Fourier Transform of Schwartz Space

Question

I am trying to read through Corollary 8.23 in Folland, p. 250, which is a proof that the Fourier transform maps the Schwartz space into itself. I do not see why the following is true

$$\|x^\alpha \partial^\beta f\|_1 \leq C \|(1 + |x|)^{n+1} x^\alpha \partial^\beta f\|_u.$$

where $f$ is in Schwartz space, $\alpha, \beta$ are arbitrary multi-indices, and $\|\cdot\|_u$ is the uniform norm.

I also do not see why it follows that

$$\|\widehat{f}\|_{(N, \beta)} \leq C_{N, \beta} \sum_{|\gamma| \leq N} \|f\|_{(\beta + n + 1, \gamma)}$$

where $\displaystyle\|f\|_{(N, \alpha)} = \sup_{x \in \mathbb{R}^n} (1 + |x|)^N |\partial^\alpha f(x)|$.

Answer 1

We can write, since $f\in\mathcal S(\mathbb R^n)$ \begin{align*} \lVert x^{\alpha}\partial^{\beta}f\rVert_1&\leqslant \int_{\mathbb R^n}|x|^{\alpha}|\partial^{\beta}f(x)|dx\\\ &=\int_{\mathbb R^n}(1+|x|)^{n+1}|x|^{\alpha}|\partial^{\beta}f(x)|\frac 1{(1+|x|)^{n+1}}dx\\\ &\leqslant C'\sup_{x\in\mathbb R^n}(1+|x|)^{n+1}|x|^{\alpha}|\partial^{\beta}f(x)| \int_{\mathbb R^n}\frac{dx}{(1+|x|)^{n+1}}\\\ &=C'\sup_{x\in\mathbb R^n}(1+|x|)^{n+1}|x|^{\alpha}|\partial^{\beta}f(x)| s_n\int_0^{+\infty}\frac{r^{n-1}}{(1+r)^{n+1}}dr, \end{align*} where $s_n$ is the area of the unit sphere in $\mathbb R^n$. The last integral is convergent, and we get the expected result putting $C:=C's_n\int_0^{+\infty}\frac{r^{n-1}}{(1+r)^{n+1}}dr$.

For the second fact, note that $\partial^{\beta}\widehat f(x)=\int_{\mathbb R^n} i^{\beta}t^{\beta}e^{it\cdot x}f(t)dt$, hence for $x\in\mathbb R^n$: \begin{align*} (1+|x|)^N|\partial^{\beta}\widehat f(x)|&= (1+|x|)^N\left|\int_{\mathbb R^n}e^{it\cdot x}t^{\beta}f(t)dt\right|\\\ &=\sum_{k=0}^N\binom Nk|x|^k\left|\int_{\mathbb R^n}e^{it\cdot x} t^{\beta}f(t)dt\right|\\\ &=\sum_{k=0}^N\binom Nk\sum_{|\gamma |=k}\left|\int_{\mathbb R^n} x^{\gamma}e^{it\cdot x} t^{\beta}f(t)dt\right|\\\ &=\sum_{|\gamma|\leqslant N}\binom Nk\left|\int_{\mathbb R^n} x^{\gamma}e^{it\cdot x}t^{\beta}f(t)dt\right|. \end{align*} Let $\displaystyle I_{\gamma}(x):=\int_{\mathbb R^n} x^{\gamma}e^{it\cdot x}t^{\beta}f(t)dt$. Integrating by parts and using Leibniz formula, we have \begin{align*} |I_{\gamma}(x)|&=\left|\int_{\mathbb R^n}e^{it\cdot x}\sum_{\alpha\leqslant \gamma}\binom{\gamma}{\alpha}\partial^{\alpha}f(t)t^{\beta-\alpha}\frac{\beta !}{(\beta-\alpha)!}dt\right|\\\ &\leqslant \beta !\sum_{\alpha\leq \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}\int_{\mathbb R^n}\left|\partial^{\alpha}f(t)t^{\beta-\alpha}\right|dt, \end{align*} and using the first point \begin{align*} |I_{\gamma}(x)|&\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\sup_{x\in\mathbb R^n} (1+|x|)^{n+1}|x|^{\beta-\alpha}|\partial^{\alpha}f(x)|\\\ &\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\sup_{x\in\mathbb R^n} (1+|x|)^{n+1}(1+|x|)^{\beta-\alpha}|\partial^{\alpha}f(x)|\\\ &\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\sup_{x\in\mathbb R^n} (1+|x|)^{n+1+\beta}|\partial^{\alpha}f(x)|\\\ &\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\lVert f\rVert_{(n+1+\beta,\alpha)}. \end{align*} Putting $A_{\gamma,\beta}=\beta\max_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}$. Then $|I_{\gamma}(x)|\leqslant A_{\gamma,\beta}\sum_{\alpha\leq\gamma}\lVert f\rVert_{(n+1+\beta,\alpha)}$. Now, put $\displaystyle B_{N,\beta}:=\max_{|\gamma|\leqslant N}A_{\gamma,\beta}\binom N{|\gamma|}$. We get \begin{align*} \lVert \widehat f\rVert_{(N,\beta)}&\leqslant B_{N,\beta}\sum_{|\gamma|\leqslant N}\: \sum_{\alpha\leq \gamma} \lVert f\rVert_{(n+1+\beta,\alpha)}\\\ &\leqslant B_{N,\beta}\sum_{|\gamma '|\leqslant N} D(\gamma')\lVert f\rVert_{(n+1+\beta,\gamma')}, \end{align*} where $D(\gamma')$ denote the number of times on which $\gamma'$ is obtained in the double sum. Finally, we get $$\lVert \widehat f\rVert_{(N,\beta)}\leqslant C_{N,\beta}\sum_{|\gamma |\leqslant N} \lVert f\rVert_{(n+1+\beta,\gamma)}$$ putting $\displaystyle C_{N,\beta}:=B_{N,\beta}\max_{|\gamma'|\leqslant N}D(\gamma')$.

Answer 2

$\newcommand{\R}{\mathbb{R}}$ I think I maybe figured out what Folland was trying to say, but it is non-intuitive if I'm right. First, notice that in the latest errata he wrote that the sum should be $$ \|\hat{f}\|_{(N,\alpha)} \leq C_{N,\alpha} \sum_{|\gamma| \leq N} \|f\|_{(|\alpha| + n + 1, \gamma)},$$ so our goal is to get something like this.

We first write out the definition to notice that $$ \|f\|_{(N,\alpha)} = \sup_{x \in \mathbb{R}^n} (1+|x|)^N |\partial^\alpha f(x)|. $$ In Proposition 8.3 of Folland, he derives that there is some $\eta > 0$ so that $$ \sum_1^n |x_j|^N > \eta |x|^N,$$ so we use this to get $$ (1+|x|)^N \leq 2^N(1+|x|^N) \leq 2^N \eta^{-1} \sum_{|\beta| \leq N} |x^\beta|.$$ Substituting this in and now using $\hat{f}$, we get the following: $$ \begin{split} \|\hat{f}\|_{(N, \alpha)} \leq \sup_{x \in \R^n}\left[ 2^N \eta^{-1} \sum_{|\beta| \leq N} |x^\beta \partial^\alpha \hat{f}| \right] \\ \text{(Use Folland 8.22 (d) on }\partial^\alpha \hat{f}\text{)} = \sup_{x \in \R^n}\left[ 2^N \eta^{-1} \sum_{|\beta| \leq N} |x^\beta \widehat{(-2\pi i x)^\alpha f}| \right] \\ \text{(Use linearity and the absolute value to pull out constants)}= \sup_{x \in \R^n}\left[ 2^N \eta^{-1} \sum_{|\beta| \leq N}(2\pi)^{|\alpha||} |x^\beta \widehat{x^\alpha f}| \right] \\ \text{(Use Folland 8.22 (e) on }x^\beta \widehat{x^\alpha f}\text{, pull constants out with linearity)} = \sup_{x \in \R^n}\left[ 2^N \eta^{-1} \sum_{|\beta| \leq N}(2\pi)^{|\alpha|-|\beta|} | \widehat{\partial^\beta x^\alpha f}| \right] \\ \text{(Use the product rule to rewrite }\partial^\beta x^\alpha f\text{)} = \sup_{x \in \R^n}\left[ 2^N \eta^{-1} \sum_{|\beta| \leq N}(2\pi)^{|\alpha|-|\beta|} \left|\left[x^\alpha \partial^\beta f + \sum c_{\gamma \delta} x^\delta \partial^\gamma f \right]^{\wedge} \right| \right] \\ \text{(Use linearity of Fourier transform)} = \sup_{x \in \R^n}\left[ 2^N \eta^{-1} \sum_{|\beta| \leq N}(2\pi)^{|\alpha|-|\beta|} \left|\widehat{x^\alpha \partial^\beta f} + \sum c_{\gamma \delta} \widehat{x^\delta \partial^\gamma f} \right| \right] \\ \text{(T.I.)}\leq 2^N \eta^{-1} \sum_{|\beta| \leq N}(2\pi)^{|\alpha|-|\beta|} \left[\|\widehat{x^\alpha \partial^\beta f}\|_u + \sum c_{\gamma \delta} \|\widehat{x^\delta \partial^\gamma f}\|_u \right] \\ \text{(Inequality from Folland)}\leq 2^N \eta^{-1} \sum_{|\beta| \leq N}(2\pi)^{|\alpha|-|\beta|} \left[C\|(1+|x|)^{n+1} x^\alpha \partial^\beta f\|_u + \sum c_{\gamma \delta} C\|(1+|x|)^{n+1} x^\delta \partial^\gamma f\|_u \right] \\ \text{(Use the fact that}|x|^\delta \leq (1+|x|)^{|\alpha|}\text{)} \leq 2^N \eta^{-1} \sum_{|\beta| \leq N}(2\pi)^{|\alpha|-|\beta|} \Big[C\|(1+|x|)^{n+|\alpha|+1} \partial^\beta f\|_u + \\ \sum c_{\gamma \delta} C\|(1+|x|)^{n+1+|\alpha|} \partial^\gamma f\|_u \Big] \\ \text{(By definition)} = 2^N \eta^{-1} \sum_{|\beta| \leq N}(2\pi)^{|\alpha|-|\beta|} \left[C\|f\|_{(n+|\alpha|+1, \beta)} + \sum c_{\gamma \delta} C\|f\|_{(n+|\alpha|+1, \gamma)} \right], \end{split} $$ From here we can absorb constants and get the same thing that Folland has.

Now, to see why the inequality itself is true, we basically use what the prior answer did. Notice that $$ \|\widehat{x^\alpha \partial^\beta f}\|_u = \|\widehat{x^\alpha \partial^\beta f}\|_\infty \leq \|x^\alpha \partial^\beta f\|_1,$$ and $$ \int_{\mathbb{R}^n} |x^\alpha \partial^\beta f| = \int_{\mathbb{R}^n} \frac{(1+|x|)^{n+1}}{(1+|x|)^{n+1}}|x^\alpha \partial^\beta f|dx \leq \left(\int_{\mathbb{R}^n} \frac{dx}{(1+|x|)^{n+1}}\right) \|(1+|x|)^{n+1} x^\alpha \partial^\beta f\|_u,$$ so setting $$ C = \int_{\mathbb{R}^n} \frac{dx}{(1+|x|)^{n+1}} < \infty $$ we have the desired result.

Personally the other answer is more intuitive and on par with the length of this proof, so I don't see why Folland opted to go this route.