If $f$ is convex, is $\sup_x f(x,y)$ convex?

Question

Suppose that $f(x,y)$ convex. Let $g(y) = \sup_x f(x,y)$. Then is $g$ convex?

I can't find a counter example but I know that this holds true if we substitute $\inf$ for $\sup$, so I feel that it should not hold true.

Answer 1

Let $f:D\subset \mathbb{R}^{2}\rightarrow \mathbb{R}$ be convex on $D$ and assume there exists $c=c(y)\in \mathbb{R}$ $$f(x,y)\leq c(y)$$ for a fixed all $(x,y)\in D$.

Now fix $(x,b)\in D\cap \big\{(x,y)\in D: y=b\big\},\;$ $\lambda \in [0,1].\;$ Since $f$ is convex on $D$ then $$ f( (1-\lambda) (x,y)+\lambda (x,b)) \leq (1-\lambda) f(x,y)+\lambda f(x,b)) $$ Simply take the supremum over $x$ to get

$$ g( (1-\lambda) (y)+\lambda g(b)) \leq (1-\lambda) g(y)+\lambda g(b) $$ for every $y\in \mathbb{R}$ such that $(x,y)\in D$. Since the line $y=b$ was chosen arbitrarily, then $g$ is convex.