A couple weeks ago I wrote about the Weierstrass approximation theorem, the theorem that says every continuous function on a closed finite interval can be approximated as closely as you like by a polynomial.
The post mentioned above uses a proof by Bernstein. And in that post I used the absolute value function as an example. Not only is |x| an example, you could go the other way around and use it as a step in the proof. That is, there is a proof of the Weierstrass approximation theorem that starts by proving the special case of |x| then use that result to build a proof for the general case.
There have been many proofs of Weierstrass’ theorem, and recently I ran across a proof due to Lebesgue. Here I’ll show how Lebesgue constructed a sequence of polynomials approximating |x|. It’s like pulling a rabbit out of a hat.
The staring point is the binomial theorem. If x and y are real numbers with |x| > |y| and r is any real number, then
Now apply the theorem substituting 1 for x and x² – 1 for y above and you get
The partial sums of the right hand side are a sequence of polynomials converging to |x| on the interval [-1, 1].
If you’re puzzled by the binomial coefficient with a top number that isn’t a positive integer, see the general definition of binomial coefficients. The top number can even be complex, and indeed the binomial theorem holds for complex r.
You might also be puzzled by the binomial theorem being an infinite sum. Surely if r is a positive integer we should get the more familiar binomial theorem which is a finite sum. And indeed we do. The general definition of binomial coefficients insures that if r is a positive integer, all the binomial coefficients with k > r are zero.
2 thoughts on “Lebesgue’s proof of Weierstrass’ theorem”
As an introductory analysis student I found this quite helpful. The only problem I noticed was that the binomial theorem only proves that the power series equals what it’s supposed to on (-1,1), but you assert that it’s true on [-1,1]. It is true that you can prove this for x=+/-1 easily because (1^2-1)^k=0 for all k>0 so the power series at x=+/-1 reduces to nCr(1/2,0), but to realize that this equals 1 you have to know that Π(i for i from m to n)=1 when m>n, so that the 0th falling factorial and therefore the generalized binomial product choosing 0 of any real number is 1. I would suggest that you add this information to make your proof clear.
Actually I misunderstood this before. The binomial theorem proves that this series converges when the absolute value of y in (x+y)^r is less than 1, i.e. |x^2-1|<1, i.e. 0<x<sqrt(2). You still need to demonstrate that the series converges at x=0, where x^2-1=-1, for the proof to work.