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Is the following claim true? It feels like it should be true, but I don't really know how to show it.

Let $X$ be a Banach space, and $x \in X$ an element of it. Then there exists a functional $\phi \in X^*$ such that $\| \phi \| = 1$ and $\| x \| = | \phi(x) |$.

If I'm not mistaken, it would suffice to say that there exists a sequence $(\phi_k)_{k = 1}^\infty$ of unit functionals for which $| \phi_k (x) | \to \| x \|$, since the unit ball in $X^*$ is compact in the weak topology. However, I don't know how to prove the former result.

EDIT: I forgot to actually define $\psi$ as $\psi(\lambda x) = \lambda$.

My intuition is that I should be able to invoke Hahn-Banach and define a linear function $\psi$ on $\mathbb{C} x \subseteq X$ bounded by the norm $\rho(x) = \| x \|$ on $X$, then extend it from $\mathbb{C} x$ to all of $X$. Is this a correct application of Hahn-Banach?

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  • $\begingroup$ @Nephanth Wouldn’t that require a Hilbert space? $\endgroup$
    – AJY
    Jul 16, 2020 at 16:27
  • $\begingroup$ Oopsie comment deleted, I read wrong $\endgroup$
    – tbrugere
    Jul 16, 2020 at 16:28

3 Answers 3

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You don't need completeness of $X$. This is true in normed spaces. Your last idea is a good one:

We may assume $x \neq 0$. Define the functional

$$\varphi: \Bbb{C}x \to \Bbb{C}: \lambda x \mapsto \lambda \Vert x \Vert$$

Then it is easily checked that $\Vert \varphi \Vert =1$ (the inequality $\leq$ is obvious, and then note that $\varphi(x/\Vert x \Vert) = 1$ so also $\Vert \varphi\Vert \geq 1$). By Hahn-Banach, we can extend to a functional $\tilde{\varphi}: X \to \Bbb{C}$ with $\Vert \tilde{\varphi} \Vert =1$ and this is the functional you are looking for.

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  • $\begingroup$ Yes, that’s the one I had in mind! I just forgot to actually say what $\psi$ was. $\endgroup$
    – AJY
    Jul 16, 2020 at 16:29
  • $\begingroup$ Yeah, so you pretty much had it proven. All you had left to do was check that the functional on the one-dimensional subspace has norm $1$ and then Hahn-Banach allows you to extend the functional to the entire space while preserving the norm. $\endgroup$
    – J. De Ro
    Jul 16, 2020 at 16:30
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Yes, this is a perfect application. You can start by defining a functional on the one-dimensional space spanned by $x$ as $\phi(ax)=a\|x\|$ which has norm $=1$ and extend using HB.

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W.l.o.g say $x \neq 0$. As you said then look at the subspace $\mathbb{C}x$ and the map $\phi(\lambda x) = \lambda \| x \|$ defined on $\mathbb{C}x$. Then $\phi$ is of course well-defined, linear, bounded and it holds \begin{align} \phi(x) = \| x \| \, \text{ and } \, \| \phi \| = \sup_{\| \lambda x \| \leq 1, \, \lambda \in \mathbb{C}} |\lambda | \| x \| = 1. \end{align} Using Hahn-Banach you can then extend $\phi$ to and element $\Phi\in X^*$ with $\| \Phi \| = 1$.

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