Let $(X,||.||)$ be an normed vector space and $X^*$ its dual. In our lecture we recently stated that $$||x||=\sup_{\lambda \in X^*,||\lambda|| \leq 1} |\lambda(x)| $$ moreover , we stated that there is always such a $\lambda \in X^*$ with $||x||=\lambda(x)$, so the supremum is taken into place by some $\lambda$.
On the other hand for the definition of $||\lambda|| =\sup_{x \in X, ||x|| \leq 1}|\lambda(x)|$ there is not necessarily a $x \in X$ that takes shape of the supremum.
Does this mean in some way, that the dual space $X^*$ of $X$ is greater or "richer" than $X$ in some way? I still have some struggles to get a good intuition of the dual space.
2026-03-28 21:50:31.1774734631
Concerning linear functionals on normed spaces
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Concerning your question there are some quite useful lemmata and comments.
Because $||i(x)||= ||x||$ holds, $i$ is an isometric (and clearly linear) map. So $i(X)$ is a subspace of $X^{**}$; By identifying any $x \in X$ with the function $x^{**}: X^* \rightarrow \mathbb{K}$ that plugs this very $x$ in every $x^* \in X^*$, you define an embedding of $X$ in $X^{**}$. Additionally, if $i$ is surjective, you call X an reflexive space and $X \cong X^{**}$.