Convexity of a homogeneous bivariate function and inequality by Beurling I am currently trying to understand the proof of Theorem 1 in this article (On the uniform convexity of $L^p$ and $l^p$ spaces). I am only interested in the left hand side of the inequality (the right hand side corresponds to Clarkson's inequality)
The author states that when $1<p<2$, for $u,v\geq 0$ $$ \xi(u,v) = (u^{\frac 1p}+ v^{\frac 1p})^p + |u^{\frac 1p}-v^{\frac 1p}|^p $$ is convex. However, I don't understand the following argument that is made in the paper: $\xi(0,0)=0$, $\xi(u,v)=\xi(v,u)$ and $\xi(tu,tv)=t\xi(u,v)$ for $t>0$ therefore convexity of $\xi(u,1)$ is sufficient.
The author of this question mentions in a comment the same paper, and a comment responded that the result is "visually trivial".
Trying to blindly apply convexity inequalities doesn't yield much: $$\xi(t(x,y)+(1-t)(u,v))\leq t(t\xi(x,y)+(1-t)\xi(x,v))+(1-t)(t\xi(u,y)+(1-t)\xi(u,v))$$
How can the statement be proven ?
 A: Suppose $u,v>0$, note that $\xi(u,v) = v \xi({u \over v},1)$.
Then ${\partial \xi(u,v) \over \partial u} = {\partial \xi ( {u \over v},1) \over \partial u}$,
${\partial \xi(u,v) \over \partial v} = \xi({u \over v},1)- {u \over v} {\partial \xi ( {u \over v},1) \over \partial u}$,
${\partial^2 \xi(u,v) \over \partial u^2} = {1 \over v} {\partial^2 \xi ( {u \over v},1) \over \partial u^2}$,
${\partial^2 \xi(u,v) \over \partial v \partial u} = -{u \over v^2} {\partial^2 \xi ( {u \over v},1) \over \partial u^2}$,
${\partial^2 \xi(u,v) \over \partial v^2} = -{u \over v^2} {\partial \xi ( {u \over v},1) \over \partial u} + {u \over v^2} {\partial \xi ( {u \over v},1) \over \partial u} + {u^2 \over v^3} {\partial^2 \xi ( {u \over v},1) \over \partial u^2} = {u^2 \over v^3} {\partial^2 \xi ( {u \over v},1) \over \partial u^2} $.
Hence
$H_\xi(u,v) = {1 \over v^3} {\partial^2 \xi ( {u \over v},1) \over \partial u^2} \begin{bmatrix} v^2 & -uv \\ -uv & u^2
\end{bmatrix}$, and if ${\partial^2 \xi ( {u \over v},1) \over \partial u^2}  \ge 0$ then $H_\xi(u,v) \ge 0 $ (as in positive semidefinite).
Convexity on $u,v \ge 0$ follows from continuity.
