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Let $\mathscr F f = \frac{1}{\sqrt{2 \pi}} \int_{\mathbb R} f e^{-ix \xi} d \xi $ be the usual Fourier transform on the real line. How can I justify the expression $$ \mathscr F (\sqrt{1- \partial_x^2} f) = \sqrt{1+ \xi^2} \mathscr Ff $$ if the lefthand fourier transform exsits?

Luma
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  • The title and the body don't match up. What did you intend? $\partial_x$ or $\partial_x^2$? – Pedro Apr 20 '13 at 20:07
  • You first have to define $\sqrt{1 - \partial_x^2} f$. That's a pseudodifferential operator which is commonly defined using the Fourier transform. So your formula is a definition, not a result: $\sqrt{1 - \partial_x^2} f := \mathcal{F}^{-1} \sqrt{1 + \xi^2} \mathcal{F} f$. – Hans Engler Apr 20 '13 at 20:17

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The comment by Hans Engler--that $\sqrt{1-\partial_{x}^{2}}f := \mathcal{F}^{-1}\sqrt{1+\xi^{2}}\mathcal{F}f$, is correct. However, why this is a good definition can be justified in the following ad hoc manner: Assume that we can expand $\sqrt{1-\partial_{x}^{2}}$ as we would expand $\sqrt{1-x^{2}}$ as a binomial series: \begin{align*} \sqrt{1-\partial_{x}^{2}}f = f- \frac{\partial_{x}^{2}}{2}f - \frac{\partial_{x}^{4}}{8}f - \cdots \end{align*} Then recall that $\mathcal{F}[\partial_{x}^{n}f](\xi) = (i\xi)^{n}\mathcal{F}[f](\xi)$, so \begin{align*} \mathcal{F}\left[\sqrt{1-\partial_{x}^{2}}f\right](\xi) &= \mathcal{F}[f](\xi) -\frac{(i\xi)^2}{2}\mathcal{F}[f](\xi) -\frac{(i\xi)^{4}}{8}\mathcal{F}[f](\xi) -\cdots \\ &= \mathcal{F}[f](\xi) +\frac{\xi^2}{2}\mathcal{F}[f](\xi) -\frac{\xi^{4}}{8}\mathcal{F}[f](\xi) -\cdots \\ &= \left[1+\frac{\xi^{2}}{2} - \frac{\xi^{4}}{8} +\cdots \right]\mathcal{F}[f](\xi)\\ &= \sqrt{1+\xi^{2}}\mathcal{F}[f](\xi), \end{align*} where again the last step follows via binomial series. The problem with this whole analysis is that we have no idea what it means for the binomial series in powers of $\partial_{x}$ to converge; hence it's better to just take $\sqrt{1-\partial_{x}^{2}}f := \mathcal{F}^{-1}\sqrt{1+\xi^{2}}\mathcal{F}f$ as a definition.

user14717
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