Decay of fractional derivative of Schwartz function

Question

Let $\phi$ be a Schwartz function and let $\alpha>0$. I want to analyze the decay as $x\rightarrow\infty$ of: $$\int_\mathbb{R}e^{2\pi i x\xi}|\xi|^\alpha\phi(\xi)\,d\xi$$Heuristically, for $\xi\ll x$, we have one oscillation from $\xi=0$ to $\xi=1/x$ at which point the amplitude is of order $|x|^{-\alpha}$ so the answer should be $|x|^{-\alpha-1}$. However, I have had trouble making this intuition precise. I have tried smoothly partitioning into $|\xi|\ll x$ and $|\xi|\gg |x|$ and separately doing a dyadic decomposition at level $|x|^{-1}$, but the details have not worked out. I am interested a precise solution as well as any heuristic ways of thinking about this sort of integral. If it would help, I can share what I've tried in more detail.

Answer 1

Your intuition seems spot on in terms of how to think about this, so I don't really have anything to add in terms of the heuristics. I'm not sure what went awry when you tried out the dyadic decomposition, but here's how to make a fairly precise argument of it to establish that there's decay like $|x|^{-\alpha - 1}$ as $x \to \infty$.

Step 1: The dyadic decomposition

First, we can smoothly partition to $|\xi| \gtrsim 1$ and then also to $|\xi| \approx 2^{-k}$ for $k \geq 1$. The first part ($|\xi| \gtrsim 1$) more straightforwardly has rapid decay in $x$ since this avoids the lack of $C^{\infty}$ smoothness at the origin, i.e. we're taking the inverse Fourier transform of a straight Schwartz function.

We're then left to analyze the dyadic decomposition $$ \sum_{k = 1}^{\infty} \int_{\mathbb{R}} e^{2 \pi i x \xi} |\xi|^{\alpha} \eta(2^k \xi) \phi(\xi) \, d\xi, $$ where $\eta$ is the smooth function being used to produce the dyadic decomposition.

Step 2: Setting up integration by parts

As is often the case in this type of set-up, integration by parts is our friend. We use the fact that $$ e^{2 \pi i x \xi} = (2 \pi i x)^{-1} D_{\xi} (e^{2 \pi i x \xi}), $$ with $D_{\xi}$ denoting differentiation with respect to $\xi$. So then we have $$ \int_{\mathbb{R}} e^{2 \pi i x \xi} |\xi|^{\alpha} \eta(2^k \xi) \phi(\xi) \, d\xi = (-2 \pi i x)^{-\ell} \int_{\mathbb{R}} e^{2 \pi i x \xi} D_{\xi}^{\ell} \bigl[ |\xi|^{\alpha} \eta(2^k \xi) \phi(\xi) \bigr] \, d\xi $$ for any integer $\ell \geq 0$.

The derivative here can essentially be written as $$ \sum_{m_1 + m_2 + m_3 = \ell} C_{m_1, m_2, m_3, \alpha} |\xi|^{\alpha - m_1} \cdot 2^{m_2 k} \eta^{(m_2)}(2^k \xi) \cdot \phi^{(m_3)}(\xi), $$ I think modulo signum terms that come from differentiating $|\xi|$.

Step 3: Estimating the integrand

From here, the idea is to use "crude" size estimates: the derivatives of $\eta$ and $\phi$ can be bounded by additional constants (i.e. independent of $\xi$, though with dependence on $m_2$ and $m_3$), while for $\xi \approx 2^{-k}$ we have $|\xi|^{\alpha - m_1} \lesssim 2^{k (m_1 - \alpha)}$. So in all, the derivative term can be bounded by $$ \sum_{m_1 + m_2 + m_3 = \ell} C_{m_1, m_2, m_3, \alpha} 2^{k (m_1 - \alpha)} \cdot 2^{m_2 k} \leq C_{\ell, \alpha} 2^{k (\ell - \alpha)}, $$ with constants different than those that appeared above.

Therefore we can bound one integral term in the dyadic decomposition as follows: \begin{align*} \left| (-2 \pi i x)^{-\ell} \int_{\mathbb{R}} e^{2 \pi i x \xi} D_{\xi}^{\ell} [ |\xi|^{\alpha} \eta{2^k \xi} \phi(\xi) ] \, d\xi \right| &\leq C_{\ell, \alpha} |x|^{-\ell} \int_{|\xi| \approx 2^{-k}} 2^{k (\ell - \alpha)} \, d\xi \\ &= C_{\ell, \alpha} |x|^{-\ell} 2^{k (\ell - \alpha - 1)}. \end{align*} Noting that this bound holds for any $\ell \geq 0$, we can see that if we don't actually use integration by parts ($\ell = 0$) we get a bound on the order of $2^{-k (\alpha + 1)}$, and that integration by parts provides an improvement if $|x| \gtrsim 2^{k}$.

Step 4: Re-assembling the dyadic pieces

So for a fixed $x$, we want to use integration by parts (with $\ell > \alpha + 1$) on the dyadic terms where $2^k \leq |x|$, but no integration by parts ($\ell = 0$) on the terms where $2^k > |x|$. This allows us to estimate as follows: \begin{align*} \left| \sum_{k = 1}^{\infty} \int_{\mathbb{R}} e^{2 \pi i x \xi} |\xi|^{\alpha} \eta(2^k \xi) \phi(\xi) \, d\xi \right| &\leq \sum_{2^k \leq |x|} C_{\ell, \alpha} |x|^{-\ell} 2^{k (\ell - \alpha - 1)} + \sum_{2^k > |x|} C' 2^{-k (\alpha + 1)} \\ &\leq C_{\ell, \alpha} |x|^{-\ell} \cdot |x|^{\ell - \alpha - 1} + C_{\alpha} |x|^{-\alpha - 1} \\ &= C_{\alpha} |x|^{-\alpha - 1}. \end{align*} At the end we've dropped the constant's dependence on $\ell$ since we can really think of $\ell$ depending on $\alpha$ (for example by taking $\ell$ to be the be the smallest integer greater than $\alpha + 1$).

This gives the claimed decay of the original (inverse) Fourier transform as $x \to \infty$. Of course, if $\phi$ vanishes to some degree at the origin we can get a correspondingly higher degree of decay at $\infty$, but in general $|x|^{-\alpha - 1}$ is the best we can do.

Answer 2

If all the derivatives $\phi^{(k)}(0)=0$ then $|\xi|^a\phi$ is Schwartz thus so is its Fourier transform.

Otherwise let $\phi^{(k)}(0)$ be the first non-vanishing derivative, wlog assume that $a\in (0,1)$ and $l\in 0,1$ then $$h=sign(\xi)^l |\xi|^a\phi-\frac{\phi^{(k)}(0)}{k!} \xi^k sign(\xi)^l |\xi|^a e^{-|\xi|} =O(|\xi|^{a+k+1})$$ is $k+2+\lfloor a\rfloor$-times differentiable in $L^1+\Bbb{C}\delta$ so that for $|x|\ge 1$ $$\mathcal{F}^{-1}(sign(\xi)^l|xi|^a\phi)$$ $$=\mathcal{F}^{-1}(\frac{\phi^{(k)}(0)}{k!} \xi^k sign(\xi)^l |\xi|^a e^{-|\xi|})+\mathcal{F}^{-1}(h)$$ $$=\frac{\phi^{(k)}(0)}{k!} ((1-2i\pi x)^{-a-k-1}+(-1)^{k+l} (1+2i\pi x)^{-a-k-1}) \Gamma(a+k+1) + (-2i\pi x)^{-k-2-\lfloor a\rfloor}\mathcal{F}^{-1}(h^{(k+2)})$$ $$ = C x^{-a-k-1}+O(x^{-k-2-\lfloor a\rfloor})$$