A short note about tempered distributions

Recently I am doing distributions theory, this notes is for filling small gaps and showing some “applications”. I should put some definitions at the top to make this article self-containing. Unfortunately I do not have much time, so I go directly to what I hope to say first and I will fill in the background later.

Proposition          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,

where |x| is the Euclidean norm)

induces a tempered distribution on \mathbb{R}^n, given by

\displaystyle\langle f,\varphi \rangle = \int f\varphi for all \varphi \in \mathcal{S}(\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

Example 1 (Linear analysis prelim, 2009)          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 I learnt. 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 three facts, which can be checked by computations (on \mathbb{R}^n):

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

\widehat{\delta} = 1; and


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

Example 2 (Linear analysis prelim, 2008)

Show that, for any tempered distribution \mu on \mathbb{R}^n, there is a unique tempered distribution \omega that solves

\Delta \omega - \omega = \mu.

Consider n=2. If \mu is the distribution \displaystyle \langle \mu,\phi \rangle = \int_{\mathbb{R}} \phi(0,y)dy, show that \omega is a function, defined at every point (x,y) and bounded by a constant. Also, for each 0\leq \theta<2\pi, calculate \displaystyle \lim_{r\rightarrow \infty} \omega (r\cos\theta,r\sin\theta).

Solution. Thanks to the instructor of linear analysis providing the main idea of this problem. (To be continued)



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