Definition: Let $X$ be a topological vector space and let $x\in X$. Then $x$ defines a linear functional $\hat{x}$ on $X^*$ via $\hat{x}(f)=f(x)$ $(f\in X^*)$.

let $X$ be a normed space and let $x\in X$. I am trying to show that $\hat{x}\in X^{**}$. And my attempts are:

  1. Let $f\in X^*$ and let $(f_i)$ be a net in $X^*$ with $f_i\overset{\|\cdot\|}{\longrightarrow} f$ in $X^*$. Then $$|\hat{x}(f_i)-\hat{x}(f)|=|f_i(x)-f(x)|=|(f_i-f)(x)|\leqslant\|f_i-f\|\cdot\|x\|\rightarrow0.$$ Thus $\hat{x}(f_i)\rightarrow\hat{x}(f)$. Hence, $\hat{x}$ is a continuous linear functional on $X^*$; that is, $\hat{x}\in X^{**}$.

  2. Using a corollary of the Hahn-Banach Theorem, $$||\hat{x}||=\sup\{|\hat{x}(x^*)|:\|x^*\|\leqslant1\}=\sup\{|x^*(x)|:\|x^*\|\leqslant1\}=\|x\|.$$ Thus $\hat{x}\in X^{**}$.

But my professor mentioned I didn't need any proof at all. It follows immediately by using definition in functional anaylsis. But I don't know what definition gives $\hat{x}\in X^{**}$? Any helps will be appreciated!!

  • 2
    $\begingroup$ For future reference, the TeX command \| (or \Vert) can be used to produce a $\Vert$. $\endgroup$ – Xander Henderson May 10 '18 at 19:12
  • $\begingroup$ Did you mean normed vector space instead of topological vector space? You've used $\|\cdot\|$ referring to a norm. $\endgroup$ – B. Mehta May 10 '18 at 19:21

From the fact that $f$ is bounded, you have $$ |\hat x(f)|=|f(x)|\leq\|f\|\,\|x\|. $$ So $\hat x$ is bounded and $\|\hat x\|\leq\|x\|$.

  • $\begingroup$ I am sorry I am a little slow. To show $\hat{x}$ is bounded we need show there exists $c>0$ such that $\|\hat{x}(f)\|\leq c\|f\|$ for all $f\in X^*$. That is the definition. Then how to get $\hat{x}$ is bounded by the inequality you give? Thank you! $\endgroup$ – Answer Lee May 11 '18 at 2:30
  • 1
    $\begingroup$ Exactly. Take $c=\|x\|$. Note that we usually don't use double vertical bars for the absolute value (which of course is the norm on $\mathbb C$). $\endgroup$ – Martin Argerami May 11 '18 at 2:33

You've correctly noticed that something needs to be shown, in particular that $\hat{x}$ is a continuous linear functional. So, you just need to show that $\hat{x}$ is a bounded linear functional, which is easy: take $\|f\|=1$, then $$|\hat{x}(f)| = |f(x)| \leq \|x\|$$ hence $\|\hat{x}\|:=\sup_{\|f\|=1} |\hat{x}(f)|\leq\|x\|$, so is bounded.

Notice that with Hahn-Banach, you can additionally show that $\|\hat{x}\| = \|x\|$ (as you did), but that's not necessary here.

  • $\begingroup$ Thanks for your answer. I have used this fact to show it. But my professor mentioned just using definition. Really don't know how to show this. $\endgroup$ – Answer Lee May 10 '18 at 19:24
  • $\begingroup$ This is just by definition. There's no theorems applied here at all, to show $\hat{x} \in V^{**}$, you just need to show it's bounded: which is immediate from the equation in the post. $\endgroup$ – B. Mehta May 10 '18 at 19:26

$V^*$ is the dual space of $V,$ and the double dual space $V^{**}$ is isomorphic to $V$. Can you take it from here?

Edit: I'm assuming $dim(V)<\infty$ here.

  • $\begingroup$ The double dual is not always isomorphic to $V$, and the OP certainly doesn't seem to be assuming this. $\endgroup$ – B. Mehta May 10 '18 at 19:07
  • $\begingroup$ true, i added my implicit assumption. $\endgroup$ – emma May 10 '18 at 19:13

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.