6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $$ f = f\ast\delta =f\ast (\Delta^m\varphi_0) + f\ast\xi = \Delta^m(f\ast\varphi_0)+f\ast\xi \in \Delta^m[\mathcal{B}_0] + \mathcal{B}_0 = \mathcal{B}_0. $$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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