0
$\begingroup$

Let $M\subseteq X$ be a subset of a normed space. I have been asked to show that the annihilator of $M$,$M^a$ is closed. To do this I assume that it isn't closed. I.e there exists some functional $f\in \bar{M^a}$ such that $f\notin M^a$. So $\exists x_f\in M$ such that $|f(x_f)|=\delta>0$. Now consider an arbitrary sequence $(f_n)$ in $M^a$. Then $\forall N\in \mathbb{N}$ we have $n>N$ implies that $|f(x)-f_n(x)|>\delta /2$ so $(f_n)$ does not converge to $f$. As $(f_n)$ was arbitrary this contradicts the fact that $f\in \bar{M^a}$ and I conclude.

My issue is with my notion of convergence, because I'm not using the norm of the dual space in defining my lack of convergence. I've seen mention of weak and strong convergence but we have not encountered these notions yet. Intuitively it seems that the convergence of functionals requiring the value of the functionals to converge to the value of the limit functional at each point makes more sense than using the norm of the dual space, but intuition has lead me astray before. If we do have to define convergence of functionals using the norm of the dual space my proof would need a lot more work. So does my proof look correct or is it way off the mark? Thanks in advance.

$\endgroup$
2
$\begingroup$

Showing that $f_n(x) \to f(x)$ for all $x \in X$ (directly or via contradiction) proves that $f_n \to f$ in the weak-$^*$ sense.

A weak-$^*$ closed set is closed in the norm topology.

To avoid concepts you haven't encountered yet, suppose that $\{f_n\}$ is a sequence in $M^a$ that converges in norm to a functional $f$. Thus $$ \lim_{n \to \infty} ||f_n - f\| = \lim_{n \to \infty} \sup_{\|x\| \le 1} |f_n(x) - f(x)| \to 0.$$

Let $x \in M \setminus \{0\}$ and let $y = x/\|x\|$. Then $$ |f(y)| \le |f_n(y) - f(y)| + |f_n(y)| \le \|f_n - f\| $$ for all $n$ because $f_n(y) = 0$. Since this last expression tends to $0$ as $n \to \infty$ you conclude $f(y) = 0$.

Thus $f(x) = 0$ too, so that $f$ vanishes on $M$. Thus $f \in M^a$, so that $M^a$ includes all its (norm) accumulation points.

$\endgroup$
  • $\begingroup$ Sorry I just realised the question stated that $M$ was a subset not subspace and I originally put subspace. Can we still say that $y$ is in $M$? That was what was giving me problems in using the norm convergence $\endgroup$ – K.Power Apr 13 '16 at 19:37
  • $\begingroup$ Nevermind I was able to slightly adapt your proof to make it work by just considering $x\in M\backslash \{0\}$ thanks $\endgroup$ – K.Power Apr 13 '16 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.