In "Real Analysis: Modern Techniques and Their Applications", p. 249 Theorem 8.22, Folland stated two statements about the smoothness and decay properties of the Fourier transform.
The first statement is the following:
If $x^\alpha f \in L^1$ for all $|\alpha| \leq k$, then $\hat{f} \in C^k$
and $\partial^\alpha\hat{f} = [(-2\pi i x)^\alpha f]^\wedge$ for any $|\alpha| \leq k$.
See also this MSE post.
It seems to me that we only need $f \in L^1$ and $|x|^k f \in L^1$. Moreover, since $\partial^\alpha\hat{f}$ is a fourier transform of an $L^1$ function, it should vanish at infinity. Thus, I propose the following strengthening:
Suppose $f \in L^1$ and $|x|^k f \in L^1$. Then:
- $x^\alpha f \in L^1$ for any $|\alpha| \leq k$
- $\hat{f} \in C^k_0$, where $C^k_0$ is the space of $C^k$ functions $\varphi$ such that $\partial^\alpha \varphi$ vanishes at infinity for
all $|\alpha| \leq k$- $\partial^\alpha\hat{f} = [(-2\pi i x)^\alpha f]^\wedge$ for any $|\alpha| \leq k$
Question 1: Is the above strengthened statement correct?
Another statement by Folland is the following:
Suppose:
- $f \in C^k$
- $\partial^\alpha f \in L^1$ for all $|\alpha| \leq k$
- $\partial^\alpha f \in C_0$ for all $|\alpha| \leq k-1$
Then $\widehat{\partial^\alpha f}(\xi) = (2\pi i \xi)^\alpha \hat{f}(\xi)$
for any $|\alpha| \leq k$.
It seems to me that the condition "$\partial^\alpha f \in C_0$ for all $|\alpha| \leq k-1$" is unnecessary.
Question 2: Can we still obtain the same result if we only assume conditions 1 and 2 in the above statement? What could go wrong if we drop condition 3?
Best Answer
Q1: You are correct; this follows from the existence of $C_k>0$ for which $$ C^{-1} (1+|x|^k) \leq \sum_{|\alpha| \leq k} |x^{\alpha}| \leq C (1+|x|^k). $$ Once this is established, we see that having $f, |x|^kf \in L^1$ is equivalent to having $x^{\alpha}f \in L^1$ for all $|\alpha|\leq k.$ To prove this estimate, we will note that $(1+t)^k \sim (1+t^k)$ for $t \geq 0$ and that the $\ell_1$ and $\ell_2$ norms are equivalent on $\Bbb R^n$ to get $$ (1+|x|^k) \sim (1+|x|)^k \sim \left( 1 + \sum_{i=1}^n |x_i|\right)^k, $$ and letting $y_i = \sqrt{|x_i|}$ the multinomial theorem gives $$ \left(1+\sum_{i=1}^n y_i^2\right)^k = \sum_{|\alpha| \leq k} {k \choose \alpha} y^{2\alpha}, $$ from which the estimate follows. The last part follows from the Riemann-Lebesgue lemma in Folland (Proposition 8.22(f)).
Q2: This is true, but one needs to be careful with the proof - note that the vanishing condition is used to integrate by parts, as one needs the endpoint terms to vanish in the limit.
One way you can do this is via duality; if $\varphi \in C^{\infty}_c(\Bbb R^n)$ we have \begin{align*} \int_{\Bbb R^n} \widehat{\partial^{\alpha}f}(\xi) \varphi(\xi) \,\mathrm d\xi &= \int_{\Bbb R^n} \partial^{\alpha}f(x) \hat{\varphi}(x) \,\mathrm dx \\ &= (-1)^{\alpha}\int_{\Bbb R^n} f(x) \partial^{\alpha}\hat{\varphi}(x) \,\mathrm dx \\ &=(-1)^{\alpha}\int_{\Bbb R^n} f(x) [(-2\pi i \xi)^{\alpha}\varphi]\hat{}(x) \,\mathrm dx \\ &= \int_{\Bbb R^n} (2\pi i \xi)^{\alpha} \hat{f}(\xi) \varphi(\xi) \,\mathrm d\xi. \end{align*} Here we used Lemma 8.25 to move the Fourier transform to $\varphi,$ integrated by parts, applied first result (Proposition 8.22(d)) to $\varphi,$ and moved the Fourier transform back. Since this holds for all $\varphi \in C^{\infty}_c(\Bbb R^n),$ the identify follows.
More generally this duality trick allows one to define the Fourier transform of less regular functions and distributions. This is discussed in Chapter 9 of the book (in particular, see exercise 17 in that chapter).