Let $E$ be a normed space. If $p\colon E\longrightarrow \mathbb{R}$ is sublinear and continuous at $0$, then $p$ is continuous. The proof is pretty straightforward, one just has to observe that sublinearity implies $$ - p(y-x)\leq p(x)-p(y)\leq p(x-y). $$ I am interested in the case when $p$ is not defined on all of $E$, but only on a closed convex cone $C\subset E$. Is it still true that continuity at $0$ implies continuity?
The proof from above clearly fails in this case as $x-y$ and $y-x$ are not necessarily in $C$. For that reason I suspect there are counterexamples, yet I could not find any. I am particularly interested in the case $E=L^2(X,\mu)$, $C=L^1_+(X,\mu)$.
No, it's not true. Let $E = \mathbb{R}^2$, $C = [0, \infty) \times \mathbb{R}$, and $$f(x, y) = \begin{cases} |y| & \text{if } x = 0 \\ 0 & \text{otherwise} \end{cases}.$$ Then, $f$ is sublinear, continuous at $0$, but not on the other points of the line $\{0\} \times \mathbb{R}$.