This is a restated version of my original very broad question.

Let $P$ be probability a measure on an interval $[a,b]$ ($-\infty<a<b<\infty$) that's dominated by Lebesgue measure. Let $\langle f, g \rangle_P=\int_{[a,b]} f \cdot g \, dP$ be an inner product for integrable functions associated with $P$ and $\| \cdot \|_{L_2(P)}$ denote the induced $L_2$-norm. Let $f: [a,b] \rightarrow \mathbb{R}$ be integrable and $f \in C^r[a,b]$ for some $r \leq \infty$. Consider approximating $f$ by a trigonometric polynomial $a_0+\sum_{k=1}^K [a_k \cos(kx) + b_k \sin(kx)]$. Let $\pi_{K,P}(f)$ denote the projection of $f$ onto the space of $K$-th order trigonometric polynomials with respect to $\langle \cdot, \cdot \rangle_P$. Consider the approximation error $\| f-\pi_{K,P}(f) \|_{L_2(P)}$. Assume $f$ is not a trigonometric polynomial itself.

I've seen an upper bound of this error that looks like $C K^{-r}$ when $r < \infty$ and is derived from the $L_\infty$ error (https://www.springer.com/us/book/9783540506270). Is there a lower bound for the $L_2(P)$ error, ideally with the same structure as the upper bound (maybe under mild conditions)? Also, what would be the upper bound and lower bound if $r=\infty$? I'm particularly interested in the case where $f(x)=x$.

You may take $[a,b]$ to be any interval you want for convenience.

I would be very happy if you know of any results of this kind, perhaps under similar settings (e.g. a different notion of smoothness for $f$ in terms of Soblev spaces).

Original question:

Let $P$ be probability a measure on an interval $[a,b]$. Let $\langle f, g \rangle_P=\int_{[a,b]} f \cdot g \, dP$ be an inner product for integrable functions associated with $P$ and $\| \cdot \|_{L_2(P)}$ denote the induced $L_2$-norm. Let $f: [a,b] \rightarrow \mathbb{R}$ be integrable. Let $\phi_1,\phi_2,\ldots: [a,b] \rightarrow \mathbb{R}$ be a basis and I want to use $\sum_{k=1}^K \beta_k \phi_k$ to approximate $f$ for some $\beta_k \in \mathbb{R}$. In particular, let $\pi_{K,P}(f)$ denote the projection of $f$ onto $Span\{\phi_1,\ldots,\phi_K\}$ with respect to $\langle \cdot, \cdot \rangle_P$. Consider the approximation error of $\pi_{K,P}(f)$: $\| f-\pi_{K,P}(f) \|_{L_2(P)}$.

I know there are results on the upper bound of this error. But is there a more accurate estimate (not just an upper bound)? Or is there a lower bound? If so, can you provide a reference?

Here I always assume $f$ is not a linear combination of $\phi_1,\phi_2,\ldots$, i.e. $\pi_{K,P}(f) \neq f$ for any $K<\infty$ and any $P$, so there might be a nontrivial lower bound.

You may consider simplified/restricted versions of this problem. For example, you may take $\phi_k=x^{k-1}$ or $\phi_{2k-1}=\cos((k-1)x), \phi_{2k}(x)=\sin(kx)$ (trigonometric polynomial); you may take $[a,b]=[0,1]$ or $[a,b]=[0,2\pi]$.

You may add other assumptions on $f$ or $P$ as long as they are not too restrictive. For example, assume $f$ is continuously differentiable up to some order or infinitely differentiable; or, assume $P$ is dominated by Lebesgue measure.

Edit 1: Let me try to be more specific. Is there a lower bound of the approximation error when $f(x)=x$ and I use a trigonometric polynomial (with both $\sin$ and $\cos$ series) to approximate? You may assume $[a,b]=[0,2\pi],[0,\pi],[-\pi,\pi]$ or whatever finite interval as you want. The probability measure $P$ has no special property. If necessary, you may assume $P$ is dominated by Lebesgue measure.

Edit 2: Just want to emphasize that I only care the behavior on a finite interval $[a,b]$ instead of the whole real line, so if I use trigonometric polynomial for approximation, I feel $f(x)=x$ not being periodic should not make the approximation behave super badly.


This is very broad, and largely depend on some details you have not provided. Usually the questions are -

  1. How smooth is $f$?
  2. Is $P$ a measure with any special properties, e.g., has an associated Sturm Liouville operator?
  3. What is the choice of the basis $\phi _1, \ldots , \phi _N, \ldots$.

All these questions change the answer considerably. For example, let $P=m$ the Lebesgue measure, then for a function $f\in C^r([0,1])$, using the right polynomial expansion, $$ \| f - \pi _{K,m} \| _2 \leq CK^{-r}$$

Hence, if $f$ is only integrable, convergence might be agonizingly small. That also means that if $f$ is analytic, but not periodic, its expansion by cosines will be very slow, since it will be discontinuous when considered as a function on $\mathbb{T}$.

Two good general reference books are "Interpolation and Approximation" by the late Philip Davis, and "Approximation Theory and Approximation Practice" by Nick Trefethen. I would advise you to be more specific in your question, because as it is, it's just too broad.

  • $\begingroup$ I've just edited my question trying to be more specific. Can you see whether you know any relevant reference? Thanks. $\endgroup$ – David Hongxiang QIU Oct 7 '18 at 17:28
  • $\begingroup$ If $P$ has no general property whatsoever, I suspect you may run into serious issues. $f(x)=x$ is not periodic, and therefore its $L^2$ error will be pretty bad for every $P$ absolutely continuous with respect to Lebesgue measure. But without any limitation, if $P= \delta(x)$, the point mass at $x=0$, a sine-only basis can give zero $L^2$ norm. $\endgroup$ – Amir Sagiv Oct 7 '18 at 17:55
  • $\begingroup$ The assumption of Edit 2 should not change the result, in my opinion - if $f(a) \neq f(b)$, then we should expect a very slow convergence rate using Fourier with any scaling I can think of. $\endgroup$ – Amir Sagiv Oct 7 '18 at 19:48
  • $\begingroup$ But you can always somehow extend the definition of f to the whole real line to make it periodic and at least continuous. Also, if we take $[a,b]=[−\pi,\pi]$, then $f(-x)=-f(x)$, so there seems to be a sine series that well approximates it. $\endgroup$ – David Hongxiang QIU Oct 7 '18 at 20:25
  • $\begingroup$ I am quite lost with your question, and the added edits don't help - restate it from the top, and state your assumptions (not what they can be, but what they are concretely). $\endgroup$ – Amir Sagiv Oct 7 '18 at 21:33

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