Solution to the Week's Problem: Opalg

[Chris L T521]

Here's this week's problem.

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Background Info: Let $X$ be a normed linear space. The linear operator $J:X\rightarrow X^{\ast\ast}$ defined by
\[J(x)[\psi] = \psi(x) \text{ for all $x\in X$, $\psi\in X^{\ast}$}\]
is called the natural embedding of $X$ into $X^{\ast\ast}$.

Problem: Let $X$ be a normed linear space. Show that the natural embedding $J:X\rightarrow X^{\ast\ast}$ is an isometry.

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[Chris L T521]

This week's problem was correctly answered by Opalg. You can his solution below.

Spoiler

This result is essentially a consequence of the Hahn–Banach theorem. If $x\in X$ then we can define a linear functional $\phi$ on the one-dimensional subspace of $X$ spanned by $x$ by $\phi(\lambda x) = \lambda\|x\|.$ Then $\phi$ is bounded, with $\|\phi\| = 1$, and $\phi(x) = \|x\|.$ By the H–B theorem, $\phi$ extends to a linear functional (still called $\phi$) on the whole of $X^*$, with $\|\phi\| = 1$.

Denote by $X^*_1$ the unit ball of $X^*$. Then $\psi \in X^*_1\;\Rightarrow\; |\psi(x)| \leqslant \|\psi\|\|x\| \leqslant \|x\|$. It follows that $$\sup_{\psi\in X^*_1}|\psi(x)| \leqslant \|x\|.$$ On the other hand, $$\sup_{\psi\in X^*_1}|\psi(x)| \geqslant |\phi(x)| = \|x\|.$$ Therefore $$\sup_{\psi\in X^*_1}|\psi(x)| = \|x\|.$$

Then $$\|J(x)\| = \sup_{\psi\in X^*_1}|J(x)[\psi]| = \sup_{\psi\in X^*_1}|\psi(x)| = \|x\|.$$