Lebesgue’s proof of Weierstrass’ theorem

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

(x+y)^r & =\sum_{k=0}^\infty {r \choose k} x^{r-k} y^k.

Now apply the theorem substituting 1 for x and x┬▓ – 1 for y above and you get

|x| = (1 + (x^2 - 1))^{1/2} =\sum_{k=0}^\infty {1/2 \choose k} (x^2-1)^k

The partial sums of the right hand side are a sequence of polynomials converging to |x| on the interval [-1, 1].

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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.

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