Let $f: \mathbb{R}^n \rightarrow \mathbb{C}$ be a smooth function, that is $f \in C^{\infty}(\mathbb{R}^n)$, such that $f$ and all its derivatives belong to the Lebesgue space $L^p(\mathbb{R}^n)$, with $1 \leq p < \infty$. I am trying to prove that f must vanish at infinity, that is it satisfies \begin{equation} \lim_{|x| \rightarrow \infty} |f(x)| = 0. \end{equation} I could only prove the statement for $p=1$ (see the note below). Any help is welcome. Thank you very much for your attention in advance.

NOTE (Motivation and Case p=1) First a matter of notation: if $\beta_i$ is a non-negative integer for $i=1,\dots,n$, we call $\beta=(\beta_1,\dots,\beta_n)$ a multi-index, we set $|\beta|=\beta_1+\dots+\beta_n$ as usual, and \begin{equation} D^{\beta}f = \frac{\partial^{|\beta|} f}{\partial x_{1}^{\beta_1} \dots \partial x_{n}^{\beta_n}}. \end{equation} We also denote by $D_i$ the partial derivative with respect to the i-th coordinate.

The relevance of the question comes from the fact that the space of functions \begin{equation} \mathcal{D}_{L^p} = \left \{ f \in C^{\infty}(\mathbb{R}^n) : D^{\beta} f \in L^p(\mathbb{R}^n) \textrm{ for each multi-index $\beta$} \right \} \end{equation} is relevant in the theory of distributions. This vector space is topologised through the family of semi-norms \begin{equation} || f ||_{p,N} = \max \left \{ || D^{\beta} f ||_p : |\beta| \leq N \right \} \quad \quad (N=0,1,2,\dots), \end{equation} where $|| . ||_p$ denotes the norm of the space $L^p(\mathbb{R}^n)$. It was introduced by Schwartz in Théories des Distributions, Chapter VI, $\S{8}$, where I found the statement I am trying to prove, which is a crucial step in a remarkable embedding theorem. To state it, let us also introduce the space $\mathcal{B_0}$ (in the notation used by Schwartz this space is denoted with the symbol $\dot{\mathcal{B}}$) which is the vector space of all $f \in C^{\infty}(\mathbb{R}^n)$ such that $f$ and all its derivatives vanish at infinity. We topologise $\mathcal{B_0}$ thought the family of semi-norms: \begin{equation} || f ||_{\infty,N} = \max \left \{ || D^{\beta} f ||_{\infty} : |\beta| \leq N \right \} \quad \quad (N=0,1,2,\dots), \end{equation} where $|| . ||_{\infty}$ denotes the norm of the space $L^{\infty}(\mathbb{R}^n)$.

Schwartz states that if $1 \leq p < \infty$ then each $f \in \mathcal{D}_{L^p}$ not only is bounded, but it also vanishes at infinity. This clearly implies that for $q \geq p$, we have $\mathcal{D}_{L^p} \subset \mathcal{D}_{L^q} \subset \mathcal{B_0}$. Moreover each inclusion is continuous.

Now let us come back to our question. The case $p=1$ can be proved as follows. Let $g \in C^{\infty}(\mathbb{R}^n)$ be a function with compact support such that $g=1$ on the unit ball with center $0$. Fix $r > 0$ and for any $x \in \mathbb{R}^n$ define $g_r(x)=g(x/r)$. Then set $\phi_r=fg_r$. If $d > 0$ is such that $[-d,d]^n$ contains the support of $\phi_r$, then by repeated integration we get for any $x=(x_1,\dots,x_n)$: \begin{equation} \phi_r(x) = \int_{-d}^{x_1} \dots \int_{-d}^{x_n} (T\phi_r)(y) dy = \int_{-\infty}^{x_1} \dots \int_{-\infty}^{x_n} (T\phi_r)(y) dy, \end{equation} where $T=D_1...D_n$ (by the way, note that the last equality just says that $\phi_r$ is the convolution of $T\phi_r$ and Heaviside function $H$ on $\mathbb{R}^n$) . For any $R > 0$ define \begin{equation} Q(R)= \{ x=(x_1,\dots,x_n) \in \mathbb{R}^n : \min\{x_1,\dots,x_n\} \leq -R \}. \end{equation} We have that for any $x \in Q(R), z \in Q(R)$: \begin{equation} \left| \phi_r(x) - \phi_r(z) \right| \leq \int_{Q(R)} |(T\phi_r)(y)| dy. \end{equation} By using Leibniz formula we get that there exists $C> 0$ such that for any $r \geq 1$ we have \begin{equation} \int_{Q(R)} |(T\phi_r)(y)| dy \leq C M(R), \end{equation} where \begin{equation} M(R) = \max \left \{ \int_{Q(R)} \left| (D^{\beta}f)(x) \right| dx : |\beta| \leq n \right \}. \end{equation} By taking $R$ large enough, we get that for any $\epsilon > 0$, there exists $R > 0$ such that for any $x \in Q(R)$, $y \in Q(R)$, we have $|f(x) - f(y)| < \epsilon$. Since $f \in L^1(\mathbb{R}^n)$, we conclude that for any $\epsilon > 0$ there exists $r > 0$ such that $|f(x)| < \epsilon$ for any $x \in Q(r)$. By applying this result to $f(-x)$, we then get the desired conclusion.

A final note. Clearly, the fact that $f$ vanishes at infinity implies that $f$ is bounded. In the case $p=1$, this fact can be directly proved by using the representation above \begin{equation} \phi_r(x) = \int_{-\infty}^{x_1} \dots \int_{-\infty}^{x_n} (T\phi_r)(y) dy. \end{equation} Actually, from this representation and Leibniz formula we get \begin{equation} ||f||_{\infty} \leq 2^{n} || g ||_{\infty,n} ||f||_{1,n}, \end{equation} so that, if we set $A=2^{n} || g ||_{\infty,n}$, we conclude that \begin{equation} ||f||_{\infty} \leq A ||f||_{1,n}. \end{equation} Clearly the same inequality applies to each derivative of f. So we conclude that $\mathcal{D}_{L^1} \subset \mathcal{B_0}$ and that the inclusion $\mathcal{D}_{L^1} \hookrightarrow \mathcal{B_0}$ is continuous. Moreover, we also get that for $1 < q < \infty$ we have $\mathcal{D}_{L^1} \subset \mathcal{D}_{L^q}$ and that the inclusion $\mathcal{D}_{L^1} \hookrightarrow \mathcal{D}_{L^q}$ is continuous. We have so proved two particular cases of the general embedding theorem stated by Schwartz.

  • $\begingroup$ Are you sure that the claim is true? Wouldn't a smooth function with unbounded support of finite measure give a counterexample? $\endgroup$ – Dirk Dec 16 '16 at 23:07
  • $\begingroup$ Dear Dirk, the claim is not mine. As I said, it is made by Laurent Schwartz in Théorie des Distributions, Hermann Paris, 1966, Chapter VI, $\S 8$, pp. 199 - 200. He claims that every $\phi \in \mathcal{D}_{L^p}$, with $1 \leq p < \infty$ is bounded, so that we also have $\phi \in \mathcal{D}_{L^q}$ with $q \geq p$, and it vanishes at infinity. $\endgroup$ – Maurizio Barbato Dec 17 '16 at 9:15

It is not that difficult, I need the following proposition on a parametrix, somewhere mentioned in Schwartz book for the specific case of a fundamental solution of the iterated Laplace equation.

Proposition: Let $k\in\mathbb{N}, K\subset\mathbb{R}^n$ compact containing $0$ as an interior point. There exist $\varphi\in\mathcal{D}_K^k, m\in\mathbb{N}$, and $\xi\in\mathcal{D}_K$ such that $$ \delta=\Delta^m\varphi+\xi,$$ where $\Delta^m$ is the $m$-times iterated Laplace operator. [$\mathcal{D}_K^k$ denotes the space of $k$-times differentiable functions with support in $K$.]

Not let $K$ be such a compact set and $f\in\mathcal{D}_{L^p}$. Consider the convolution map $C_f\colon \mathcal{D}_K\to \mathcal{B}_0, C_f(\varphi)=f\ast\varphi$. Note that for $\varkappa\in\mathbb{N}^n$ we have (here $q$ is the conjugate exponent of $p$, $\lambda$ is the Lebesgue measure on $\mathbb{R}^n$ and $B_{K}(x) = \{ w \in \mathbb{R}^n: w = x - y, \text{ with } y \in K \} $): $$ |\partial^\varkappa(f\ast\varphi)(x)| \leq [\lambda(K)]^{1/q} ||\varphi||_\infty\bigl{(}\int_{B_K(x)} |\partial^\varkappa(f)|^p\bigr{)}^{1/p} \longrightarrow 0 \text{ as } |x|\to\infty, $$ and $$ ||\partial^\varkappa(f\ast\varphi)||_\infty \leq [\lambda(K)]^{1/q}||\varphi||_\infty ||\partial^\varkappa f||_p, $$ showing that $C_f$ is well-defined and continuous. Choose now for $k=0$ an $m$ and a $\varphi_0\in\mathcal{D}_K^0$ according to the proposition.

Now let $(\psi_i)_i$ be a sequence in $\mathcal{D}_K$ with $||\psi_i-\varphi_0||_\infty\to 0$, which is possible as $C_c^{\infty}$ is dense in $\mathcal{D}_K^0$. By the formula above, $(C_f(\psi_i))_i$ is a Cauchy sequence in $\mathcal{B}_0$, a Frechet space, so the limit $f\ast\varphi_0$ (considered in the space of distributions) is in fact an element of $\mathcal{B}_0$ too. So you conclude $$ f = f\ast\delta =f\ast (\Delta^m\varphi_0) + f\ast\xi = \Delta^m(f\ast\varphi_0)+f\ast\xi \in \Delta^m[\mathcal{B}_0] + \mathcal{B}_0 = \mathcal{B}_0. $$

  • $\begingroup$ I just realize, that the argumentation via Cauchy sequences is not necessary. One may directly apply the estimates to see $f\ast\varphi_0\in\mathcal{B}_0$. $\endgroup$ – Vobo Dec 18 '16 at 14:28
  • $\begingroup$ Dear Vobo, thank you very very much for your invaluable help: I could never solve the problem by myself! There is only a little correction to make in your formulas, I think: the constant $const_{K,f}$ actually does not depend on $f$. If we call it $C$, then we have $|| \partial^{\beta} f ||_{\infty} = || (\partial^{\beta} \Delta^{m}f) \ast \phi_0 + (\partial^{\beta}f) \ast \xi ||_{\infty} \leq C || \phi_0 ||_{\infty} ||\partial^{\beta} \Delta^{m}f ||_{p} + C || \xi ||_{\infty} ||\partial^{\beta} f ||_{p}$, which proves that $\mathcal{D}_{L^p} \hookrightarrow \mathcal{B}_{0}$ is continuous. $\endgroup$ – Maurizio Barbato Dec 18 '16 at 15:59
  • $\begingroup$ I also found the proposition about the "parametrix" you stated: it is at p. 191 of Schwartz, Theories des distributions, Hermann Paris, 1966, just after Theorem XIX of Chapter VI. $\endgroup$ – Maurizio Barbato Dec 18 '16 at 16:03
  • $\begingroup$ A final comment: from the inequality I stated in my previous comment it immediately follows that also the inclusion $\mathcal{D}_{p} \hookrightarrow \mathcal{D}_{L^q}$, with $q \geq p$ is continuous, so completing the proof of the embedding theorem stated by Schwartz (see the note in my post). $\endgroup$ – Maurizio Barbato Dec 18 '16 at 16:09
  • $\begingroup$ I have replaced $const_{K,f}$ in Vobo's answer (which as I noted before does not depend actually on $f$ and which I called simply $C$ in my previous comment) by its exact value $[\lambda(K)]^{1/q}$. $\endgroup$ – Maurizio Barbato Dec 19 '16 at 18:02

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