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I'm currently learning about dual spaces. There's a small step in a proof about "abundance" of elements in the dual space, that I don't get:

Lemma. Consider a normed space $(X, ||\cdot||)$ and its dual space $X^*$. For every $x \in X$ there is a $x^* \in X^*$ with $${\langle x^*,x\rangle}_{{X^*} \times X} = {||x||}^2_X = {||x^*||}^2_{X^*}.$$

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Proof. Consider $M=span\{x\}$. Set $f(tx)=t{||x||}^2_X$, with $\, f \in L(M,\mathbb R)$, $\, t \in \mathbb R$. $$ {||f||}_{L(M,\mathbb R)} = \sup_{{||tx||}_X \leq 1} |f(tx)| = {||x||}_X $$ By Hahn-Banach's Thm. we can extend $f$ to $x^* \in L(M,\mathbb R)$. Then we have $$ {||x^*||}_{X^*} = {||f||}_{L(M,\mathbb R)} = {||x||}_X, \quad {\langle x^*,x\rangle}_{{X^*}\times X} = f(x) = {||x||}_X^2 .$$

Now my humble question, is how to see this equality: $$\sup_{{||tx||}_X \leq 1} |f(tx)| = {||x||}_X $$

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Because $$ \sup_{\|t\,x\|\le 1} |f(tx)| = \sup_{\|t\,x\|\le 1} t \,\|x\|^2 = \|x\| \, \sup_{\|t\,x\| \le 1} \|t\,x\| = \|x\|.$$ Note that the supremum ranges over $t$ while $x$ is fixed.

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