Fourier transform of tempered distributions

This is an extension to my previous post A short note about tempered distributions. Our objective is to discuss the Fourier transform of distributions and give some “little” applications. Before that let us review some basics.

We shall use the multi-index notation. Denote by C^\infty_c (X) the space of compactly supported smooth functions on X which is an open set in \mathbb{R}^n.

Definition 1          A linear map u:C^\infty_c(X)\rightarrow \mathbb{C} is called a distribution if for any compact set K\subset X there are C>0 and a nonnegative integer N such that

\displaystyle|\langle u,\varphi \rangle| \leq C\sum_{|\alpha|\leq N}\sup_x|\partial^\alpha \varphi|

for any \varphi \in C^\infty_c(X) and supp(\varphi)\subset K. Denote by \mathcal{D}'(X) the space of distributions.

Example 1 (Delta distribution on \mathbb{R}^n)         \langle\delta,\varphi\rangle = \varphi(0).

Example 2       Let L^1_{loc}(\mathbb{R}^n) be the space of locally integrable functions, that is, functions which are integrable on any compact subset of \mathbb{R}^n. If f\in L^1_{loc}(\mathbb{R}^n), then f induces a distribution given by

\displaystyle \langle f,\varphi \rangle = \int_{\mathbb{R}^n} f\varphi.

For example, \log|x| is locally integrable on \mathbb{R} so it defines a distribution.

Definition 2 (Restriction of a distribution)          If X'\subset X is open, then given any u\in \mathcal{D}'(X) one can define the restriction of u to X' by: given any \varphi\in C^\infty_c(X'), we view \varphi\in C^\infty_c(X) by zero extension, then define

\langle u\mid_{X'} ,\varphi\rangle = \langle u,\varphi\rangle.

Example 3  (p.v. 1/x on \mathbb{R})          Define

\displaystyle \langle p.v.1/x,\varphi \rangle = \lim_{\varepsilon \rightarrow 0} \int_{|x|\geq \varepsilon} \frac{\varphi(x)}{x}dx.

By mean value theorem one can check p.v.1/x defines a distribution. Clearly its  restriction to \mathbb{R}\setminus\{0\} is the distribution 1/x.

Definition 4  (distributional derivative)        Given u\in C^\infty_c(X) one defines

\langle \partial^\alpha u,\varphi\rangle= (-1)^{|\alpha|}\langle u,\partial^\alpha \varphi \rangle.

One can see that this definition comes from integration by parts formula: \int f\partial g = - \int g\partial f whenever one of f,g has compact support in \mathbb{R}^n.

Example 4          \partial \log|x| = p.v.1/x.

Example 5          Define the Heaviside function H by H(x)=0 if x\leq 0 and H(x)=1 if x>0. Then \partial H = \delta.

Definition 5 (support of a distribution)          Given u\in \mathcal{D}'(X) one defines the support of u by

supp(u) = complement of the largest open set such that the restriction of u to that set is 0.

Theorem 1 (characterization of zero support distribution)          If u\in \mathcal{D}'(\mathbb{R}^n) be such that supp(u)=\{0\}, then there are a nonnegative integer N and complex numbers c_\alpha such that

\displaystyle u = \sum_{|\alpha|\leq N}c_\alpha \partial^\alpha \delta.

Definition 6          Given f\in L^1 (\mathbb{R}^n). Define the Fourier transform of f by \displaystyle \hat{f}(\xi) = \int f(x)e^{-ix\cdot \xi}d\xi.

Example 6          \displaystyle \widehat{ \exp\left(-\frac{1}{2}|x|^2\right)} = (2\pi)^{n/2}\exp\left(-\frac{1}{2}|\xi|^2\right).

Definition 7 (Schwartz class)          A function \varphi\in C^\infty(\mathbb{R}^n) is called rapidly decreasing if for all \alpha,\beta, \displaystyle\|\varphi\|_{\alpha,\beta} = \sup |x^\alpha \partial^\beta \varphi|<\infty. Denote by \mathcal{S}(\mathbb{R}^n) the space of such functions, and call this collection the Schwartz class. The above seminorms \|\varphi\|_{\alpha,\beta} generate a complete metrizable topology (see Frechet space) of \mathcal{S}(\mathbb{R}^n).

Theorem 2          The space C^\infty_c(\mathbb{R}^n) is dense in \mathcal{S}(\mathbb{R}^n).

Definition 8 (tempered distribution)          A linear map u:\mathcal{S}(\mathbb{R}^n)\rightarrow \mathbb{C} is called a tempered distribution if there is a constant C\geq 0 and a nonnegative integer N such that

\displaystyle |\langle u,\varphi \rangle|\leq C\sum_{|\alpha|,|\beta|\leq N}\sup |x^\alpha \partial^{\beta} \varphi|

for all \varphi \in \mathcal{S}(\mathbb{R}^n). Denote by \mathcal{S}(\mathbb{R}^n) the space of tempered distributions.

Proposition 3        A locally integrable function f on \mathbb{R}^n of polynomial growth (that is, there exist C\geq 0 and non-negative integer M such that

                    |f(x)|\leq C(1+|x|)^M on \mathbb{R}^n),

induces a tempered distribution on \mathbb{R}^n.

Proof. Since f is locally integrable, it induces a distribution. Next we check f\varphi is integrable for \varphi\in \mathcal{S}(\mathbb{R}^n). Indeed for such \varphi, |f\varphi| = |f|(1+|x|)^{-M}(1+|x|)^M |\varphi| \leq C (1+|x|)^M |\varphi| which is integrable as \varphi is rapidly decreasing. So f\varphi is integrable. Thus \langle f,\cdot \rangle is well defined on \mathcal{S}(\mathbb{R}^n).

It remains to establish a semi-norm estimate for \langle f,\cdot \rangle in order to show it is continuous on \mathcal{S}(\mathbb{R}^n). For each \varphi \in \mathcal{S}(\mathbb{R}^n),

\displaystyle |\langle f,\varphi \rangle|\leq C \int (1+|x|)^M |\varphi| = C \int (1+|x|)^{M+2n} |\varphi| (1+|x|)^{-2n} \leq C_1 C_2 \sum_{|\alpha|,|\beta|\leq N} \|\varphi\|_{\alpha,\beta}

where C_1\geq 0 is some constant, \displaystyle C_2 = \int (1+|x|)^{-2n} dx <\infty and N is some nonnegative integer.                \square

Definition 9          The Fourier transform \hat{u} of u\in \mathcal{S}'(\mathbb{R}^n) is given by \langle \hat{u},\varphi \rangle = \langle u,\hat{\varphi}\rangle.

One idea of defining Fourier transform like this comes from the fact that

\displaystyle \int f(x)\hat{g}(x)dx = \int \hat{f}(\xi)g(\xi) d\xi for all f,g\in L^1(\mathbb{R}^n). This fact can be proved by Fubini’s Theorem.

Theorem 4          The Fourier transform \mathcal{F}:u\mapsto \hat{u} is a continuous isomorphism \mathcal{S}(\mathbb{R}^n)\rightarrow \mathcal{S}(\mathbb{R}^n), so does its inverse.

Example 7          \widehat{\delta} = 1 and  \widehat{1}=(2\pi)^{n}\delta. Also \widehat{x^\alpha} = (-1/i)^{|\alpha|}(2\pi)^n \partial^\alpha \delta.

Example 8         Show that

\displaystyle \langle u_1, \varphi \rangle = \int_{\mathbb{R}} \varphi(x,0)dx and

\displaystyle \langle u_2, \varphi \rangle = \int_{\mathbb{R}} \varphi(0,y)dy

are tempered distributions on \mathbb{R}^2. Also show that \widehat{u_1} = 2\pi u_2, where \widehat{} denotes the Fourier transform.

Solution. For the first part, a direct check would work. Or it is just an application to some general results. Let \delta be the Dirac delta distribution. For any two distributions u and v on \mathbb{R}^n, one can define the tensor product

\langle u \otimes v, \varphi \rangle = \langle u(x), \langle v(y),\varphi(x,y) \rangle \rangle = \langle v(y), \langle u(x),\varphi(x,y) \rangle \rangle for \varphi \in C^{\infty}_c (\mathbb{R}^{2n}).

Using semi-norm estimates one can check u \otimes v is a tempered distribution on \mathbb{R}^{2n} whenever u and v are tempered distributions on \mathbb{R}^n. In our case, u_1 = 1\otimes \delta while u_2 = \delta \otimes 1. By the above proposition 1 is a tempered distribution on \mathbb{R}. It is known that \delta is tempered too. Thus u_1 = 1\otimes \delta is a tempered distribution on \mathbb{R}^2, and so is u_2. The second part follows from the following facts which can be checked by computations (on \mathbb{R}^n):

\widehat{u\otimes v} = \widehat{u} \times \widehat{v};

In our case, \widehat{u_1}=\widehat{1\otimes\delta}=(2\pi)\delta\otimes 1 = 2\pi u_2.                        \square

Proposition 5           Let \displaystyle P=P(\partial)=\sum_{|\alpha|\leq m}c_{\alpha}(i)^{-|\alpha|}\partial^\alpha be a linear differential operator with constant coefficients. Denote \displaystyle P(\xi) = \sum_{|\alpha|\leq m}c_\alpha \xi^\alpha. If P(\xi)\neq 0 whenever \xi\in \mathbb{R}^n\setminus\{0\} and u\in \mathcal{S}'(\mathbb{R}^n) satisfies Pu=0, then u is a polynomial.

Proof. Taking Fourier transform on both sides of Pu=0 one gets P(\xi)\hat{u}(\xi)=0 (Note \widehat{\partial^\alpha u}=(i)^{|\alpha|}\xi^\alpha \hat{u}). The assumption on P implies supp(\hat{u})=\{0\}. So by Theorem 1,

\displaystyle \hat{u} = \sum_{|\alpha|\leq N}b_\alpha \partial^\alpha \delta

which is the Fourier transform of \displaystyle \sum_{|\alpha|\leq N}b_\alpha' x^\alpha for some constants b_\alpha' (cf. Example 7). Then by Theorem 4, u is a polynomial.                      \square

Corollary 6 (Liouville’s Theorem in complex variables)           Let f be an entire function satisfying

|f(x)|\leq C(1+|x|)^N

for some constant C>0 and nonnegative integer N. Then f is a polynomial of degree at most N.

Proof. By Proposition 3 one can view f\in \mathcal{S}'(\mathbb{R}^2). Let P(\xi)=\xi_1 + i \xi_2. Clearly $P$ is nonzero if (\xi_1,\xi_2)\neq (0,0). And P(\partial) = (-i) (\partial_x +i\partial_y). Since f is entire, it satisfies the Cauchy-Riemann equations so \partial_x f+ i \partial_y f=0. Thus Pf=0. By the previous proposition f is a polynomial. By the given estimate f has degree at most N.                  \square

Corollary 7          Let u\in C^2(\mathbb{R}^n) be such that

|u(x)|\leq C(1+|x|)^N

for some constant C>0 and nonnegative integer N. If \Delta u is a polynomial then u is a polynomial.

Proof. Let \displaystyle \Delta u = \sum_{|\alpha|\leq m}c_\alpha x^\alpha. Taking Fourier transform on both sides and following the same argument as the proof of Proposition 5, one sees that supp(\hat{u})=\{0\} so u is a polynomial.                        \square

The following is a direct consequence.

Corollary 8          If u is a harmonic function on \mathbb{R}^n satisfying

|u(x)|\leq C(1+|x|)^N

for some constant C>0 and nonnegative integer N, then u is a polynomial of degree at most N.

For the case n=2, this result can be proved using tools in complex analysis say Poisson integral formula and Cauchy estimate.

(To be continued…)

This entry was posted in Analysis, Functional analysis. Bookmark the permalink.

One Response to Fourier transform of tempered distributions

  1. paullailai says:

    An easy exercise, If \Lambda is a lattice on ${\Bbb R}^d$, then

    \widehat{\delta_{\Lambda}}  = \frac{1}{|\det(\Lambda)|}\delta_{\Lambda^{o}}

    where \Lambda^{o} is the dual lattice of \Lambda. i.e.

    \Lambda^{o} = \{x: \langle\lambda,x\rangle\in{\Bbb Z} \ \mbox{for all } \ \lambda\in\Lambda\}.

    (Hint: Use Poisson Summation Formula)

    In general, what can we say about \widehat{\delta_{\Lambda}} when \Lambda is not a lattice. Is it a measure? That’s not known.

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