# Error of Discrete Fourier Transform on Finite Domain (vs. Continuous FT) in terms of Sobolev order

My question is about quantifying the error that occurs by approximating the continuous Fourier transform on a finite domain by using a discretised version with resolution $N$ for a function of a given Sobolev order $k$. I'm interested in the $L^2$-norm on the full (but finite) discrete grid, which I can prove is at most $\mathcal{O}(N^{-(k-1)})$, but numerically, I see $\mathcal{O}(N^{-k})$. I'm looking for ways to prove this better bound. First let me introduce some notation to formulate the question more clearly.

Let $$\Omega:=[0,L]^2, \qquad \hat\Omega=\mathbb{Z}^2,$$ as well as the finite discretisations (taking $N\in 2\mathbb{N}$ for convenience) $$\Omega_\mathrm{fin}:=\left[0:\frac{L}{N}:L\right)^2 \quad \text{and} \quad \hat\Omega_\mathrm{fin}:=\left[-\frac{N}{2}:\frac{N}{2}\right)^2.$$ The notation $[a:b:c]=\{a,a+b,\ldots,c-b,c\}$ is Matlab-inspired (missing $b$ defaults to $1$), with round brackets indicating exclusion of the respective endpoint as for open/closed intervals.

Define $$\mathcal{F}[f](\hat x, \hat y):= \frac{1}{L^2}\int_\Omega f(x,y) \mathrm{e}^{-2\pi\mathrm{i}(\hat x \frac{x}{L}+\hat y \frac{y}{L})} \mathrm{d} x\,\mathrm{d} y$$ and $$\mathcal{F}_\mathrm{fin}:=\frac{1}{N^2} \sum_{(x,y)\in\Omega_\mathrm{fin}} f(x,y) \mathrm{e}^{-2\pi\mathrm{i}(\hat x \frac{x}{L}+\hat y \frac{y}{L})},$$ as well as the usual Sobolev Space $$H^k(\Omega):=\{f\in L^2(\Omega) : \int_\Omega \left\lvert\frac{\partial^{k_1+k_2} f}{\partial^{k_1} x \, \partial^{k_2}y}\right\rvert^2\mathrm{d} x\,\mathrm{d} y < \infty, \; k_1+k_2\le k\}.$$ This space can be characterised on the Fourier side as $$H^k(\hat\Omega):=\{\hat f\in L^2(\hat\Omega) : \langle (\hat x,\hat y) \rangle^k \hat f(\hat x,\hat y)\in L^2(\hat\Omega)\},$$ where $\langle z \rangle:= \sqrt{1+\lvert z\rvert^2}$ is the regularised absolute value.

The norm I'm trying to bound is $$\left\lVert \mathcal{F}[f]-\mathcal{F}_\mathrm{fin}[f\bigr|_{\Omega_\mathrm{fin}}] \right\rVert_{L^2(\hat\Omega_\mathrm{fin})},$$ which should be $\mathcal{O}(N^{-k})$ for $f\in H^k$, ideally.

I'd also be happy with a 1d-proof or a reference (all the references I found so far deal with fft-errors as opposed to the error of discretising the continuous transform).

This finishes the actual question, below are my attempts/reasoning for the rate I claim to prove, resp. what a proof for the better rate will likely have to show.

I started by analysing the pointwise difference $$\left\lvert \mathcal{F}[f](\hat x_0, \hat y_0)-\mathcal{F}_\mathrm{fin}[f\bigr|_{\Omega_\mathrm{fin}}](\hat x_0, \hat y_0) \right\rvert.$$ Basically, the sum in $\mathcal{F}_\mathrm{fin}$ is the trapezoidal rule approximation to the integral in $\mathcal{F}$. An analysis in terms of Fourier coefficients seems necessary, because the trapezoidal rule in terms of differentiability only achieves $\mathcal{O}(N^{-2})$, regardless of higher $k$. The following idea is from [J. Waldvogel, Towards a general error theory of the trapezoidal rule; in Approximation and Computation, 2011].

By introducing $h(x,y):= f(x,y) \mathrm{e}^{-2\pi\mathrm{i}(\hat x_0 \frac{x}{L}+\hat y_0 \frac{y}{L})}$ and the trapezoidal quadrature operator $$T[h]:= \frac{L^2}{N^2} \sum_{(x,y)\in\Omega_\mathrm{fin}} h(x,y),$$ we see that $\mathcal{F}[f](\hat x_0, \hat y_0)=\mathcal{F}[h](0,0)$. By inserting $h=\mathcal{F}^{-1}[\mathcal{F}[h]]$ into T, we see that $$T[h]=\frac{L^2}{N^2} \sum_{(x,y)\in\Omega_\mathrm{fin}} \sum_{(\hat x,\hat y)\in\hat\Omega} \mathcal{F}[h](\hat x, \hat y)\mathrm{e}^{2\pi\mathrm{i}(\hat x \frac{x}{L}+\hat y \frac{y}{L})}\\ =L^2 \sum_{(\hat x,\hat y)\in\hat\Omega} \mathcal{F}[h](\hat x, \hat y) \delta(\hat x \mathrm{mod} N) \delta(\hat y \mathrm{mod} N)=L^2 \sum_{(\hat x,\hat y)\in\hat\Omega} \mathcal{F}[h](N\hat x, N\hat y),$$ by virtue of the summation property of the roots of unity (after interchanging the sums) $$\sum_{\ell=0}^{N-1} \mathrm{e}^{2\pi\mathrm{i} j \frac{\ell}{N}} = N\delta( j \mathrm{mod} N) \quad \forall j\in\mathbb{Z}.$$

In particular, we have that $$\left\lvert \mathcal{F}[f](\hat x_0, \hat y_0)-\mathcal{F}_\mathrm{fin}[f\bigr|_{\Omega_\mathrm{fin}}(\hat x_0, \hat y_0)] \right\rvert = \biggl\lvert \mathcal{F}[h](0,0)-\sum_{(\hat x,\hat y)\in\hat\Omega} \mathcal{F}[h](N\hat x, N\hat y)\biggr\rvert \\ = \biggl\lvert \sum_{(\hat x,\hat y)\in\hat\Omega\setminus(0,0)} \mathcal{F}[f](N\hat x+\hat x_0, N\hat y+\hat y_0)\biggr\rvert.$$

By inserting the representation $\hat f(\hat x,\hat y)=\langle (\hat x, \hat y) \rangle^{-k}\hat g(\hat x, \hat y)$ for a $\hat g\in L^2(\hat\Omega)$, we can pull out $N^{-k}$ while the sum is still finite (both $\lvert \hat x_0 \rvert$ and $\lvert \hat y_0 \rvert$ are less than $\frac{N}{2}$). Summing this over the $N^2$ points in $\hat\Omega_\mathrm{fin}$ gives $N^{-(k-1)}$ for the $L^2$-norm.

Any further gain (using this avenue of proof) would have to come from bounding $$\sum_{(\hat x_0,\hat y_0)\in\hat\Omega_\mathrm{fin}} \biggl\lvert \sum_{(\hat x,\hat y)\in\hat\Omega\setminus(0,0)} \langle (\hat x+\frac{\hat x_0}{N}, \hat y+\frac{\hat y_0}{N}) \rangle^{-k}\hat g(N\hat x+\hat x_0, N\hat y+\hat y_0)\biggr\rvert^2$$ with a lower power of $N$ than $2$. There is certainly some decay there, but harnessing it is not obvious. In particular, the knowledge that $\hat g \in L^2(\hat\Omega)$ is not easily applied.

Thanks for reading through this long question!

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Thanks a lot for the reference! Unfortunately, this only deals with the pointwise error of classically differentiable functions (e.g. Corollary 3.2). I also checked the reference from the introduction about $L^2$-functions (Auslander & Grünbaum), but the only additional properties they consider are bandlimited functions. What I'm looking for are $L^2$-errors for functions from a Sobolev space... –  Axel Mar 20 '14 at 16:13