Given in $\mathcal S(\mathbb R)$ the following functional: $$F_n: <F_n|\phi> = \int_0^\infty dx \sin^2(nx) \phi(x)$$
I have to prove that this is a tempered distribution, to get the derivate, and to determine the limit.
I can't find exercises of this kind so if somebody has suggestions, links or material to share on these topics that would be great.
To prove that this is a tempered distribution I've tried: $$|<F_n|\phi>| \le \int_0^\infty dx \sin^2(nx)|\phi(x)| \le 1 \sup_x|\phi(x)| $$
And, since $\phi \in S(\mathbb R)$ this should be limited. I'm not sure that this proves that $F_n$ is a distribution though, and I was looking for a way to apply directly the theorem that states that $$f \in S(\mathbb R) \iff \exists \xi \in \mathcal C^0 : |\xi(x)| \le C(1+|x|^n)$$ for some $C, n, k \ge 0 : $ $$<f|\phi>=<\xi^{(k)}|\phi>=(-1)^k \int dx \,\xi(x)\phi^{(n)}(x) $$ Also, I have no clue on how to solve the other two questions of the problem...
\begin{align*} \left<F_{n},\phi\right>&=\int_{0}^{\infty}\dfrac{|\sin(nx)|}{(1+|x|)^{2}}(1+|x|)^{2}|\phi(x)|dx\\ &\leq\int_{0}^{\infty}\dfrac{1}{(1+|x|)^{2}}[\rho_{0,0}(\phi)+2\rho_{1,0}(\phi)+\rho_{2,0}(\phi)]dx\\ &\leq C[\rho_{0,0}(\phi)+2\rho_{1,0}(\phi)+\rho_{2,0}(\phi)], \end{align*} where $C=\displaystyle\int_{0}^{\infty}\dfrac{1}{(1+|x|)^{2}}dx<\infty$, $\rho_{\alpha,\beta}(\phi)=\sup_{x\in{\bf{R}}}|x|^{\alpha}|\phi^{(\beta)}(x)|$, so $F_{n}$ is a tempered distribution.
\begin{align*} \left<F_{n}',\phi\right>&=-\left<F_{n},\phi'\right>\\ &=-\int_{0}^{\infty}\sin^{2}(nx)\phi'(x)dx\\ &=-\dfrac{1}{2}\int_{0}^{\infty}(1-\cos(2nx))\phi'(x)dx\\ &=-\dfrac{1}{2}[\phi(\infty)-\phi(0)]+\dfrac{1}{2}\left[\cos(2nx)\phi(x)\bigg|_{x=0}^{x=\infty}+2n\int_{0}^{\infty}\sin(2nx)\phi(x)dx\right]\\ &=\dfrac{1}{2}\phi(0)-\dfrac{1}{2}\phi(0)+n\int_{0}^{\infty}\sin(2nx)\phi(x)dx\\ &=n\int_{0}^{\infty}\sin(2nx)\phi(x)dx. \end{align*}
And \begin{align*} \left<F_{n},\phi\right>&=\dfrac{1}{2}\int_{0}^{\infty}(1-\cos(2nx))\phi(x)dx\\ &=\dfrac{1}{2}\int_{0}^{\infty}\phi(x)dx-\dfrac{1}{2}\int_{0}^{\infty}\cos(2nx)\phi(x)dx\\ &=\dfrac{1}{2}\int_{0}^{\infty}\phi(x)dx-\dfrac{1}{2}\int_{-\infty}^{\infty}\cos(2nx)\xi(x)dx, \end{align*} where $\xi(x)=\phi(x)$ for $x\geq 0$ and $\xi(x)=0$ for $x<0$, then $\xi\in L^{1}({\bf{R}})$, then by Riemann-Lebesgue Lemma, one has \begin{align*} \int_{-\infty}^{\infty}\cos(2nx)\xi(x)dx\rightarrow 0 \end{align*} as $n\rightarrow\infty$, so \begin{align*} \left<F_{n},\phi\right>\rightarrow\dfrac{1}{2}\int_{0}^{\infty}\phi(x)dx \end{align*} as $n\rightarrow\infty$.