Show that $\left|\inf_{x\in A} f(x)-\inf_{x\in A} g(x)\right| \leq \sup_{x\in A} |f(x)-g(x)|$ Let $f,g:A\to \mathbb{R}$ be convex functions defined on a convex subset $A\subset\mathbb{R}$. How can I prove the following inequality:
$$\left|\inf_{x\in A} f(x)-\inf_{x\in A} g(x)\right| \leq \sup_{x\in A} |f(x)-g(x)| \text{ ?}$$
My plan was to look first at the case where $\inf_{x\in A} f(x)$ and $\inf_{x\in A} g(x)$ are attained, but even this case gives me trouble.
Any help on this is very appreciated.
 A: This doesn't really have much to do with convexity. I will assume the functions are bounded below.
Let $\epsilon > 0$ be given and start with a point $x_0 \in A$ where
$$g(x_0) < \inf_{x \in A} g(x) + \epsilon.$$ Then $$\inf_{x \in A} f(x) \le f(x_0) \le |f(x_0) - g(x_0)| + g(x_0) < \sup_{x \in A} |f(x) - g(x)| + \inf_{x \in A} g(x) + \epsilon$$ This gets you to
$$\inf_{x \in A} f(x) - \inf_{x \in A} g(x) < \sup_{x \in A} |f(x) - g(x)| + \epsilon$$ but with $x_0$ no longer in the picture you can let $\epsilon \to 0^+$ to discover
$$\inf_{x \in A} f(x) - \inf_{x \in A} g(x) \le \sup_{x \in A} |f(x) - g(x)|.$$
Now do the same thing starting with $f$ rather than $g$.
A: In the case where the two infima are attained, then they are both attained either at the same point in the domain, or at two different points.
In the former case you have
\begin{align}
& \left| \inf_{x\,\in\,A} f(x) - \inf_{x\,\in\,A} g(x) \right| \\[8pt]
= {} & |f(x_0)-g(x_0)| \\
& \text{where $x_0$ is the point where the infima are attained} \\[8pt]
\le {} & \sup_{x\,\in\,A}|f(x)-g(x)|.
\end{align}
In the other case you have
\begin{align}
& \left| \inf_{x\,\in\,A} f(x) - \inf_{x\,\in\,A} g(x) \right| \\[8pt]
= {} & |f(x_0)-g(x_1)| \\
& \text{where $x_0,x_1$ are the points where} \\
& \text{the respective infima are attained.} \\ & \text{No generality is lost be assuming} \\
& f(x_0)\le g(x_1). \text{ Then we have:} \\[8pt]
\le {} & |f(x_0)-g(x_0)| \\[8pt]
\le {} & \sup_{x\,\in\,A}|f(x)-g(x)|.
\end{align}
