Interpretation of a set characterized by epigraph of a convex function

Question

This may sound like a silly question but somehow I have been struggling with this for hours. Let $f: X\times Y\to \mathbb{R}$ be a convex function and let $C$ be a convex subset of $Y$, where $X$ and $Y$ are vector spaces. Then, regarding the following set, $$ E = \{(x, t)\in X\times \mathbb{R}: f(x, y)\leqslant t, \;\text{for some}\;y\in C\}, $$ can we interpret $E$ as the projection of a convex set $F$ onto its first and third elements? If so, is it that $$ F = \{(x, y, t)\in X\times Y\times \mathbb{R}:f(x, y)\leqslant t,\; y\in C\} = \mathrm{epi\;}f\cap (X\times C\times \mathbb{R})? $$ Thanks for any clarification!

Answer 1

Essentially yes. If we define $E$ slightly differently: $$E = \{(x, 0, t)\in X\times Y \times \mathbb{R}: f(x, y)\leqslant t, \;\text{for some}\;y\in C\},$$ then $E$ is the image of $F$ under the (linear) projection map $P : X \times Y \times \Bbb{R} \to X \times Y \times \Bbb{R}$ onto $X \times \{0\} \times \Bbb{R}$. If we identify $X \times \{0\} \times \Bbb{R}$ with $X \times \Bbb{R}$, then we get the formulation of $E$ you've written down.

Proving that $P(F) = E$ is quite straightforward. First, suppose that $(x, y, t) \in P(F)$. Then there exists some $(x', y', t') \in F$ such that $P(x',y',t') = (x', 0, t') = (x, y, t)$. Since $(x', y', t') \in F$, we get $f(x', y') \le t$ and $y' \in C$. Thus, $(x, y, t) = (x', 0, y')$ belongs to $E$, by definition of $E$, and so $P(F) \subseteq E$.

Now, suppose that $(x, y, t) \in E$. By definition of $E$, we have $y = 0$ and there exists some $y' \in C$ such that $f(x, y') \le t$. This says, by definition of $F$, that $(x, y', t) \in F$. We also can see that $P(x, y', t) = (x, 0, t) = (x, y, t)$, hence $(x, y, t) \in P(F)$, completing the proof of the set equality.