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?

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


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