How can prove that $T$ is a distribution?

Question

Let $f_t: \mathbb R^2 \to \mathbb R$ define by $f_t(x,y) = t \sin(t|x^2+y^2-1|)$.

For all $\phi\in D(\mathbb R^2-\{(0,0)\})$, we denote by $T$ the map given by $$T(\phi)=\lim_{t\to +\infty}\int_{\mathbb R^2} f_t(x,y) \, \phi(x,y)\, \mathrm{d}x\, \mathrm{d}y$$

How can prove that $T$ is a distribution of order $0$ on $\mathbb R^2-\{(0,0)\}$?

Remark: I have prove that $T: D(\mathbb R^2-\{(0,0)\}) \to \mathbb R $ is a linear map. It remains to show $T$ is continuous. In fact, I started by using the polar coordinate system, then we get $$T(\phi)=\lim_{t\to +\infty}\int_{0}^{2\pi} \int_{0}^{\infty}\, t \sin(t|r^2-1|) \, \phi(r,\theta)\,r\, \mathrm{d}r\, \mathrm{d}\theta=... ??$$

Thank you in advance

Answer 1

First, we note that for $\mathrm{supp}(\phi)\subset \Omega = B_{r_2}(0)\setminus B_{r_1}(0)$ for some $0 < r_1 < r_2$, \begin{align*} I_t(\phi) &:= \int_0^{2\pi}\int_{r_1}^{r_2} t\sin(t\lvert r^2-1\rvert)\phi(r, \theta)\,r\,\mathrm{d}r\,\mathrm{d}\theta \\ &= \int_{r_1}^{r_2} t\sin(t\lvert r^2-1\rvert)\left[\int_0^{2\pi} \phi(r, \theta)\,\mathrm{d}\theta\right]r\,\mathrm{d}r \\ &= \Im\left[\int_{r_1}^{r_2} te^{it\lvert r^2-1\rvert}\left[\int_0^{2\pi} \phi(r, \theta)\,\mathrm{d}\theta\right]r\,\mathrm{d}r\right] \end{align*} We let $\Phi(r) := \int_0^{2\pi} \phi(r, \theta)\,\mathrm{d}\theta$. First, we note that for $r_i > 1$, $$\int_1^{r_i} te^{it\lvert r^2-1\rvert}\Phi(r)\,r\,\mathrm{d}r = \int_1^{r_i} te^{it(r^2-1)}\Phi(r)\,r\,\mathrm{d}r = \frac{t}{2}\int_0^{\sqrt{1+r_i}} e^{itu}\Phi\left(\sqrt{1+u}\right)\,\mathrm{d}u$$ for $u = r^2-1$. Recall from the proof of the Riemann-Lebesgue lemma that $$\left\lvert\int_0^{\sqrt{1+r_i}} e^{itu}\Phi\left(\sqrt{1+u}\right)\,\mathrm{d}u\right\rvert\leq \frac{1}{t}\int_0^{\sqrt{1+r_i}} \left\lvert\frac{\Phi'(\sqrt{1+u})}{2\sqrt{1+u}}\right\rvert\,\mathrm{d}u$$ and $\Phi'(r) = \int_0^{2\pi} \partial_r\phi(r, \theta)\,\mathrm{d}\theta$, so \begin{align*} \left\lvert\int_1^{r_i} te^{it\lvert r^2-1\rvert}\Phi(r)\,r\,\mathrm{d}r\right\rvert &\leq \frac{1}{2}\int_0^{\sqrt{1+r_i}} \frac{1}{2\sqrt{1+u}}\left\lvert\int_0^{2\pi} \partial_r\phi(\sqrt{1+u}, \theta)\,\mathrm{d}\theta\right\rvert\,\mathrm{d}u \\ &= \frac{1}{2}\int_1^{r_i} \left\lvert\int_0^{2\pi} \partial_r\phi(r, \theta)\,\mathrm{d}\theta\right\rvert\,\mathrm{d}r \end{align*} We can show a similar bound for the case $r_i\leq 1$, which allows us to write \begin{align*} \lvert I_t(\phi)\rvert &\leq \frac{1}{2}\int_{r_1}^{r_2} \left\lvert\int_0^{2\pi} \partial_r\phi(r, \theta)\,\mathrm{d}\theta\right\rvert\,\mathrm{d}r \\ &\leq \frac{1}{2}\int_{r_1}^{r_2}\int_0^{2\pi} \lvert \partial_r\phi(r, \theta)\rvert\,\mathrm{d}r\,\mathrm{d}\theta \\ &\leq \pi(r_2-r_1)\|\phi\|_{C^1} \end{align*} for all $t$, which implies $\lvert T(\phi)\rvert\leq \pi(r_2-r_1)\|\phi\|_{C^1}$. Therefore, $T$ is a distribution and has order at most 1. I'm not sure how to improve this to order 0.

Answer 2

Lemma. Let $\varphi \in C_c^1(\mathbb{R})$. Then

$$ \lim_{t\to\infty} \int_{\mathbb{R}} \frac{t \sin(t|u|)}{2} \varphi(u) \, du = \varphi(0). $$

Proof of Lemma. We have

\begin{align*} \int_{\mathbb{R}} \frac{t \sin(t|u|)}{2} \varphi(u) \, du &= \int_{0}^{\infty} t \sin(tu) \cdot \frac{\varphi(u) + \varphi(-u)}{2} \, du \\ &= \varphi(0) + \int_{0}^{\infty} \cos(tu) \left( \frac{\varphi'(u) - \varphi'(-u)}{2} \right) \, du \\ &\xrightarrow[t\to\infty]{} \varphi(0) \end{align*}

by the Riemann-Lebesgue lemma. ////

Now we return to the original problem. Let $\phi \in \mathcal{D}(\mathbb{R}^2\setminus\{0\})$ and write

$$ \varphi(r) = \int_{\partial B_1(0)} \phi(r\omega) \, \sigma(\mathrm{d}\omega), $$

where $\sigma$ is the surface measure on $\partial B_1(0)$. Then applying polar coordinates,

\begin{align*} \int_{\mathbb{R}^2} f_t(x, y) \phi(x, y) \, \mathrm{d}x\mathrm{d}y &= \int_{0}^{\infty} rt \sin(t|r^2-1|)\varphi(r) \, \mathrm{d}r \\ &= \int_{-1}^{\infty} \frac{t\sin(t|u|)}{2} \varphi\left(\sqrt{u+1}\right) \end{align*}

where $u = r^2 - 1$. Now taking $t\to\infty$ and applying the lemma together with the observation $\varphi \in C_c^{1}((-1,\infty))$, we have

$$ T\phi = \varphi(1) = \int_{\partial B_1(0)} \phi(\omega) \, \sigma(\mathrm{d}\omega). $$

That is, $T\phi$ is the $2\pi$ times the average value of $\phi$ on the unit circle. This gives

$$|T\phi| \leq 2\pi \|\varphi\|_{\sup}$$

and hence $T$ is a distribution of order $0$.