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Take a pseudodifferential operator $T$ of order $m$ with symbol $t(x,\xi)$ and $a=a(x)$ a smooth function with bounded derivatives of all orders (then $a$ is a symbol of order 0). Then with $R_{m-2} $ …

A preliminary remark. The operator $(d/dx)^\gamma$ is never continuous on the Schwartz space (and thus on tempered distributions) except if $\gamma$ is a non-negative integer, since you introduce a s …

Too long for a comment. Your spelling of the name of the mathematician Joseph Fourier is incorrect. Also your formula is almost impossible to read: your fractional derivative on the lhs is the Fourier …

The only local pseudo-differential operators are the differential operators and this entails that the only local Fourier multipliers are polynomials. It is a classical result due to J. Peetre, Math. S …

The fractional Laplacean $(-\Delta)^{\alpha/2}$ in $\mathbb R^n$ is, up to some constant the Fourier multiplier $\vert\xi\vert^\alpha$. So its inverse is, at least formally, the Fourier multiplier $\v …

With $q=1/p$, let me write your equation as $$ \vert D\vert u+ q u= h_0, \quad u(\pm 1)=0. $$ Multiplying the equation by $u$, we get $$\Vert{u}\Vert_{H^{1/2}_0}^2\le \Vert{u}\Vert_{H^{1/2}_0}^2+\unde …

Assuming the function $f$ of class $C^1$, you find that, using Taylor's formula with integral remainder, $$ f(t)-f(s)=(t-s) f_1(t,s),\quad \text{with $f_1$ continuous},$$ so that $ (T_\alpha f)(t)= …

No. The operator $(-∆)^s$ is the Fourier multiplier $\vert \xi\vert^{2s}$ so that, say for $f$ in the Schwartz space whose Fourier transform vanishes near the origin, we have $$ \Vert(-∆)^{-s} f\Vert …

Up to some normalization, the Fourier transform of the characteristic function $\mathbf 1_I$ of a compact interval $I$ is $$ \widehat{\mathbf 1_I}(\xi)=\frac{\sin \xi}{\xi}. $$ Obviously the function …