Jun 2, 2016 18:19
A question about why $M^\perp$ is closed in $E^*$. In my notes it says it's closed because $\ell \mapsto \ell(x)$ is continuous. But I'm not seeing it?
I can find a quick and dirty $\epsilon$-$N$ proof, but I feel that's just me ignoring the important piece I'm supposed to understand to save me from that trouble.
Proof: Let $(\ell_n) \subseteq M^\perp$ be a sequence of elements converging to $\ell \in E^*$, we want to show $\ell \in M^\perp$. Let $0 \neq x \in M$ be arbitrary.
Since $\ell_n \to \ell$ in the operator norm of the dual space, we have: for all $\varepsilon > 0$, there exists $N_\varepsilon > 0$ such that if $n > N_\varepsilon$, then $\|\ell_n - \ell\| < \varepsilon/\|x\|$.
In particular, $\|\ell_n - \ell\| = \displaystyle \sup_{\|y\| \neq 0} \frac{\vert (\ell_n - \ell)(y)\vert}{\|y\|} < \varepsilon/\|x\|$.
So it holds for $x$: $\frac{\vert (\ell_n - \ell)(x)\vert}{\|x\|} = \frac{\vert \ell(x) \vert}{\|x\|} < \varepsilon/\|x\|$ which implies $l(x) = 0$, as desired.
Similarly, $L_\perp$ is closed in $M$. The reasoning given is that $x \mapsto \ell(x)$ is continuous, but I don't see it. Again, I can figure out a quick $\epsilon$-$N$ proof.
So I'm not seeing why both of these mappings are continuous: $\ell \mapsto \ell(x)$ and $x \mapsto \ell(x)$ and I'm not seeing why them being continuous implies that the two sets above are closed.
