# The Unapologetic Mathematician

## 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. We can pick a $c$ between them and points $x$ and $y$ so that $a. 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