# Prove that $\|x\|=\sup\{|f(x)|:f\in X^*, \,\|f\|=1\},$ where $x\in X$ and $X^*$ denotes the dual space of $X$.

Let $$x$$ be an element of a normed linear space $$X$$ and let $$X^*$$ denote the dual space of $$X$$. Prove that \begin{align} \|x\|=\sup\{|f(x)|:f\in X^*, \,\|f\|=1\} \end{align}

MY TRIAL

It suffices to show that \begin{align} \forall\;\epsilon>0,\;\exists\;|f(x_{\epsilon})|\in \{|f(x)|:f\in X^*, \,\|f\|=1\}\;\;\text{such that}\end{align} \begin{align} \|x\|-\epsilon< |f(x_{\epsilon})|\leq \|x\|.\end{align}

Let $$x\in X$$ such that $$x\neq 0.$$ Otherwise, $$\|f\|=0$$. Then, by Hanh-Banach Theorem, there exists a linear functional $$f$$ on $$X$$ such that \begin{align} \|f\|=1 \;\;\;\text{and}\;\;\;|f(x)|= \|x\|\leq \|x\|.\end{align}

Please, I'm I right thus far? If yes, I am stuck here as I don't know how to arrive at \begin{align} \|x\|-\epsilon< |f(x)|.\end{align} If no, can you help fix my wrong(s)?

• Well $\|x\|-\epsilon<\|x\|$ – SmileyCraft Jan 10 at 21:18
• @SmileyCraft: Yes, I agree with you! Okay, do you mean $\| x\|-\epsilon<\| x\|=|f(x)|$? – Omojola Micheal Jan 10 at 21:19

Your approach is flawed for two reasons that I see: first, you talk about an $$x_\epsilon$$, not sure what you expect by that; it's the $$f$$ that varies here, the $$x$$ is fixed. Second, you say

"by the Hahn-Banach theorem, there exists a linear function $$f$$ on $$X$$ with $$\|f\|=1$$ and $$|f(x)|=\|x\|$$".

That's not wrong: but it's precisely what you are supposed to prove.

For any $$f$$ with $$\|f\|\leq1$$, you have $$|f(x)|\leq\|f\|\,\|x\|=\|x\|$$. So $$|f(x)|\leq\|x\|$$ for any $$f$$ in your set. And now, we use Hahn-Banach as you say. On the one-dimensional subspace $$\mathbb C\,x$$, define $$f_0(\lambda x)=\lambda\,\|x\|$$. Then $$|f_0(\lambda x)|=|\lambda\,\|x\|\,|=\|\lambda x\|,$$ so $$\|f_0\|=1$$. Now, by Hahn-Banach, there exists $$f\in X^*$$, with $$\|f\|=\|f_0\|=1$$, and $$f(x)=f_0(x)=\|x\|$$. So $$\|x\|=\max\{|f(x)|:\ f\in X^*,\ \|f\|=1\}.$$

Credits to Martin Argerami. I present the full proof for future readers.

Let $$x\in X$$ be fixed, such that $$x\neq 0.$$ Consider the subspace spanned by $$x$$, defined by $$Z=\{ \alpha x:\;\alpha\in \Bbb{R} \}.$$

Define $$f$$ arbitrarily by \begin{align} f:\,&Z\to \Bbb{R}\\& z\mapsto \alpha \| x \|. \end{align} Let $$z\in Z$$, then $$|f(z)|= |\alpha \| x \||= \|\alpha x \|= \|z \|.\,$$ This implies that $$|f(z)|=\left|\|z \| \right|\leq \left|\|z \|\right|=\|z \|$$ which gives boundedness of $$f$$ and $$\|f \|=1.$$ Let $$\gamma,\lambda\in \Bbb{R}$$ and $$y,z\in Z,$$ then $$\exists\,\alpha,\beta \in \Bbb{R}$$ such that $$y=\alpha x$$ and $$z=\beta x.$$ Then, $$f$$ is linear, since

\begin{align}f\left(\gamma y+\lambda z\right)&= f\left[(\gamma \alpha+\lambda \beta)x\right] \\&= (\gamma \alpha+\lambda \beta)\| x \|\\&= \gamma (\alpha\| x \|)+\lambda( \beta\| x \|)\\&= \gamma f\left( y\right)+\lambda f\left(z\right).\end{align} Since $$f$$ is a linear functional, $$\;\alpha f\left( x\right)=f\left(\alpha x\right)=f\left(z \right)=\alpha \| x \|$$ implies that $$f\left( x\right)=\| x \|.$$ By the implication of $$f$$ being a bounded linear functional, we have by the Hanh-Banach Theorem, that there exists $$F\in X^{*}$$ s.t. $$F(x)=f(x)$$ for all $$x\in Z$$ and \begin{align}F(x)=f(x)=\| x \|\;\;\text{and}\;\;\|F \|=\| f \|=1.\end{align} So, \begin{align}\| x \|=\left|\| x \| \right|=\left| F(x) \right|\;\;\text{for some}\;\;F\in X^{*}\;\;\text{and}\;\;\|F \|=1.\end{align} Taking $$\sup$$ over such $$\;F\in X^{*}\;\text{and}\;\|F \|=1,$$ we get \begin{align}\| x \|=\sup\{\left| F(x) \right|:\,F\in X^{*},\;\|F \|=1\}.\end{align}

Let’s denote $$S_x =\sup\{|f(x)|:f\in X^*, \,\|f\|=1\}$$.

We have to prove that $$\Vert x \Vert =S_x$$.

If $$\Vert f \Vert =1$$, then $$\vert f(x)\vert \le \Vert x\Vert$$. Hence $$S_x \le \Vert x\Vert$$. Conversely for $$x\neq0$$, on the subspace $$\mathbb Rx$$, the linear form defined by $$f(y) = \lambda \Vert x \Vert$$ for $$y = \lambda x$$ is such that $$f(y) \le \Vert y \Vert$$. Using Hahn Banach theorem we can extend $$f$$ to a linear form on $$X$$ such that $$\vert f(y)\vert \le \Vert y \Vert$$ for $$y \in X$$ and $$\vert f(x) \vert = \Vert x \Vert$$. Hence $$S_x \ge \Vert x \Vert$$, allowing us to conclude.