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The Fourier transform of a radially symmetric function is a radially symmetric function. Basic scaling shows that the Fourier transform of $\|\xi\|$ must be $\|x\|^{-n-1}$ in some sense. Some care must be taken in interpreting this, since this function is not integrable. A good reference for this is Gelfand and Shilov, Generalized Functions.

Edit by Tom Leinster Here I'll try to flesh out Michael's answer. Michael: please feel free to edit.

We begin by finding the Fourier transform of $\|\xi\|$. A priori this doesn't make sense as $\|\xi\|$ isn't integrable, but I'll proceed formally anyway, hoping that there's a world in which this is all kosher. Write $g(\xi) = \|\xi\|$ and $g_a(\xi) = g(a\xi)$ (for $a > 0$). For completely general reasons, $$\widehat{g_a}(x) = a^{-n} \hat{g}(x/a).$$ Also, for this particular $g$ we have $g_a = a g$, so $\widehat{g_a} = a\hat{g}$. Hence $\hat{g}(x) = a^{-(n+1)} \hat{g}(x/a)$, giving $$\hat{g}(x) = \|x\|^{-(n+1)} g(x/\|x\|).$$ On the other hand, $g$ is spherically symmetric, so $\hat{g}$ is too; hence $\hat{g}$ has constant value $C$ on the unit sphere. So the Fourier transform of $\|\xi\|$ is $C\|x\|^{-(n+1)}$.

Now we take the Fourier transform of each side of the equation $(\widehat{(-\Delta)^{1/2}}f)(\xi) = \|\xi\|\hat{f}(\xi)$. This gives $$((-\Delta)^{1/2}f)(x) = C\|x\|^{-(n+1)} * f = C\int_{\mathbb{R}^n} \frac{f(y)}{\|x-y\|^{n+1}} \ dy.$$ That seems good, but now I have three problems. First, this isn't the integral formula I was looking for. (Maybe I'm missing a simple trick.) Second, I don't know the value of $C$. Third, I don't know how to make this all rigorous: what are some sufficient conditions on $f$ guaranteeing that $(-\Delta)^{1/2}\bigl((-\Delta)^{1/2}f\bigr)$ is defined (in the sense of the integrals existing) and equal to $f$?

Edit by Michael Renardy:

The problem is that $\|x\|^{-n-1}$ is not integrable at zero. Therefore it needs to be replaced by a regularization. The theory of such regularizations is developed in detail in Section 3 of Chapter I in Gelfand and Shilov.

2 Details added by Tom Leinster

Edit by Tom Leinster Here I'll try to flesh out Michael's answer. Michael: please feel free to edit.

We begin by finding the Fourier transform of $\|\xi\|$. A priori this doesn't make sense as $\|\xi\|$ isn't integrable, but I'll proceed formally anyway, hoping that there's a world in which this is all kosher. Write $g(\xi) = \|\xi\|$ and $g_a(\xi) = g(a\xi)$ (for $a > 0$). For completely general reasons,\widehat{g_a}(x) = a^{-n} \hat{g}(x/a).Also, for this particular $g$ we have $g_a = a g$, so $\widehat{g_a} = a\hat{g}$. Hence $\hat{g}(x) = a^{-(n+1)} \hat{g}(x/a)$, giving\hat{g}(x) = \|x\|^{-(n+1)} g(x/\|x\|). On the other hand, $g$ is spherically symmetric, so $\hat{g}$ is too; hence $\hat{g}$ has constant value $C$ on the unit sphere. So the Fourier transform of $\|\xi\|$ is $C\|x\|^{-(n+1)}$.

Now we take the Fourier transform of each side of the equation $(\widehat{(-\Delta)^{1/2}}f)(\xi) = \|\xi\|\hat{f}(\xi)$. This gives((-\Delta)^{1/2}f)(x) = C\|x\|^{-(n+1)} * f = C\int_{\mathbb{R}^n} \frac{f(y)}{\|x-y\|^{n+1}} \ dy.That seems good, but now I have three problems. First, this isn't the integral formula I was looking for. (Maybe I'm missing a simple trick.) Second, I don't know the value of $C$. Third, I don't know how to make this all rigorous: what are some sufficient conditions on $f$ guaranteeing that $(-\Delta)^{1/2}\bigl((-\Delta)^{1/2}f\bigr)$ is defined (in the sense of the integrals existing) and equal to $f$?

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The Fourier transform of a radially symmetric function is a radially symmetric function. Basic scaling shows that the Fourier transform of $\|\xi\|$ must be $\|x\|^{-n-1}$ in some sense. Some care must be taken in interpreting this, since this function is not integrable. A good reference for this is Gelfand and Shilov, Generalized Functions.