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"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).

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I think you should add a weight to your question. The Hermite polynomials are an orthonormal basis of $L^2(\exp(-x^2/2) dx)$ and not of $L^2$. So assuming that $f \in L^2(\exp(-x^2/2) dx)$ would be sufficient to have an expansion in Hermite polynomials converging in $L^2$ (with respect to that weight). So e.g. the assumption $f(x) \exp(x^2/2)$ is bounded would be more naturall than just $f$ bounded... (in some sense ...) –  Helge Jun 24 '10 at 15:08
    
Your last sentence doesn't make sense to me since you're proposing a much stronger condition than boundedness. In any case, I didn't mention the weight because (1) for reasons external to this question, bounded functions are what I'm really interested in; and (2) I wanted to keep the statement of my question simple. –  Mark Meckes Jun 24 '10 at 15:28
    
Now, I'm confused. Do you want to approximate with the Hermite polynomials $H_n(x)$ or with $H_n(x) \exp(- x^2/2)$ ? In the first case uniform convergence is trivally wrong! One might hope for uniform convergence on compact sets. In the second case, the condition $\liminf_{|x|\to\infty} |f(x)| > 0$ should imply that the expansion coefficients are unbounded. (This statement is a feeling, I didn't try proving it). This again destroys all hopes for a nice sense of convergence... –  Helge Jun 24 '10 at 15:39
    
Ack, what I really wanted was uniform boundedness on compact sets, and I forgot to include that crucial last point. I've edited to correct that. I did want to approximate $f$ by $\sum c_n H_n$, but now you've got me thinking that maybe it would work better to approximate $f(x)\exp(-x^2/2)$ by $\sum c_n H_n(x) \exp(-x^2/2)$; I'll think about that some more. –  Mark Meckes Jun 24 '10 at 15:54
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There is a chapter (9.1 ff p368) in the book of G. Sansone "Orthogonal functions" about uniform convergence of series of Hermite polynomials. There is a preview of the book here books.google.com/…. –  user6129 Jun 24 '10 at 18:07

3 Answers 3

up vote 3 down vote accepted

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).

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Am I missing something ? Shouldn't the condition be Hf in L^2 ? With $f_n=1/(n \log n)$, I don't see how to get the uniform convergence of $\sum f_n \psi_n$ from the finiteness of <f,Hf>. –  user6129 Jun 24 '10 at 20:58
    
I think you are right, and one indeed needs $\| H f\|$ to be finite. I'll add a comment ... –  Helge Jun 24 '10 at 22:11
    
Well, yeah, clearly. But I accepted this answer because it gives me all the tools I need to prove the kinds of results I can use. –  Mark Meckes Jun 24 '10 at 23:30

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 ?

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Quite possibly something along the same lines, but necessarily more complicated, works. I'd need to know how to simply relate the coefficients for $f$ to those for $f'$, and also uniform bounds on the Hermite polynomials themselves. These are the kinds of "well-known basic facts about Hermite polynomials" that I'm missing. –  Mark Meckes Jun 24 '10 at 15:32
    
That is, assuming these are "well-known basic facts" at all. –  Mark Meckes Jun 24 '10 at 15:34

I too was looking into this recently and found good citations difficult to find. Here are two I found which I can recommend:

.1. In the book "Special Functions and their Applications" by Lebedev, he gives the following clear theorem statement (with a moderately technical but not too bad-looking proof): "Assume $f \in L^2(\gamma)$ (i.e., is square-integrable w.r.t. the standard Gaussian measure) and is piecewise-$\mathcal{C}^1$ on every finite interval $[-a,a]$. Then the Hermite expansion of $f$ converges pointwise at every point of continuity of $f$ (and further, converges to $(f(x+)-f(x-))/2$ at any jump discontinuity $x$).

It would be nice to also know that one has uniform convergence on any interval $[-a,a]$, but I didn't immediately see how to read that out.

.2. The book "Gaussian measures" by Bogachev gives pretty careful statements about the domains under which formal operations (e.g., differentiation of Hermite expansions) hold.

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Oh, I guess the desired uniform convergence statement from in #1 probably follows from Egorov's theorem. –  Ryan O'Donnell Oct 19 '13 at 17:46

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