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As the title says, I would like to understand and get intution for fourier expanstion converging pointwise for nice functions.

In particular, I might start by asking how often (of course this isn't well defined), when I pick an orthonormal system with respect to the usual integral inner product, does it give pointwise convergence for nice functions.

Let's define nice functions as analytic (of course I'll be much happer if things hold for just diffrentiable functions for instance).

I'm not sure how to define an arbtiary orthonormal system, so I leave this open to the answerer that is more knowledgeable than me.

Thanks

Andy
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  • https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem is one of the more general results of this type. – Chappers Jul 21 '17 at 20:05
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    @Chappers Thank you, that is indeed a relevant result (although I happen to be aware of it). However, notice that I'm choosing the coefficients here with regard to the integral inner product, which doesn't have to concide with the convergence one would think of with Weierstrass – Andy Jul 21 '17 at 20:07
  • In which case, there's the Sturm–Liouville theory angle. Are you familiar with that? – Chappers Jul 21 '17 at 20:10
  • @Chappers Nope, although I may have heard the name regarding boundry conditions of ODE's? – Andy Jul 21 '17 at 20:10
  • There's https://math.stackexchange.com/questions/2186354/proof-for-sturm-liouville-eigenfunction-expantion-pointwise-convergence-theorem , for example. Wikipedia's article on Sturm–Liouville theory has most of the background. – Chappers Jul 21 '17 at 20:14
  • @Chappers I skimmed the wikipedia article, I don't quite see how it relates to my question, can you elaborate? – Andy Jul 21 '17 at 20:45
  • The point is that the eigenfunctions of a Sturm–Liouville system form an orthonormal basis, which can be shown to converge pointwise to functions it converges to in $L^2$ norm in much the same way as Fourier series do. – Chappers Jul 21 '17 at 22:41
  • But thinking about it, perhaps a more useful result is that if you have an orthonormal basis, Bessel's equality tells you that the inner-product–constructed series converges to the function in $L^2$. If you also have convergence of the derivative to the derivative of the function (i.e. convergence in the Sobolev space $H^1$, this is sufficient for pointwise convergence, by an appropriate embedding theorem: in particular, there is a one-dimensional Sobolev inequality that bounds the sup norm by the $H^1$-norm: see https://math.stackexchange.com/q/2049461/221811 , for example. – Chappers Jul 21 '17 at 22:45
  • I can write that up in more detail as an answer if you think that's what you want. – Chappers Jul 21 '17 at 22:45
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    @Chappers umm, this seems very interesting, but my ultimate goal is an orthonormal with respect to the integral inner product that has pointwise convergence. Can you provide classes of such orthonormal systems? So I don't mind if you prove it with the Sobolev space, I'm interested in seeing what kind of orthonormal systems satisfy what I want. – Andy Jul 22 '17 at 08:13

2 Answers2

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This is a very interesting question. First, it is not at all obvious how (or whether it is possible) to extend the relatively elementary Fourier-Dirichlet result on pointwise convergence to general orthonormal systems (let's say even for (the simplest, "non-singular") Sturm-Liouville problems with smooth coefficients). Yes, of course, the $L^2$ convergence is the main point, and for utilitarian purposes not so many people ask further. And, yes, everyone imagines that for very smooth functions, the spectral expansion in terms of the eigenfunctions should converge pointwise, etc.

It cannot be quite this simple, unfortunately/interestingly enough. For example, the eigenfunctions for the Dirichlet problem $u''=\lambda u$ and $u(0)=0=u(2\pi)$ are $\sin(nx/2)$. So of course things like the constant function $1$ are in their $L^2$ span. But that $L^2$ expansion cannot possibly converge pointwise to $1$, because $1$ is just $1$, at the endpoints, while all the eigenfunctions are $0$ there.

Yes, this failure can be viewed as essentially irrelevant. And, indeed, for functions in the corresponding $H^1$ Sobolev space (attached to the operator + boundary conditions), since we have Sobolev imbedding $H^1\subset C^o$, convergence is in $C^o$.

But, still, the eigenfunctions themselves are only in $H^1$, not $H^\infty$. So there will inevitably be troubles... understandably, at the endpoints, but such troubles propagate in spectral expansions (for the same reasons that Fourier transforms interchange local smoothness and global decay).

paul garrett
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  • What if I compromise with pointwise convergence almost everywhere? Does it get easier? – Andy Jul 24 '17 at 11:12
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An orthogonal set $\mathbf A $ in an inner product space is one such that for all distinct $ x,y \in \mathbf A $, we have $\langle\,x,y\rangle = 0$. An orthogonal set $\mathbf A $ is called orthonormal whenever we have $\lVert x\rVert = 1$ for all $x \in A$. Furthermore, an othonormal set is called complete (or an orthonormal basis) whenever it is not contained in another orthonormal set. By using Zorn's Lemma, we can extend any orthonormal set to a complete orthonormal set. In any Hilbert space, such as that of the square integrable functions $L^2[\pi,-\pi]$ over a closed interval in which we do ordinary Fourier analysis, we can give an expansion of any point in terms of any complete orthonormal set. In particular, if $A=\{a_1, a_2, a_3...\}$ is a complete orthonormal set and if $x$ is any point we have $x =\sum_{i=1}^\infty \langle x,a_i \rangle a_i $.

BMcNally
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  • I don't understand, are you claiming every orthonormal system can be extended to an orthonormal system that also has pointwise convergece? This isn't true I believe. – Andy Jul 24 '17 at 06:11
  • Yes you're right. I think I was trying to help you in understanding what it meant " to define an arbtiary orthonormal system".An "orthonormal system" is a result of the inner product structure of the space, and here is convergence in the norm induced by the inner product when norm is complete.Out of all the different modes of convergence, we can prove implications between them, but it is in general impossible to show that norm convergence implies pointwise .For counterexamples look here https://math.stackexchange.com/questions/138043/does-convergence-in-lp-implies-convergence-almost-everywhere – BMcNally Jul 24 '17 at 13:04