Let $(X, \|\cdot\|_X)$ be a normed vector space and $(X^{\ast},\|\cdot\|_{X^{\ast}})$ its dual space. I have to prove, that

$$ \forall x\in X:\quad \|x\|_X = \sup_{T\in X^{\ast}}\{|T(x)| : \|T\|_{X^{\ast}}=1\}$$

How can I show this using Hahn-Banach theorem?


2 Answers 2


There's a neat corollary of Hahn-Banach saying that for every $x$ there exists a linear operator $T$ with unitary norm such that $T(x)=||x||$.

Using this and the definition of the norm on the dual space, you have your proof.


I will elaborate on the previous poster's answer by proving the aforementioned corollary, which is often referred to as "$X^*$ norms $X$." In particular:

Claim: For any $x\in X$ there exists $T\in X^*$ such that $\vert\vert T\vert\vert=1$ and $\vert T(x)\vert=\vert\vert x\vert\vert$.

Pf.: Let $L$ be the one dimensional subspace defined by $x$, that is $L=\{\alpha x\mid \alpha\in\mathbb{R}\}$. Let $T_0:L\rightarrow\mathbb{R}$ send $\alpha x\mapsto\alpha\vert\vert x\vert\vert$. It's easy to check that $T_0$ is bounded, linear, and has norm $1$. Thus, we may use Hahn-Banach to extend this functional to $X$, as $T$, while preserving its norm. This $T$ works.

This immediately answers your question because it shows us that $\sup_{T\in X^{\ast}}\{|T(x)| : ||T||_{X^{\ast}}=1\}\geq \vert\vert x\vert\vert$ and it is obvious that $\sup_{T\in X^{\ast}}\{|T(x)| : ||T||_{X^{\ast}}=1\}\leq \vert\vert x\vert\vert$ because if $\vert\vert T\vert\vert=1$ then $\vert T(x)\vert\leq \vert\vert x\vert\vert$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.