# Exercise in Hahn-Banach Theorem; Finding linear functional $-p(-x)\leq f(x)\leq p(x)$

(The following exercises are in Kreyszig's book 218 page; EXE 10) I want to solve the following exercise : If $$X=l^\infty$$, let $$p(x)=\lim\sup x_i$$, which is sublinear. Then find a linear functional $$f(x)$$ s.t. $$-p(-x)\leq f(x) \leq p(x)$$

Background : If $$p$$ is a sublinear function on a real vector space, i.e., $$p(x+y)\leq p(x) + p(y),\ p(cx)=cp(x),\ c\geq 0$$

then there exists linear functional $$f$$ s.t. $$-p(-x)\leq f(x) \leq p(x)$$

Proof : By Hahn-Banach theorem we have $$f(x)\leq p(x)$$ so that $$f(-x)\leq p(-x)$$

That is $$-p(-x)\leq f(x) \leq f(x)$$

Now, we return to original question : $$X=l^\infty$$, let $$p(x)=\limsup x_i$$. So $$-p(-x)=-\limsup (-x_i)=\liminf x_i$$ Hence if such functional $$f$$ exists, then $$f(x)=\lim x_i$$ when $$\lim x_i$$ exists. If $$x_{2i+1}=2,\ x_{2i}=1$$ then $$v_{2i+1}=1,\ v_{2i}=2$$ then $$f(x+v)=3$$. That is, the problem is how determine $$f$$ on $$x$$ where $$\lim x_i$$ does not exist.

Try : If we let $$f(x)=\frac{\lim\inf x_i + \lim\sup x_i}{2}$$, then note that $$f(x+v)\neq f(x)+f(v)$$. Thank you in advance.

I think that the exercise is expecting you to do exactly what you did - to prove that there exists a functional with the given properties. (Not to explicitly write down such $$f$$.)
You can notice that in this way you get a functional $$f\in \ell_\infty^* \setminus \ell_1$$. (A few proofs of the fact that $$\ell_\infty^*\ne\ell_1$$ are also collected here: Dual of $$l^\infty$$ is not $$l^1$$.) Existence of such functional cannot be proved in ZF, so any proof has to use some non-constructive step at some point. Some posts where the fact that this is not provable in ZF is mentioned: Nonnegative linear functionals over $$l^\infty$$ and $$\ell^1$$ vs. continuous dual of $$\ell^{\infty}$$ in ZF+AD.