I need prove, that a distribution $$\langle F_f,\phi \rangle= p.v. \int\limits_{-1/2}^{1/2}\frac{\phi(t)}{t\cdot \ln{|t|}}\mathrm{d}t$$ belongs to $S^\prime$ (adjoint to Schwartz space) and I need find, Fourier transform of the distribution.
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|||||||||||||
|
|
If $\phi\in\mathcal{S}$, $\psi(t)=\frac{\phi(t)-\phi(0)}{t}\in C^\infty$, and by the Mean Value Theorem, $\|\psi\|_{L^\infty}\le\|\phi'\|_{L^\infty}$ Thus, $$ \begin{align} \lim_{\epsilon\to0}\left(\int_{-1/2}^{-\epsilon}\frac{\phi(t)}{t\log|t|}\,\mathrm{d}t+\int_{\epsilon}^{1/2}\frac{\phi(t)}{t\log|t|}\,\mathrm{d}t\right) &=\lim_{\epsilon\to0}\left(\int_{-1/2}^{-\epsilon}\frac{\psi(t)}{\log|t|}\,\mathrm{d}t+\int_{\epsilon}^{1/2}\frac{\psi(t)}{\log|t|}\,\mathrm{d}t\right)\\ &=\int_{-1/2}^{1/2}\frac{\psi(t)}{\log|t|}\,\mathrm{d}t\tag{1} \end{align} $$ Therefore $$ \begin{align} \left|\text{p.v.}\int_{-1/2}^{1/2}\frac{\phi(t)}{t\log|t|}\,\mathrm{d}t\right|&=\left|\int_{-1/2}^{1/2}\frac{\psi(t)}{\log|t|}\,\mathrm{d}t\right|\\ &\le\left|\int_{-1/2}^{1/2}\frac{1}{\log|t|}\,\mathrm{d}t\right|\;\|\psi\|_{L^\infty}\\ &\le\frac{1}{\log(2)}\|\phi'\|_{L^\infty}\tag{2} \end{align} $$ Inequality $(2)$ says that $\langle F_f,\cdot\rangle\in\mathcal{S}'$. The Fourier Transform of the distribution would be $$ \begin{align} \text{p.v.}\int_{-1/2}^{1/2}\frac{e^{-2\pi i\xi t}}{t\log|t|}\,\mathrm{d}t &=-i\int_{-1/2}^{1/2}\frac{\sin(2\pi\xi t)}{t\log|t|}\,\mathrm{d}t\\ &=-2i\int_0^{1/2}\frac{\sin(2\pi\xi t)}{t\log|t|}\,\mathrm{d}t\tag{3} \end{align} $$ I can not find a closed form for $(3)$. |
|||||||||||
|
