it might be too simple but I couldnt show the second part
L linear real space
$P:L\rightarrow \Bbb R$ is called positive homogenious functional if for every $x\in L$ and $\alpha\ge 0$ , $P(\alpha x)=\alpha P(x)$
$P(x+y)=P(2\frac{(x+y)}2)=2P(\frac x2+\frac y2)\le2(\frac12P(x)+\frac 12P(y))=P(x)+P(y)$
so $P(x+y)\le P(x)+ P(y)$
(should'nt it be homogenious only? I mean for all $\alpha \in \Bbb R, P(\alpha x)=\alpha P(x) $)
edit: my second trying is:
$P(x+y)=-2\frac 12P(-x+(-y))$
$\frac 12P(-x+(-y))+\frac 12P(y)\ge \frac12 P(-x)$ (because it is convex function)
when we choose $-x-y=x_1$ and $y=x_2$
it follows that
$\frac 12P(x_1)+\frac 12P(x_2)\ge \frac12 P(x_1+x_2)$
so
$P(x+y)=-2\frac 12P(-x-y))\ge -2\Bigg(\frac 12 P(-x)+\frac 12 P(-y)\Bigg)=-\Bigg(P(-x)+P(-y)\Bigg)=P(x)+P(y)$
thus
$P(x+y)\ge P(x)+P(y)$ and $P(x+y)=P(x)+P(y)$
I'm not sure $-x-y=x_1$ and $y=x_2\in L $ is this correct?