TITLE: Prove that there exist linear functionals $L_1, L_2$ on $X$
QUESTION [11 upvotes]: Let $X$ be a linear space, $p, q$ sublinear functionals on $X$, and $L$ a linear functional on $X$ such that $|L(x)| ≤ p(x) + q(x),$ for all $x ∈ X$. Prove that there exist linear functionals $L_1, L_2$ on $X$ such that $L(x) = L_1(x) + L_2(x),$ and $|L_1(x)| ≤ p(x), |L_2(x)| ≤ q(x),$ for all $x ∈ X.$

My Work:
First I thought to use Hahn Banach Theorem. But since there is no known subspace it was useless. Then I tried to make $L(x)$ as $L(x)=\frac{L(x+\lambda)+L(x-\lambda)}{2}$ for some scalar $\lambda$ but failed to find suitable $L_1$ and $L_2$. I think this problem is little bit tricky. I want to try it myself and I only need a hint to start. Can somebody please give me a hint?

REPLY [2 votes]: In full generality, this is false: take $X=\mathbb{R}$, $L(x)=x$,
$p(x)=x^+=\max\{x,0\}$ and $q(x)=x^-=p(-x)$. In this case $p(x)+q(x)=|x|$,
so $|L|\le p+q$. But $L_1$ should satisfy $|L_1(x)|\le x^+$, so
$L_1(x)=0$ if $x\le 0$, i.e. $L_1\equiv 0$, and in the same way $L_2\equiv 0$, contradicting $L=L_1+L_2$.
I will prove that $L_1$ and $L_2$ exist assuming that $p(-x)=p(x)$ and $q(-x)=q(x)$ (i.e. $p$ and $q$ are seminorms).
We can endow $X$ with a norm such that $p,q$ are continuous: set $V:=\{x\in X:p(x)=q(x)=0\}$ (which is a linear subspace)
and if $V$ is bigger than $\{0\}$ we can solve the problem in the quotient $X/V$, so wlog we may assume $V=\{0\}$ (tell me if this is not clear). Now $\|x\|:=p(x)+q(x)$ defines a norm on $X$ and $p,q$ are continuous wrt this norm.
Call $Y:=X\times \mathbb{R}$ and $C:=\{(x,t)\in Y:t>p(x)\}$, $D:=\{(x,t)\in Y:t<L(x)-q(x)\}$. Since $p$ is continuous and convex (why?), its epigraph $C$ is open and convex. In the same way (since $L-q$ is continuous and concave) $D$ is open and convex.
By the first geometric form of Hahn-Banach, there exists a linear functional $f:Y\to\mathbb{R}$ such that $f(y)>f(y')$ for any $y\in C$, $y'\in D$.
Write $f(x,t)=g(x)+\alpha t$ and observe that $\alpha>0$: in fact, since $(x,p(x)+1)\in C$ and $(x,L(x)-q(x)-1)\in D$, we have
$$ g(x)+\alpha \Big(p(x)+1\Big)>g(x)+\alpha\Big(L(x)-q(x)-1\Big), $$
which gives $\alpha>0$ as $p(x)+1>L(x)-q(x)-1$.
Set $\gamma:=\inf_{y\in C}f(y)\ge \sup_{y'\in D}f(y')$. Since
$p(0)=0$ we have $(0,\epsilon)\in C$ for any $\epsilon>0$, so we get $\gamma=0$, thus $f(x,t)=g(x)+\alpha t>0$ for any $(x,t)\in C$. This gives $g(x)+\alpha p(x)\ge 0$, i.e.
$$ -\frac{g}{\alpha}\le p. $$
In the same way
$$ -\frac{g}{\alpha}\ge L-q. $$
Finally put $L_1:=-\frac{g}{\alpha}$ and $L_2:=L-L_1$.
We have $L_1(x)\le p(x)$ and $L_2(x)\le q(x)$ for any $x\in X$.
Applying this with $-x$ and using our additional assumption we obtain
$-L_1(x)=L_1(-x)\le p(-x)=p(x)$ and similarly $-L_2(x)\le q(x)$.
Thus $|L_1(x)|\le p(x)$ and $|L_2(x)|\le q(x)$.