What definition gives $\hat{x}\in X^{**}$?

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!!

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

• 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! – Answer Lee May 11 '18 at 2:30
• 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 $\mathbb C$). – 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.

• 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. – Answer Lee May 10 '18 at 19:24
• 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. – 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.

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