madhadron

Measuring pointiness of functions

Status: Done

Naively, it seems unlikely that an approximation by smooth functions like polynomials or complex exponentials can reproduce points and jags in the original function. As it turns out, it can, if the points aren’t too pointy. How do we measure “pointiness?”

Consider the family of functions fk(x)=1|2x1|1/kf_{k}(x)=1-|2x-1|^{1/k}. The first few of these look like:

Point functions

At x=12x=\frac{1}{2}, these functions are obviously not differentiable, that is

limε0fk(x+ε)f(x)ε\lim_{\varepsilon\rightarrow0}\frac{f_{k}(x+\varepsilon)-f(x)}{\varepsilon}

does not exist. The difference in the numerator doesn’t shrink fast enough to keep up with ε\varepsilon in the denominator. But what if we slow down the denominator? The easiest way is to raise ε\varepsilon to some power less than 1. So take

limε0fk(x+ε)f(x)εα\lim_{\varepsilon\rightarrow0}\frac{f_{k}(x+\varepsilon)-f(x)}{\varepsilon^{\alpha}}

where 0α10\leq\alpha\leq 1. If we put in a sufficiently small α\alpha, then the limit may exist.

The fkf_{k}’s are smnooth everywhere except at x=12x=\frac{1}{2}. Consider our new, slowed limit there.

limε0fk(12+ε)fk(12)εα=limε0(2ε)1/kεα\lim_{\varepsilon\rightarrow0}\frac{f_{k}(\frac{1}{2}+\varepsilon)-f_{k}(\frac{1}{2})}{\varepsilon^{\alpha}} = \lim_{\varepsilon\rightarrow0}\frac{(2\varepsilon)^{1/k}}{\varepsilon^{\alpha}}

which exists when α<1k\alpha < \frac{1}{k}. To see more clearly what this means, consider the box [x,x+ε]×[f(x),f(x+ε)][x,x+\varepsilon]\times[f(x),f(x+\varepsilon)]:

Shrinking boxes on a function

How much must we change ε\varepsilon to half the height of the box? If the function is differentiable, the answer (eventually, for a sufficiently small box) is to halve the horizontal dimensions of the box as well, since all differentiable can be well approximated as linear on sufficiently small regions. For our fkf_{k}, we have to shrink ε\varepsilon faster and faster as kk increases.

More generally, a function is said to be α\alpha-Hölder continuous if α\alpha is the largest exponent in the denominator for which the limit above exists on the whole interval under consideration. The fkf_{k}’s provide a family of examples illustrating α\alpha’s throughout the admissable range. α=0\alpha=0 is discontinuous, and corresponds to ff_{\infty}, which is a Dirac δ\delta function. We can’t make the limit exist. Differentiable functions have α=1\alpha=1.

This mostly comes up when you’re looking at Fourier analysis, where the theorem I learned is that partial Fourier sums converge at least where α12\alpha \geq \frac{1}{2}. For rougher creatures, the series may behave itself or not.