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Consider $\Omega = (0,2 \pi)^d$ and the negative Sobolev space $H^{-s}(\Omega)$, defined as the dual of $H^s_0(\Omega)$ for the $L^2$ inner product. Due to the simple shape of $\Omega$, we can see $H^{-s}(\Omega)$ as a Banach/Hilbert space with the inner product: $$ \langle f, g \rangle_{H^{-s}} := \sum_{n \in \mathbb{Z}^d} f_n g_n (1 + | n |^2)^{-s}, $$ where $(f_n)$ and $(g_n)$ are the Fourier coefficients of $f$ and $g$.

Question: Is $C^\infty_0(\Omega)$ dense in $H^{-s}(\Omega)$?

dgontier
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  • Partial answer: By truncating the Fourier series you can show that trigonometric polynomials and hence $C^{\infty}(\overline\Omega)$ is dense in $H^{-s}(\Omega)$ for all $s.$ I think you can also use $C^{\infty}_c$ functions, but I can't think of an easy argument at the moment. – ktoi Dec 28 '20 at 20:51

1 Answers1

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The answer is affirmative. In a previous version of this post I claimed the opposite, but I was wrong. (The previous version is below. Following my usual policy, I am NOT deleting it).

Claim. The space $C^\infty(\mathbb T)$ of smooth and $2\pi$ periodic functions is dense in $H^{-s}(\mathbb T)$ for all $s\in \mathbb R$.

Proof. We will prove that $C^\infty(\mathbb T)^\bot$ is reduced to $\{0\}$. So, let $g\in H^{-s}(\mathbb T)$ be such that $$ \sum_{-\infty}^\infty \frac{\hat{g}(n)\overline{\hat\phi(n)}}{(1+|n|^2)^s}=0, \qquad \forall \phi \in C^\infty(\mathbb T).$$ In particular, letting $\phi(x)=e^{imx}$, so that $\hat\phi(n)=0$ for $n\ne m$ and $\hat\phi(m)=1$, we see that $$\frac{\hat{g}(m)}{(1+|m|^2)^s}=0, $$ which implies $\hat{g}(m)=0$. Since $m$ is arbitrary, we conclude that $g=0$, as claimed. $\Box$

Remark. This is exactly the proof suggested by the OP dgontier in comments below, and also by ktoi in comments to the main question.

Remark. I find this result a bit surprising. If $f\in H^{-s}(\mathbb T)$, its Fourier coefficients $\hat{f}(n)$ are allowed to grow as $n\to \infty$. If $s>1/2$, the space $H^{-s}(\mathbb T)$ contains singular distributions such as the Dirac comb $\delta$. On the other hand, if $\phi\in C^\infty(\mathbb T)$, then its Fourier coefficients decay faster than any polynomial. So, I find it surprising that $C^\infty$ functions suffice to approximate arbitrary $H^{-s}$ elements.

PREVIOUS VERSION OF THIS ANSWER.

WARNING. There is a mistake, marked below. Now I have to go, but I am not deleting this answer, maybe someone will point out a solution, or it will be useful in some other way.

The answer is negative. To ease typing, let me assume $d=1$. Let $f_n\in C^\infty_0(0, 2\pi)$ and suppose that there is a $f\in H^{-s}$ such that $\lVert f_n-f\rVert_{H^{-s}}\to 0$. This means that $$ \sum_{k=-\infty}^\infty \frac{\lvert \hat{f}_n(k)-\hat{f}(k)\rvert^2}{1+k^{2s}}\to 0, \qquad n\to \infty.$$ In particular, for each $k_0\in\mathbb Z$, $$\tag{*} \lim_{n\to \infty} \frac{\lvert \hat{f}_n(k_0)-\hat{f}(k_0)\rvert^2}{1+k_0^{2s}} =0.$$ Now, integration by parts shows that, since $f_n$ is smooth, $\hat{f}_n(k)$ decays as $\lvert k \rvert\to 0$ (faster than any power of $k$, but we won't need this quantification). For sufficiently big $n$, we have from (*) that $$ \frac{\lvert \hat{f}_n(k_0)-\hat{f}(k_0)\rvert^2}{1+k_0^{2s}} \le 1.$$

(WARNING: this seems to be wrong). So, taking the limit as $k_0\to \infty$ shows that $$ \lim_{k\to \infty} \hat{f}(k)=0.$$

However, not all $f\in H^{-s}$ satisfy this. For example, if $s>1/2$ then the Dirac comb is in $H^{-s}$, and $\hat{\delta}(k)=1$ for all $k$.

Giuseppe Negro
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  • Thanks for the comment. I think I now have a proof that it is correct: I think it is enough to say that $C^\infty_0$ is dense in $L^2$ (for the $L^2$ inner product, hence also for the $H^{-s}$ one), and that $L^2$ is dense in $H^{-s}$. – dgontier Dec 29 '20 at 13:37
  • @dgontier: So you say it is true? I have a concern, which I tried to express in this answer, but rather perfunctorily I am afraid. In negative order Sobolev spaces, the Fourier coefficients are not required to decay at infinity, and they can even grow, if $s$ is big enough. For example, the distribution $\delta$, which is such that $\hat{\delta}(n)=1$ for all $n$, belongs to $H^{-s}(\mathbb R^d)$ for all $s<d/2$. On the other hand, smooth functions have very fast decaying Fourier coefficients. I am not sure you can approximate distributions such as $\delta$ with such smooth functions. – Giuseppe Negro Dec 29 '20 at 15:04
  • @dgontier: However, your reasoning seems correct to me. To reinforce it, I checked the book of Bahouri, Chemin and Danchin "Fourier analyisis and nonlinear PDEs", and Theorem 1.61 states that $C^\infty_0(\mathbb R^d)$ is dense in $H^s(\mathbb R^d)$ for all $s\in\mathbb R$ (this is NOT the space you want, but it is very similar). The proof seems to be the exact transposition of the argument you suggest. I am thus inclined to think that my intuition was wrong. – Giuseppe Negro Dec 29 '20 at 15:26
  • Thx for the reference. Indeed, I know the result to be correct in the whole space $\mathbb{R}^d$. I was concerned about bounded sets $\Omega$. I now believe the answer to be positive. Indeed, the coefficients of $H^{-s}$ functions can grow, but this is taken into account in the $H^{-s}$ norm. – dgontier Dec 30 '20 at 16:49
  • @dgontier: Yes, I agree, the answer is affirmative in every case. The proof I wrote above is for $\Omega=\mathbb T$, that is $(0, 2\pi)$, but the exact same proof works for $\Omega=(0, 2\pi)^d$ for any $d\in\mathbb N$, and actually for any bounded open set, but in that case you have to change the definition of the Fourier series; instead of complex exponentials $e^{ix\cdot n}$ you will need a complete system of eigenfunctions of the Laplacian. – Giuseppe Negro Dec 30 '20 at 17:15
  • I think the point is that $\mathcal{C}^\infty_0$ is dense in $H^{-s}$ for the $H^{-s}-norm$. If you simply truncate your Fourier serie, the coefficients in the remaining part might indeed diverge, but they do tend to zero when you multiply them by $k^{-2s}$. – Isao Oct 26 '21 at 11:26
  • @Isao: right, that's a good way of seeing this question. – Giuseppe Negro Oct 26 '21 at 11:50
  • I think it also gives you a proof: for $f\in H^{-s}$, let $g\in L^2$ be the serie with Fourier coefficients $ (1+|k|^{2s})^{-1} \hat{f}k$. Let then $g_n$ be a sequence in $\mathcal{C}^\infty$ approximating $g$ (in the $L^2$-sense), and $f_n$ the serie with Fourier coefficient $\hat{f}{n,k}=(1+|k|^{2s})\hat{g}_{n,k}$. Then the $f_n$ are smooth and approximate $f$ in the $H^{-s}$ sense. I didn't check the details. – Isao Oct 28 '21 at 11:29
  • Probably that it works because of the commutative diagram $\Delta\circ \iota=\iota' \circ \Delta$, where $\Delta$ is a fractional laplacian and $\iota,\iota'$ are two Sobolev embeddings (or someting like that...). – Isao Oct 28 '21 at 11:36