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?
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 $-x$.
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 $\langle x_0\rangle$.