I'm working with continuous, increasing, convex functions (call the set $\mathcal{U}$) and viewing them as a subspace of $\mathcal{C}=\mathcal{C}([0,1])$ (perhaps this is not the best topology, recommendations welcome). Since $\mathcal{C}$ is Polish I sense that $\mathcal{U}$ is as well. I have a proof sketch (follows) and I hope someone can either give me a reference, or tell me I'm wrong, before I get too far into it.
Sketch: These functions can be recovered by their subdifferentials, which perhaps can be viewed as cadlag functions since each point will have a closed, convex interval of subderivatives. So, if I can find a homeomorphism between $\mathcal{U}$ and the space of cadlag functions with the Skorokhod topology, I get my answer by showing that non-decreasing functions form a closed subset in the Skorokhod topology (hopefully true).
I'm aware that cadlag with Skorokhod is not a TVS, which seems to rule out finding an isomorphism, but it doesn't seem essential that the algebraic structure is mapped . . .
Thanks for any help.
New Answer: As a closed subset of a Polish space, $\mathcal{U}$ is also Polish. Note that the limit of a sequence of convex, nondecreasing functions will also be non-decreasing and (weakly) convex.
Aside: Just trying to tie this off. I don't want to leave it hanging as unanswered.
Old Answer: Obvious: $\mathcal{U}$ is closed.
Whoops. Thanks to "freakish" in comments above.