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Let $p$ be a sublinear function on $X$. Show that for $x_0 \in X$ there exists a linear functional $F$ such that $F(x_0)=p(x_0)$ and $F(x) \leq p(x)$ for all $x \in X$.

So this clearly seems to be an problem requiring Hanh-Banach Theorem, I considered the subspace $Span[x_0]$ and if I can find a linear functional on this space satisfying these conditions then I can extend this function to all of $X$ by the theorem.

Finding such a functional is proving harder than expected, my first two guesses were that $p(x)$ would be a linear function on the span, but this doesn't seem to be the case when we consider $p(-x_0)$ since this value has to be negative (unless it's zero) but this can't be r since $p\geq0$. My second guess was to map $x \rightarrow \frac{p(x_0)}{x_0}x$ but then I cant guarantee that $F \leq p$. I don't think this question is supposed to be difficult so I anchor either wrong or I'm missing something fundamental.

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You can think of the function $p$ restricted to $Span\{x_0\}$ as a one-dimensional function $a \mapsto p(ax_0)$ that is convex and piecewise linear (with one knot at $a=0$), by sublinearity. Then it is easy to find a linear function $F$ on $Span\{x_0\}$ such that $F \le p$ and $F(x_0)=p(x_0)$, and then use Hahn-Banach to extend to $X$.


More detail: For $a>0$, $p(ax_0)=a p(x_0)$. For $a < 0$, $p(ax_0) = -ap(-x_0)$. So the one-dimensional function has slope $p(-x_0)$ on $(-\infty,0)$ and slope $-p(x_0)$ on $(0,\infty)$. Furthermore, $p(x_0) \ge -p(-x_0)$ by sublinearity. So choosing $F$ to satisfy $F(ax_0):= a p(x_0)$ for all $a \in \mathbb{R}$ works.

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  • $\begingroup$ This is a bit beyond me, why is it piecewise linear by sublinearity? How does this imply anything? If $a$ is negative, isn't nothing guaranteed since $p$ is only positive homogenous? $\endgroup$ Commented Mar 31, 2017 at 4:09

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