# How to prove a norm identity for a Banach space and its dual

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?

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

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.

• Yes, that’s the one I had in mind! I just forgot to actually say what $\psi$ was.
– AJY
Jul 16, 2020 at 16:29
• 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. Jul 16, 2020 at 16:30

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.

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 $$\| \Phi \| = 1$$.