Suppose J is a convex and p-homogeneous (p>0) functional. That means $J(\lambda u +(1-\lambda)v) \leq \lambda J(u) + (1-\lambda) J(v)$ and $J(\lambda u)=|\lambda|^p J(u)$ for $u,v$ in the domain and $\lambda$ a real number.
I need to show $J(u+v)=J(u)$ for all $v \in \mathcal{N}(J)$ (the nullspace of $J$, i.e. $J(v)=0$). I have managed to prove this for $p\leq 1$ but am having trouble with the case $p>1$. Here is what I've done: ($p \leq 1$)
$J(u+v) = 2^{p} J(\frac{u}{2} +\frac{v}{2}) \leq 2^{p}(\frac{1}{2} J(u)+ \frac{1}{2} J(v))=2^{p-1} J(u)\leq J(u)$;
$J(u)=J(u+v-v)=2^p J(\frac{u+v}{2}-\frac{v}{2})\leq 2^{p} (\frac{1}{2} J(u+v)+ \frac{1}{2} J(-v))=2^{p-1} (J(u+v)+J(v))=2^{p-1} J(u+v) \leq J(u+v)$.
I've actually just figured this out.
Take any $\lambda \in (0,1)$, then
$J(u+v) = J(\lambda \frac{u}{\lambda} +(1-\lambda) \frac{v}{1-\lambda}) \leq \lambda J(\frac{u}{\lambda})+ (1-\lambda) J(\frac{v}{\lambda})=\lambda^{1-p} J(u) + (1-\lambda)^{1-p} J(v)= \lambda^{1-p} J(u)$.
Take the limit as $\lambda \rightarrow 1$ and we get $J(u+v) \leq J(u)$. The other direction is similar.