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Let $X$ be a real or complex vector space, and let $p \colon X \to \mathbb{R}$ be a real-valued function satisfying $$p(x+y) \leq p(x) + p(y) \ \mbox{ for all } \ x, y \in X$$ and $$p(\alpha x) = \vert \alpha \vert p(x) \ \mbox{ for all } \ x \in X, \ \mbox{ and for all scalars } \ \alpha.$$ Let $x_0$ be a point of $X$. Then how to show that there is a linear functional $\tilde{f}$ on $X$ such that $\tilde{f}(x_0) = p(x_0)$ and $\vert \tilde{f}(x) \vert \leq p(x)$ for all $x \in X$.

My effort:

First of all, here is the (generalized) Hahn Banach Theorem for (real or complex) vector spaces.

Let $X$ be a real or complex vector space; let $Z$ be a subspace of $X$; let $p \colon X \to \mathbb{R}$ satisfy $$p(x+y) \leq p(x) + p(y) \ \mbox{ for all } \ x, y \in X$$ and $$p(\alpha x ) = \vert \alpha \vert p(x) \ \mbox{ for all } \ x \in X \ \mbox{ and for all scalars } \ \alpha;$$ and let $f$ be a linear functional defined on $Z$ such that $$\vert f(x) \vert \leq p(x) \ \mbox{ for all } \ x \in Z.$$ Then there is a linear functional $\tilde{f}$ defined on all of $X$ such that $$\tilde{f}(x) = f(x) \ \mbox{ for all } \ x \in Z$$ and $$\vert \tilde{f}(x) \vert \leq p(x) \ \mbox{ for all } \ x \in X.$$

This is Theorem 4.3-1 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig.

Let $Z$ be the subspace of $X$ spanned by $x_0$, and let the functional $f$ be defined on $Z$ by $$f(\alpha x_0) = \alpha p(x_0) \ \mbox{ for all scalars } \ \alpha .$$

We can show that $p(x) \geq 0$ for all $x \in X$, and $f$ satisfies all the conditions in the hypothesis of the above theorem. So there is a linear functional $\tilde{f}$ on $X$ such that $\tilde{f}(x) = f(x)$ for all $x \in Z$ and $\vert \tilde{f}(x) \vert \leq p(x)$ for all $x \in X$.

Thus, $\tilde{f}(\alpha x_0) = \alpha p(x_0)$ for all scalars $\alpha$ and hence $\tilde{f}(x_0) = p(x_0)$.

Is this solution correct?

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