The Unapologetic Mathematician

Convex Functions are Continuous

Yesterday we defined a function $f$ defined on an open interval $I$ to be “convex” if its graph lies below all of its secants. That is, given any $x_1 in $I$, for any point $x\in\left[x_1,x_2\right]$ we have

$\displaystyle f(x)\leq f(x_1)+\frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)$

which we can rewrite as

$\displaystyle\frac{f(x)-f(x_1)}{x-x_1}\leq\frac{f(x_2)-f(x_1)}{x_2-x_1}$

or (with a bit more effort) as

$\displaystyle\frac{f(x_2)-f(x_1)}{x_2-x_1}\leq\frac{f(x_2)-f(x)}{x_2-x}$

That is, the slope of the secant above $\left[x_1,x\right]$ is less than that above $\left[x_1,x_2\right]$, which is less than that above $\left[x,x_2\right]$. Here’s a graph to illustrate what I mean:

The slope of the red line segment is less than that of the green, which is less than that of the blue.

In fact, we can push this a bit further. Let $s$ be the function with takes a subinterval $\left[a,b\right]\subseteq I$ and gives back the slope of the secant over that subinterval:

$\displaystyle s(\left[a,b\right])=\frac{f(b)-f(a)}{b-a}$

Now if $\left[x_1,x_2\right]$ and $\left[x_3,x_4\right]$ are two subintervals of $I$ with $x_1\leq x_3$ and $x_2\leq x_4$ then we find

$s(\left[x_1,x_2\right])\leq s(\left[x_1,x_4\right])\leq s(\left[x_3,x_4\right])$

by using the above restatements of the convexity property. Roughly, as we move to the right our secants get steeper.

If $\left[a,b\right]$ is a subinterval of $I$, I claim that we can find a constant $C$ such that $\left|s(\left[x_1,x_2\right])\right|\leq C$ for all $\left[x_1,x_2\right]\subseteq\left[a,b\right]$. Indeed, since $I$ is open we can find points $a'$ and $b'$ in $I$ with $a' and $b. Then since secants get steeper we find that

$s(\left[a',a\right])\leq s(\left[x_1,x_2\right])\leq s(\left[b,b'\right])$

giving us the bound we need. This tells us that within $\left[a,b\right]$ we have $|f(x_2)-f(x_1)|\leq C|x_2-x_1|$ (the technical term here is that $f$ is “Lipschitz”, which is what Mr. Livshits kept blowing up about), and it’s straightforward from here to show that $f$ must be uniformly continuous on $\left[a,b\right]$, and thus continuous everywhere in $I$ (but maybe not uniformly so!)

April 15, 2008 - Posted by | Analysis, Calculus