## 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

$\widehat{1}=(2\pi)^{n}\delta$.

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)