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(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.

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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.

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