# Modulus of convex functional is convex functional

I'm having a hard time trying to prove the following: show that if $$p$$ is a convex functional then $$|p|$$ is also a convex functional. Here a "convex functional" is a function $$p:X\to \mathbb{R}$$, where $$X$$ is a vector space, such that

1. For all $$x,y\in X$$ the triangle inequality $$p(x+y)\leqslant p(x)+p(y)$$ holds.
2. For every $$\lambda >0$$ the identity $$p(\lambda x)=\lambda p(x)$$ holds.

Its easy to see that $$|p(\lambda x)|=\lambda |p(x)|$$ for all $$\lambda >0$$, and if $$p(x+y)\geqslant 0$$ then $$|p(x+y)|\leqslant |p(x)|+|p(y)|$$. However I don't see an argument to complete the proof when $$p(x+y)<0$$. Can someone give me an argument to complete the proof, or in the case show that the statement to be proved is false (so the exercise would be wrong)?

P.S.: this is an exercise in page 149 of the book of functional analysis of Vladimir Kadets.

• The inequality $0=p(0)\le p(x)+p(-x)$ should help. If $p(x)<0,$ then $|p(x)|\le p(-x).$ Commented Jul 20 at 11:59
• @RyszardSzwarc Why is $p(0)=0$? Commented Jul 20 at 12:00
• @RyszardSzwarc I already used that, trying to find a proof, with no luck. I'm starting to suspect that the statement is false, and it is a typo in the book, instead of say "linear functional" it says "convex functional" Commented Jul 20 at 12:01
• @geetha290krm $p(0)=2p(0).$ Commented Jul 20 at 12:01
• Usually, such functions $X\to\mathbb R$ are called sublinear. I find it quite strange to redefine convexity. Commented Jul 20 at 15:10

If there is no mistake below, the statement to be proved is false: let $$h:\mathbb{R}^2 \to \mathbb{R},\, (x,y)\mapsto x+|y|$$, then $$h$$ is a convex functional in $$\mathbb{R}^2$$. However $$-2=h(-2,0)\leqslant h(-1,-1/2)+h(-1,1/2)=-\frac1{2}-\frac1{2}=-1$$ so it cannot be the case that $$|h(-2,0)|\leqslant |h(-1,-1/2)|+|h(-1,1/2)|$$.∎