Convergence of orthogonal polynomial expansions - MathOverflow

Convergence of orthogonal polynomial expansions

2010-06-24T14:41:45Z

"Everyone" knows that for a general $f\in L^2[0,1]$, the Fourier series of $f$ converges to $f$ in the $L^2$ norm but not necessarily in most other senses one might be interested in; but if $f$ is reasonably nice, then its Fourier series converges to $f$, say, uniformly.

I'm looking for similar results about orthogonal polynomial expansions for functions on the whole real line. What I specifically want at the moment is sufficient conditions on a bounded function $f:\mathbb{R} \to \mathbb{R}$ so that the partial sums of its Hermite polynomial expansion are uniformly bounded on compact sets, but I'm also interested in learning what's known about pointwise/uniform/etc. convergence results for Hermite and other classical orthogonal polynomials.

Possibly such results follow trivially from well-known basic facts about Hermite polynomials, but I'm not familiar with that literature and I'm having trouble navigating it. So in addition to precise answers, I'd appreciate literature tips (but please don't just tell me to look at Szegő's book unless you have a specific section to recommend).

Answer by coudy for Convergence of orthogonal polynomial expansions

2010-06-24T15:21:58Z

Let me recall a quick $L^2$ proof of the uniform convergence of the Fourier series of a $C^1$ function $f$. Let $c_n$ be its Fourier coefficients. Then

$$|f(x)| \leq |c_0|+ \Sigma|c_n|\ n \ {1\over n}\ \ \leq |c_0| + \ \ \sqrt{\Sigma \ n^2 |c_n|^2}\ \ \sqrt{\Sigma\ 1/n^2}$$

Replacing f by f minus its partial sum, and noting that $\Sigma \ n^2|c_n|^2 = ||f'||_2^2 \ $ is finite, you get uniform convergence.

So maybe you can use a similar computation in case of a family of orthogonal polynomials ?

Answer by Helge for Convergence of orthogonal polynomial expansions

2010-06-24T18:11:54Z

Define $\psi_n(x) = c_n H_n(x) e^{-x^2/2}$ as in http://en.wikipedia.org/wiki/Hermite_polynomials . Also define the differential operator $H u = - u'' + x^2 u$. Then the $\psi_n$ form an othonormal basis of $L^2$ and $H \psi_n = (2n + 1) \psi_n$.

Warning: As coudy points out below: one needs $\|H f\| < \infty$ and not just $\langle f, Hf\rangle < \infty$. So the computations below need to be changed.

Rest of original post

Given now $f$ such that $$ A = \langle f,Hf \rangle =\int \overline{f(x)} (Hf)(x) dx $$ is finite. Then by writing $f(X) = \sum_{n \geq 0} f_n \psi_n(X)$, we obtain $$ A = \langle f,Hf \rangle =\langle f, \sum_{n \geq 0} f_n H\psi_n(X) \rangle = \langle \sum_{n \geq 0} f_n \psi_n(X) , \sum_{n \geq 0} f_n (2 n + 1)\psi_n(X) \rangle $$ Now using orthonormality of the $psi_n$, we conclude that $$ A = \sum_{n \geq 0} |f_n|^2 (2n + 1). $$ Now using that the $\psi_n(x)$ are all bounded by $2$ it follows that the sequence converges uniformly!

Now, what does $\langle f,Hf \rangle < \infty$ mean for $f(x) = e^{-x^2/2} g(x)$. This can be computed to mean $$ \int |g'(x) + \frac{x}{2} g(x)|^2 e^{-x^2} dx. $$

On a philosophical level, this is not about the $H_n$ being orthogonal polynomials, but about them being eigenfunctions of a self-adjoint operator. (well the $\psi_n$ are).