Function whose graph is a cone

Question

I hope your day is going well.

Let $u : \Omega \rightarrow \mathbb{R}$ a convex function on a open convex bounded set.

I read "Let $v$ be the convex function whose graph is the cone with vertex $(x_{0},u(x_{0}))$ and the base $\Omega$ with $v =0$ on $\partial{\Omega}$. We have $v \in \mathcal{C}^{0}(\text{adh}(\Omega),\mathbb{R})$".

I'm sorry but this function exists ? And how does he know that it is continuous. I don't say it don't exist, of course. I trust the author but may anyone give me a expression of that function ? Because it is not quite obious for me. Maybe we can use convex conjuguate I don't know.

Thank you for your help !

Answer 1

An explicit expression for the function may be impossible to write down without describing the set $\Omega$ in more detail, which is why the definition of the function is given as it is. What you know about $v$ is:

1. $v$ is a convex function, so given $x,y\in {\cal C}$ and $\lambda \in [0,1]$ we have $v(\lambda x + (1-\lambda )y) \leq \lambda v(x) + (1-\lambda)v(y)$.
2. $v=0$ on $\partial \Omega$
3. $v\in C^0(\mathrm{adh}(\Omega), {\mathbb R})$

and what you know about the cone ${\cal C}$ is that it has a vertex at $(x_0, u(x_0))$ and base $\Omega$. So, if you take any point on $\Omega$ and connect it with a straight line to the vertex $x_0, u(x_0))$ then than line is part of the cone ${\cal C}$. A straight line can be mapped to an interval (say $[0,1]$ for the sake of argument) and intervals are convex, so you know that $v$ exists at every point of the domain ${\cal C}$.

So you have enough information to know that such a function $v$ exists, and if you need to calculate explicit values you'll need to define $\Omega$ explicitly first. If you do, then the convexity of $v$ and the boundary conditions will constrain what expression $v$ can have, but you may still need to make some choices (of constants or additional constraints) to find a single $v$.

You haven't provided any reference to where this comes from, but this way of starting things is common when the aim of the analysis is to understand what classes of functions $v$ are admissible for a given domain. As you continue reading you may find that the author narrows down the type of function $v$ can be further.

Lastly: for this kind of thing, try sketching the domain for simple (and complicated) $\Omega$ to help build intuition. If you let $\Omega$ be the boundary of the unit ball in ${\mathbb R}^2$ first and draw ${\cal C}$ it should be much easier to see how such a convex $v$ could be found.