Yesterday we defined a function defined on an open interval to be “convex” if its graph lies below all of its secants. That is, given any in , for any point we have
which we can rewrite as
or (with a bit more effort) as
That is, the slope of the secant above is less than that above , which is less than that above . 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 be the function with takes a subinterval and gives back the slope of the secant over that subinterval:
Now if and are two subintervals of with and then we find
by using the above restatements of the convexity property. Roughly, as we move to the right our secants get steeper.
If is a subinterval of , I claim that we can find a constant such that for all . Indeed, since is open we can find points and in with and . Then since secants get steeper we find that
giving us the bound we need. This tells us that within we have (the technical term here is that is “Lipschitz”, which is what Mr. Livshits kept blowing up about), and it’s straightforward from here to show that must be uniformly continuous on , and thus continuous everywhere in (but maybe not uniformly so!)