Oliver-Taleb Phase Shift

Euly Oliver, Loïc Taleb

Definition

\phaseshift\alpha f(x) = {1/2\pi} \int_{-\infty}^\infty\!\hat f(\xi) e^{i\xi x}e^{-i\sgn\(\xi\)\alpha}\,d\xi
\hat f(\xi) =\int_{-\infty}^{\infty}\!f(t) e^{-i\xi t}\,dt
\sgn\(x\)=\begin{cases}-1&\text{if }x\lt0\\\ \,0&\text{if }x=0\\\ \,1&\text{if }x\gt0\end{cases}

Alt Definition

\phaseshift{\alpha}f = \FT^{-1}(\xi\mapsto\FT\!f(\xi)e^{-i\sgn\(\xi\)\alpha})
\FT(f)(\xi) = \int_{-\infty}^{\infty}\!f(x) e^{-i\xi x}\,dx
\FT^{-1}(f)(x) = {1/2\pi}\int_{-\infty}^{\infty}\!f(\xi) e^{i\xi x}\,d\xi
\sgn\(x\)=\begin{cases}-1&\text{if }x\lt0\\\ \,0&\text{if }x=0\\\ \,1&\text{if }x\gt0\end{cases}

Properties

Identity: \phaseshift{2\pi n} f(x) = f(x)
Proof: Euly
\begin{aligned} \phaseshift{2\pi n} f(x) &= {1/2\pi} \int_{-\infty}^\infty\!\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)\,2\pi n}\,d\xi &= {1/2\pi} \int_{-\infty}^\infty\!\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)}(e^{2\pi})^n\,d\xi &= {1/2\pi} \int_{-\infty}^\infty\!\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)}(1)^n\,d\xi &= {1/2\pi} \int_{-\infty}^\infty\!\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)}\,d\xi &= f(x) \end{aligned} {} n\in\Z


Additivity: \phaseshift\alpha ({f+g})(x)=\phaseshift\alpha f(x)+\phaseshift\alpha g(x)
Proof: Euly
\begin{aligned} \phaseshift\alpha ({f+g})(x) &\overset{\*^1}= {1/2\pi} \int_{-\infty}^\infty\!(\hat f\(\xi\)+\hat g\(\xi\))e^{i\xi x}e^{-i\sgn\(\xi\)\alpha}\,d\xi &= {1/2\pi} \int_{-\infty}^\infty\!\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)\alpha}+\hat g\(\xi\)\ e^{i\xi x}e^{-i\sgn\(\xi\)\alpha}\,d\xi &= {1/2\pi} \int_{-\infty}^\infty\!\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)\alpha}\,d\xi+{1/2\pi} \int_{-\infty}^\infty\!\hat g\(\xi\)\ e^{i\xi x}e^{-i\sgn\(\xi\)\alpha}\,d\xi &= \phaseshift\alpha f(x)+\phaseshift\alpha g(x) \end{aligned}

\*^1\ \ Implied by linearity of Fourier transform.

Homogeneity: \phaseshift\alpha c*\!f\(x\) = c*\phaseshift\alpha f\(x\)
Proof: Euly
\begin{aligned} \phaseshift\alpha c*\!f\(x\) &\overset{\*^1}= {1/2\pi} \int_{-\infty}^\infty\!c*\!\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)\alpha}\,d\xi &= c*{1/2\pi} \int_{-\infty}^\infty\!\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)\alpha}\,d\xi &= c*\phaseshift\alpha f\(x\) \end{aligned}

\*^1\ \ Implied by linearity of Fourier transform.

Linearity of Composition: \phaseshift{\alpha} \phaseshift{\beta} f\(x\) = \phaseshift{\alpha+\beta} \!f\(x\)
Proof: Euly
\begin{aligned} \phaseshift{\alpha}\phaseshift{\beta}f &= \FT^{-1}(\xi\mapsto\FT\Big\(\phaseshift{\beta}f\Big\)(\xi)*e^{-i\sgn\(\xi\)\alpha}) &= \FT^{-1}(\xi\mapsto\FT\FT^{-1}(\zeta\mapsto\FT(f)*e^{-i\sgn\(\zeta\)\beta})(\xi)*e^{-i\sgn\(\xi\)\alpha}) &= \FT^{-1}(\xi\mapsto\FT(f)*e^{-i\sgn\(\xi\)\beta}*e^{-i\sgn\(\xi\)\alpha}) &= \FT^{-1}(\xi\mapsto\FT(f)*e^{-i\sgn\(\xi\)\beta-i\sgn\(\xi\)\alpha}) &= \FT^{-1}(\xi\mapsto\FT(f)*e^{-i\sgn\(\xi\)\(\alpha+\beta\)}) &= \phaseshift{\alpha+\beta}f \end{aligned}


Inverse: \phaseshift\alpha {}^{-1}=\phaseshift{-\alpha}
Proof: Euly
\begin{aligned} \phaseshift\alpha f &= \FT ^{-1}\Big\(\xi\mapsto\FT f\(\xi\)*e^{-i\sgn\(\xi\)\alpha}\Big\) \FT\Big\(\phaseshift\alpha f\Big\) &= \FT\FT ^{-1}\Big\(\xi\mapsto\FT f\(\xi\)*e^{-i\sgn\(\xi\)\alpha}\Big\) \FT\Big\(\phaseshift\alpha f\Big\) &= \xi\mapsto\FT f\(\xi\)*e^{-i\sgn\(\xi\)\alpha} \xi\mapsto\FT\Big\(\phaseshift\alpha f\Big\)\(\xi\)*e^{-i\sgn\(\xi\)\(-\alpha\)} &= \xi\mapsto\FT f\(\xi\)*e^{-i\sgn\(\xi\)\alpha}*e^{-i\sgn\(\xi\)\(-\alpha)} \xi\mapsto\FT\Big\(\phaseshift\alpha f\Big\)\(\xi\)*e^{-i\sgn\(\xi\)\(-\alpha\)} &= \FT f \FT ^{-1}\bigg\(\xi\mapsto\FT\Big\(\phaseshift\alpha f\Big\)\(\xi\)*e^{-i\sgn\(\xi\)\(-\alpha\)}\!\bigg\) &= \FT ^{-1}\FT f \phaseshift{-\alpha} \Big\(\phaseshift\alpha f\Big\) &= f\qquad\text{let }\beta=-\alpha \phaseshift{-\beta} \Big\(\phaseshift{\beta} f\Big\) &= f \phaseshift{-\(-\alpha\)} \Big\(\phaseshift{-\alpha} f\Big\) &= f \phaseshift{\alpha} \Big\(\phaseshift{-\alpha} f\Big\) &= f \phaseshift{-\alpha} &= \phaseshift\alpha {}^{-1} \end{aligned}


Identity: \phaseshift{\pi} f\(x)=-f\(x)
Proof: Euly
\begin{aligned} \phaseshift\pi f\(x\) &= {1/2\pi} \int_{-\infty}^\infty\!\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)\pi}\,d\xi &= {1/2\pi}(\int_{-\infty}^0\!\hat f\(\xi\)\,e^{i\xi x}e^{i\pi}\,d\xi + \int_0^\infty\!\hat f\(\xi\)\,e^{i\xi x}e^{-i\pi}\,d\xi) &= {1/2\pi}(\int_{-\infty}^0\!\hat f\(\xi\)\,e^{i\xi x}(-1)\,d\xi + \int_0^\infty\!\hat f\(\xi\)\,e^{i\xi x}(-1)\,d\xi) &= -{1/2\pi}(\int_{-\infty}^0\!\hat f\(\xi\)\,e^{i\xi x}\,d\xi + \int_0^\infty\!\hat f\(\xi\)\,e^{i\xi x}\,d\xi) &= -{1/2\pi}\int_{-\infty}^\infty\!\hat f\(\xi\)\,e^{i\xi x}\,d\xi &= -f(x) \end{aligned}


Lemma: \phaseshift\alpha f\(x) = {1/2\pi}\!(e^{i\alpha}\!\int_{-\infty}^0\hat f\(\xi\) \ e^{i\xi x} \ d\xi+e^{-i\alpha}\!\int_0^\infty\hat f\(\xi\) \ e^{i\xi x} \ d\xi)
Proof: Loïc
\begin{aligned} \phaseshift\alpha f\(x) &= {1/2\pi}\int_{-\infty}^{\infty}\hat f\(\xi\) \ e^{-i\sgn\(\xi\)\alpha} e^{i\xi x} \ d\xi &= {1/2\pi}\!(\int_{-\infty}^0\hat f\(\xi\) \ e^{i\alpha}e^{i\xi x} \ d\xi+\!\int_0^\infty\hat f\(\xi\) \ e^{-i\alpha}e^{i\xi x} \ d\xi) &= {1/2\pi}\!(e^{i\alpha}\!\int_{-\infty}^0\hat f\(\xi\) \ e^{i\xi x} \ d\xi+e^{-i\alpha}\!\int_0^\infty\hat f\(\xi\) \ e^{i\xi x} \ d\xi) \end{aligned}


Identity: \phaseshift{\alpha} f\(x)+\phaseshift{-\alpha} f\(x)=2\cos(\alpha)f\(x)
Proof: Loïc
\begin{aligned} \phaseshift\alpha f\(x) &= {1/2\pi}\!(e^{i\alpha}\!\int_{-\infty}^0\hat f\(\xi\) \ e^{i\xi x} \ d\xi+e^{-i\alpha}\!\int_0^\infty\hat f\(\xi\) \ e^{i\xi x} \ d\xi) \phaseshift{-\alpha} f\(x) &= {1/2\pi}\!(e^{-i\alpha}\!\int_{-\infty}^0\hat f\(\xi\) \ e^{i\xi x} \ d\xi+e^{i\alpha}\!\int_0^\infty\hat f\(\xi\) \ e^{i\xi x} \ d\xi) \end{aligned} \begin{aligned} \phaseshift{\alpha} f\(x)+\phaseshift{-\alpha} f\(x) &= {1/2\pi}\!(e^{i\alpha}\!\int_{-\infty}^0\hat f\(\xi\) \ e^{i\xi x} \ d\xi+e^{-i\alpha}\!\int_0^\infty\hat f\(\xi\) \ e^{i\xi x} \ d\xi) & \ \ \ \ \ + {1/2\pi}\!(e^{-i\alpha}\!\int_{-\infty}^0\hat f\(\xi\) \ e^{i\xi x} \ d\xi+e^{i\alpha}\!\int_0^\infty\hat f\(\xi\) \ e^{i\xi x} \ d\xi) &= {1/2\pi}\!((e^{i\alpha}\!+\!e^{-i\alpha})\!\int_{-\infty}^0\hat f\(\xi\) \ e^{i\xi x} \ d\xi+(e^{i\alpha}\!+\!e^{-i\alpha})\!\int_0^\infty\hat f\(\xi\) \ e^{i\xi x} \ d\xi) &= {e^{i\alpha}+e^{-i\alpha}/2\pi}\!(\int_{-\infty}^0\hat f\(\xi\) \ e^{i\xi x} \ d\xi+\int_0^\infty\hat f\(\xi\) \ e^{i\xi x} \ d\xi) &\overset{\*^1}= {\cos(\alpha)/\pi}\int_{-\infty}^{\infty}\hat f\(\xi\) \ e^{i\xi x} \ d\xi &= 2\cos(\alpha)f\(x) \end{aligned}

\*^1\ \ Identity: {e^{i\alpha}+e^{-i\alpha}/2}=\cos(\alpha)

Identity: \phaseshift{\alpha} f\(x\)-\phaseshift{-\alpha} f\(x\) = 2\sin(\alpha) \phaseshift{\pi/2} f\(x\)
Proof: Loïc
\begin{aligned} \phaseshift\alpha f\(x) &= {1/2\pi}\!(e^{i\alpha}\!\int_{-\infty}^0\hat f\(\xi\) \ e^{i\xi x} \ d\xi+e^{-i\alpha}\!\int_0^\infty\hat f\(\xi\) \ e^{i\xi x} \ d\xi) -\phaseshift{-\alpha} f\(x) &= {1/2\pi}\!(-e^{-i\alpha}\!\int_{-\infty}^0\hat f\(\xi\) \ e^{i\xi x} \ d\xi-e^{i\alpha}\!\int_0^\infty\hat f\(\xi\) \ e^{i\xi x} \ d\xi) \end{aligned} \begin{aligned} \phaseshift{\alpha} f\(x\)-\phaseshift{-\alpha} f\(x\) &= {1/2\pi} \! ((e^{i\alpha}-e^{-i\alpha})\int_{-\infty}^{0}\hat f\(\xi\)\,e^{i\xi x}\,d\xi + (e^{-i\alpha}-e^{i\alpha})\int_{0}^{\infty}\hat f\(\xi\)\,e^{i\xi x}\,d\xi) &= {e^{i\alpha}-e^{-i\alpha}/2\pi} \! (\int_{-\infty}^{0}\hat f\(\xi\)\,e^{i\xi x}\,d\xi - \int_{0}^{\infty}\hat f\(\xi\)\,e^{i\xi x}\,d\xi) &\overset{\*^1}= {i\sin(\alpha)/\pi} \! (\int_{-\infty}^{0}\hat f\(\xi\)\,e^{i\xi x}\,d\xi - \int_{0}^{\infty}\hat f\(\xi\)\,e^{i\xi x}\,d\xi) &= {\sin(\alpha)/\pi} \! (\int_{-\infty}^{0}\hat f\(\xi\)\,e^{i\xi x}e^{i{\pi/2}}\,d\xi + \int_{0}^{\infty}\hat f\(\xi\)\,e^{i\xi x}e^{-i{\pi/2}}\,d\xi) &= {\sin(\alpha)/\pi} \int_{-\infty}^{\infty}\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\){\pi/2}}\,d\xi &= 2\sin(\alpha) \phaseshift{\pi/2} f\(x\) \end{aligned}

\*^1\ \ Identity: {e^{i\alpha}-e^{-i\alpha}/2}=i\sin(\alpha)

Identity: \phaseshift{\alpha} f\(x\) = \cos(\alpha)f\(x\) + \sin(\alpha) \phaseshift{\pi/2} f\(x\)
Proof: Loïc
\begin{aligned} \phaseshift{\alpha} f\(x\) &= {\phaseshift{\alpha} f\(x\)+\phaseshift{\alpha} f\(x\) / 2} + 0 &= {\phaseshift{\alpha} f\(x\)+\phaseshift{\alpha} f\(x\) / 2} + {\phaseshift{-\alpha} f\(x\)-\phaseshift{-\alpha} f\(x\) / 2} &= {\phaseshift{\alpha} f\(x\)+\phaseshift{-\alpha} f\(x\) + \phaseshift{\alpha} f\(x\)-\phaseshift{-\alpha} f\(x\) / 2} &= {2\cos(\alpha)f\(x\) + 2\sin(\alpha) \phaseshift{\pi/2} f\(x\) / 2} &= \cos(\alpha)f\(x\) + \sin(\alpha) \phaseshift{\pi/2} f\(x\) \end{aligned}


Lemma: \phaseshift{\pi/2} f\(x\) = -\phaseshift{-{\pi/2}} f\(x\)
Proof: Euly
\begin{aligned} \sin(\alpha) \phaseshift{\pi/2} f\(x\) &= {\sin(\alpha)/2\pi} \! (\int_{-\infty}^{0}\hat f\(\xi\)\,e^{i\xi x}e^{i{\pi/2}}\,d\xi + \int_{0}^{\infty}\hat f\(\xi\)\,e^{i\xi x}e^{-i{\pi/2}}\,d\xi) &= -{\sin(\alpha)/2\pi} \! (\int_{-\infty}^{0}\hat f\(\xi\)\,e^{i\xi x}e^{i{\pi/2}}e^{-i\pi}\,d\xi + \int_{0}^{\infty}\hat f\(\xi\)\,e^{i\xi x}e^{-i{\pi/2}}e^{i\pi}\,d\xi) &= -{\sin(\alpha)/2\pi} \int_{-\infty}^{\infty}\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\){\pi/2}}e^{i\sgn\(\xi\)\pi}\,d\xi &= -{\sin(\alpha)/2\pi} \int_{-\infty}^{\infty}\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)\,({\pi/2}-\pi)}\,d\xi &= -{\sin(\alpha)/2\pi} \int_{-\infty}^{\infty}\hat f\(\xi\)\,e^{i\xi x}e^{-i\sgn\(\xi\)\,(-{\pi/2})}\,d\xi \sin(\alpha) \phaseshift{\pi/2} f\(x\) &= -\sin(\alpha) \phaseshift{-{\pi/2}} f\(x\) \phaseshift{\pi/2} f\(x\) &= -\phaseshift{-{\pi/2}} f\(x\) \end{aligned}


Linearity of Composition: \phaseshift{\alpha} \phaseshift{\beta} f\(x\) = \phaseshift{\alpha+\beta} \!f\(x\)
Proof: Euly
\begin{aligned} \phaseshift{\alpha} \phaseshift{\beta} f\(x\) &= \cos(\alpha)\phaseshift{\beta} f\(x\) + \sin(\alpha) \phaseshift{\pi/2} \phaseshift{\beta} f\(x\) &= \cos\(\alpha\)\!(\cos\(\beta\)f\(x\)+\sin(\beta)\phaseshift{\pi/2} f\(x\)) + \sin(\alpha) \phaseshift{\pi/2} \!(\cos(\beta)f\(x\)-\sin(\beta)\phaseshift{-{\pi/2}} \!f\(x\)) &= \cos(\alpha)\cos\(\beta\)f\(x\)+\cos(\alpha)\sin(\beta)\phaseshift{\pi/2} f\(x\) + \sin(\alpha)\cos(\beta)\phaseshift{\pi/2} f\(x\)-\sin(\alpha)\sin(\beta)\phaseshift{\pi/2} \phaseshift{-{\pi/2}} \!f\(x\) &= \cos(\alpha)\cos\(\beta\)f\(x\)+\cos(\alpha)\sin(\beta)\phaseshift{\pi/2} f\(x\) + \sin(\alpha)\cos(\beta)\phaseshift{\pi/2} f\(x\)-\sin(\alpha)\sin\(\beta\)f\(x\) &= \Big\(\!\cos(\alpha)\cos\(\beta\)-\sin(\alpha)\sin\(\beta\)\Big\)f\(x\) + \Big\(\!\cos\(\alpha\)\sin\(\beta\)+\sin\(\alpha\)\cos\(\beta\)\Big\)\phaseshift{\pi/2} f\(x\) &\overset{\!\!\!\*^1\*^2\!\!\!}= \cos(\alpha+\beta)f\(x\) + \sin(\alpha+\beta)\phaseshift{\pi/2} f\(x\) &= \phaseshift{\alpha+\beta} \!f\(x\) \end{aligned}

\*^1\ \ Identity: \cos(x)\cos(y)-\sin(x)\sin(y)=\cos(x+y)
\*^2\ \ Identity: \cos(x)\sin(y)+\sin(x)\cos(y)=\sin(x+y)