain is infinite-dimensional and the codomain isn't. So $\ker T\neq\{0\}$. Let $v_0\in\ker T\smallsetminus\{0\}$. But then $tv_0\in\ker T$ for all $t\in\mathbb{R}$. But this means $x_0+tv_0\in V$ for all $t$, as we claimed.

In our case, our function $\phi$ has the property that it is always nonpositive, so if we consider a strict superlevel, 0 (the zero vector in $\phi$'s domain, which is mapped to 0) will be in it, or it will be empty. That is, if $\alpha>0$, then $\phi^{-1}((\alpha,+\infty))\neq\varnothing$ will contain 0. But then it contains a line through 0, hence it contains $tv$ for all real $t$ and some $v$ in our space. But this is not possible, since our $\phi$ sends $tv\mapsto t^2\phi(v)$, which tends to $-\infty$ as $t\to\infty$ since $\phi(v)<0$.

Naturally, it could be that $\phi(v)$, at least in general. But then as soon as one component of $v$ is nonzero we have a nonzero term in the sum which is $\phi(v)$, and all terms in that sum have the same sign, hence $\phi(v)=0\iff v=0$, but then $\{tv\}=\{0\}$ is not a line.

This completes this exercise.