Let $\mathbb{T}$ denote the 1-d torus and $H^s(\mathbb{T})$ the Sobolev space of order $s\geq0$ of complex-valued functions on $\mathbb{T}$ with the identification $H^0 (\mathbb{T}) = L^2 (\mathbb{T})$. Let $\mathcal{T}_N$ denote the span of the first $N \in \mathbb{N}$ trigonometric polynomials and $P_N : L^2(\mathbb{T}) \to \mathcal{T}_N$ the standard Fourier projection. It is easily shown, by passing to sequences, that, for any $f \in H^s (\mathbb{T})$ with $s \geq 1$,

$$\|f-P_Nf\|_{L^2} \leq \bar{N}^{-s}\|f\|_{H^s}$$

where $\bar{N} = \sqrt{1 + N^2}$. Consider now the following simple argument. Let $f \in H^1(\mathbb{T})$ and fix $0 < \epsilon < 1$. Since $C^\infty (\mathbb{T})$ is dense in any $H^s (\mathbb{T})$ space, there exists a function $g_\epsilon \in C^\infty (\mathbb{T})$ such that

$$\|f - g_\epsilon\| < \frac{\epsilon}{2}.$$

Triangle inequality implies

$$\|g_\epsilon \|_{H^1} < \frac{\epsilon}{2} + \|f\|_{H^1}.$$

Since $g_\epsilon \in C^\infty (\mathbb{T})$, we have $g_\epsilon \in H^m (\mathbb{T})$ for any $m \in \mathbb{R}$, and, just to keep this as simple as possible, let's pick $m=2$. Then

$$\|g_\epsilon - P_N g_\epsilon\|_{L^2} \leq \bar{N}^{-2} \|g_\epsilon \|_{H^1} \leq \bar{N}^{-2} ( \frac{\epsilon}{2} + \|f\|_{H^1} ) \leq \frac{\epsilon}{2}$$

simply by picking the smallest $N \in \mathbb{N}$ such that $\bar{N} \geq \left ( \frac{4 \|f\|_{H^1}}{\epsilon} \right )^{1/2}$.

By triangle inequality, we obtain $\|f - P_N g_\epsilon\|_{L^2} \leq \epsilon$.

So, we have shown that, given any $f \in H^1 (\mathbb{T})$ then, for any $N \in \mathbb{N}$, there exists $g_N \in \mathcal{T}_N$ such that

$$\|f - g_N\|_{L^2} \leq 4 \bar{N}^{-2} \|f\|_{H^1}.$$

In fact, we could have done much better and gotten a super-algebraic rate of convergence and, if instead working on $\mathbb{T}^d$, avoided the curse of dimensionality. My immediate question about this (apart from is it correct) is how does one reconcile it with that fact that $P_N$ is the optimal $L^2$-projector onto $\mathcal{T}_N$. In particular, $P_N f$ should be the best one can do when approximating from $\mathcal{T}_N$. This suggests one can do a lot better, provided more regularity is available and is perhaps related to the fact that $H^1$ is relatively compact in $L^2$? Another point of view is that method approximation, particularly $f \mapsto g_\epsilon$ is discontinuous (there are lower bounds that would imply this but they are for non-linear methods and what bothers me most is that $\mathcal{T}_N$ is a linear space). Is the operator of best approximation from $\mathcal{T}_N$ in $L^2$ linear or continuous? Any thoughts on this would be helpful. Thanks.


1 Answer 1


I see the glaring error in this now. I applied the bound with the $H^1$ norm on $g_\epsilon$ when it should have been with the $H^2$ norm but then, there’s no way to relate that to the $H^1$ norm of $f$. Things make sense again. Please close/delete this.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.