The Unapologetic Mathematician

Mathematics for the interested outsider

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<x_2 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'<a and b<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 John Armstrong | Analysis, Calculus | | No Comments