Is the $\sup$ and $\inf$ of two convex functions also convex?

Question

Let $f$ and $g$ be two convex functions. I’m supposed to give informations on $\sup(f,g)$ and $\inf(f,g)$ but I have no idea what that means.

Could anyone enlighten me please?

Answer 1

Let $x,y\in\mathbb R$ and $t\in[0,1]$. We know that $$t\sup(f,g)(x)+(1-t)\sup(f,g)(y)=t\max(f(x),g(x))+(1-t)\max(f(y),g(y)).$$ But observe that the right-hand side is greater than or equal to $tf(x)+(1-t)f(y)\ge f(tx+(1-t)y)$. Similarly, it is greater than or equal to $g(tx+(1-t)y)$. Thus it is at least the maximum of those two expressions. In other words, we have that \begin{align*}t\sup(f,g)(x)+(1-t)\sup(f,g)(y)&\ge\max(f(tx+(1-t)y),g(tx+(1-t)y))\\&=\sup(f,g)(tx+(1-t)y)).\end{align*} Thus $\sup(f,g)$ is convex.

Regarding $\inf(f,g)$, we cannot say that is is either convex or concave. For instance, if $f(x)<g(x)$ for all $x$, clearly, $\inf(f,g)$ is convex. On the other hand, consider $f(x)=x^2$ and $g(x)=x^4$. Both are convex, but taking $x=0.9$, $y=1.1$, and $t=\frac12$ gives us that $$t\inf(f,g)(x)+(1-t)\inf(f,g)(y)=0.93305<1=\inf(f,g)(tx+(1-t)y).$$ So in this case $\inf(f,g)$ isn't convex. (In fact, it isn't concave either; just take $x,y\ge1$.)