# Existence of a linear functional. Hahn-Banach.

Given a sublinear functional $$p$$ in a real vector space $$X$$, show that there exists a linear functional $$f$$ in $$X$$ such that $$-p(-x)\leq f(x)\leq p(x)$$.

I am trying to use the Hahn-Banach theorem taking the vector subspace $$G_{x_0} = \{ tx_0 : t\in \mathbb{R}\}$$ for any $$x_0\in X$$ and properties of sublinear functional ($$p(\lambda x) = \lambda p(x)$$ for $$\lambda \geq 0$$ and $$p(x+y)\leq p(x) + p(y)$$ for every $$x,y\in X$$) to find a relation between $$g: G_{x_0} \to \mathbb{R}$$ and $$p$$.

Any hint?

• How are we supposed to know what you're doing wrong if you don't show us what you're doing? – Bananach Apr 28 at 12:53
• This is a lemma usually used in the proof of Hahn-Banach. – Lord Shark the Unknown Apr 28 at 13:17
• @Bananach telling me what would you do to prove it, please. The computations are not useful. – Rubén Fernández Fuertes Apr 28 at 14:19
• @LordSharktheUnknown Which lemma? The proof I have does not use it. In fact, I wanna use Hahn Banach to prove it. – Rubén Fernández Fuertes Apr 28 at 14:20

Notice that you can apply Hahn-Banach to a functional defined on the trivial subspace $$Y =\{0\}$$ of $$X$$. There is exactly one such functional given by $$f(0) = 0$$. We must have $$p(0) = 0$$ by positive homogeneity so $$f(x) \leq p(x)$$ for all $$x \in Y$$.
Then Hahn-Banach yields an extension of $$f$$ to a functional on all of $$X$$ satisfying $$f(x) \leq p(x)$$ for all $$x \in X$$. Hence $$f(-x) \leq p(-x)$$ for all $$x$$, which implies $$f(x) \geq -p(-x)$$ for all $$x$$.