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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)?

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  • $\begingroup$ Well $\|x\|-\epsilon<\|x\|$ $\endgroup$ – SmileyCraft Jan 10 at 21:18
  • $\begingroup$ @SmileyCraft: Yes, I agree with you! Okay, do you mean $\| x\|-\epsilon<\| x\|=|f(x)|$? $\endgroup$ – Omojola Micheal Jan 10 at 21:19
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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 $$ \|x\|=\max\{|f(x)|:\ f\in X^*,\ \|f\|=1\}. $$

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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}

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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.

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