Let $f$ be a linear functional on the Hilbert space $X$. Let $N$ be the null space of $f$. Show that if $f$ is not continuous, then $\bar{N}=X$

I try to show the contrapositive of the statement, i.e. if $\bar{N} \neq X$, then $f$ is continuous. Then I try to show $f$ is continuous at one point.

Since $\bar{N} \neq X$, there exists a point $x_0 \in X \backslash \bar{N}$. Then $x_0 \neq 0$. Otherwise, $x_0=0 \Rightarrow f(x_0)=0 \Rightarrow x_0 \in \bar{N}$, contradiction.

Then by a consequence of Hahn-banach theorem, there exists a bounded linear functional $F$ which extends $f$ such that $||F||=1$ and $F(x_0)=||x_0||$. Clearly $F=f$

Suppose $x_n \rightarrow x_0$. Then my aim is to show that $f(x_n) \rightarrow f(x_0)$. I stuck at here.

Can anyone guide me?

  • $\begingroup$ First, why $\;F=f\;$..."clearly"? But even more important: $\;F\;$ extends $\;f\;$...from where to where?! The l.f. $\;f\;$ is defined in the whole space $\;X\;$ , so where could it be extended to? $\endgroup$ – DonAntonio Sep 30 '13 at 11:40
  • $\begingroup$ I see. Maybe I should delete the word 'extend', then everything is fine. $\endgroup$ – Idonknow Sep 30 '13 at 11:42
  • $\begingroup$ I don't think so unless you want to give away $\;F\;$, which seems to solve your problem. Perhaps you could talk of $\;f\;$ defined on $\;N<X\;$ and then apply H-B to get a bounded, and thus continuous, l.f. on the whole of $\;X\;$ extending $\;f\;$ ... $\endgroup$ – DonAntonio Sep 30 '13 at 11:44

Hint: "Clearly $F = f$" ... why? Be more explicit here, as then you are done, as $F$ is continuous since it is bounded with $\|F\| = 1$.

  • $\begingroup$ Opps, now only I realised $F=f$ is not trivial. But i guess it has something to do with Hilbert space as I didn't use the property of Hilbert space $\endgroup$ – Idonknow Sep 30 '13 at 11:36
  • 1
    $\begingroup$ @Idonknow It has nothing to do with Hilbert spaces. For any topological vector space $E$, a linear functional on $E$ is continuous if and only if its null space is closed. If you have a discontinuous linear functional, its kernel is a $1$-codimensional subspace that isn't closed, hence its closure is the entire space. $\endgroup$ – Daniel Fischer Sep 30 '13 at 13:47

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