Prove that there exist linear functionals $L_1, L_2$ on $X$

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?

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Note that specifying a pair of linear functionals on $X$ is the same as specifying a single linear functional $X\oplus X$, by writing $M(x\oplus y)= L_1(x)\oplus L_2(y)$. Also, if $p,q$ are sublinear functionals on $X$ then $r(x\oplus y)=p(x)+q(y)$ defines a sublinear functional on $X\oplus X$.

Consider now applying Hahn-Banach to extend the functional $M$ on the diagonal of $X\oplus X$ given by $M(x\oplus x)=L(x)$ to all of $X\oplus X$.


Update: Using the Hahn-Banach theorem as suggested above is only enough to obtain linear functionals $L_1,L_2$ satisfying $L_1(x)\le p(x)$ and $L_2(x)\le q(x)$. Note the absence of the absolute value on the left hand side of these inequalities. Once you put the absolute values in, then the result no longer holds in general, as shown by Mizar’s counterexample. If, however, we restrict to the case where $p(-x)=p(x)$ and $q(-x)=q(x)$ then the inequalities remain true when the absolute values are introduced, as can be seen by applying them to both $x$ and $-x$.

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 $|L_2(x)|\le q(x)$.

As you wish only a hint, here it is. Let $X_0=\ker L$ and $x_0\in X\setminus X_0$ be an arbitrary element. Then each element $x\in X$ has a unique representation $x=x’+\lambda_x x_0$, where $x’\in X_0$ and $\lambda_x\in\Bbb R$. Then $L(x)=L(\lambda_x x_0)\le p(\lambda_x x_0)+q(\lambda_x x_0)$. So the problem can be reduced to one-dimensional case and a space $\langle x_0\rangle$.