The Unapologetic Mathematician

Mathematics for the interested outsider

Differentiable Convex Functions

We showed that all convex functions are continuous. Now let’s assume that we’ve got one that’s differentiable too. Actually, this isn’t a very big imposition. It turns out that a result called Rademacher’s Theorem will tell us that any Lipschitz function is differentiable “almost everywhere”.

Okay, so what does differentiability mean? Remember our secant-slope function:

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

Differentiability says that as we shrink the interval \left[a,b\right] down to a single point c the function has a limit, and that limit is f'(c).

So now take a<b. We can pick a c between them and points x and y so that a<x<c<y<b. Now we compare slopes to find

s(\left[a,x\right])\leq s(\left[a,c\right])\leq s(\left[c,b\right])\leq s(\left[y,b\right])

so as we let x approach a and y approach b we find

f'(a)\leq s(\left[a,c\right])\leq s(\left[c,b\right])\leq f'(b)

And so the derivative of f must be nondecreasing.

Let’s look at the statement f'(a)\leq s(\left[a,x\right]) a little more closely. We can expand this out to say

\displaystyle f'(a)\leq\frac{f(x)-f(a)}{x-a}

which we can rewrite as f(a)+f'(x)(x-a)\leq f(x). That is, while the function lies below any of its secants it lies above any of its tangents. In particular, if we have a local minimum where f'(a)=0 then f(a)\leq f(x), and the point is also a global minimum.

If the derivative f'(x) is itself differentiable, then the differential mean-value theorem tells us that f''(x)\geq0 since f'(x) is nondecreasing. This leads us back to the second derivative test to distinguish maxima and minima, since a function is convex near a local minimum.

April 16, 2008 Posted by | Analysis, Calculus | 4 Comments

   

Follow

Get every new post delivered to your Inbox.

Join 366 other followers