Let $X$ be a non-empty set and $f:X \to \mathbb R \cup \{\pm\infty\}$. Then its domain is defined as $D(f) := \{x\in E : f(x) \neq \pm\infty\}$. Let $X,Y$ be vector spaces and $X \times Y$ their direct product which is a vector space with induced component-wise operation.
I'm trying to prove below result, i.e.,
Theorem Let $F:X \times Y \to \mathbb R \cup \{+\infty\}$ and $f:X \to \mathbb R \cup \{+\infty\}$ such that $f(x) := \inf_{y\in Y}F(x, y)$. If $F$ is convex, then so is $f$.
Could you have a check on my attempt?
My attempt Let $x_1, x_2 \in D(f)$ and $t \in (0, 1)$. Fix $\varepsilon>0$. There are $y_1, y_2 \in Y$ such that $F(x_1, y_1) < f(x_1) + \varepsilon$ and $F(x_2, y_2) < f(x_2) + \varepsilon$. We have $$ \begin{align} f(tx_1 + (1-t)x_2) &\le F(tx_1 + (1-t)x_2, ty_1 + (1-t)y_2) \\ &\overset{(\star)}{\le} t F(x_1, y_1)+(1-t) F(x_2, y_2) \\ &\le t(f(x_1) + \varepsilon) + (1-t)(f(x_2) + \varepsilon) \\ &= tf(x_1)+(1-t)f(x_2) +\varepsilon. \end{align} $$
Here $(\star)$ is due to the convexity of $F$. We take the limit $\varepsilon \downarrow 0$ and get $$ f(tx_1 + (1-t)x_2) \le tf(x_1)+(1-t)f(x_2). $$
This completes the proof.