# Fourier Transform of Schwartz Space

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

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I'm not looking for other proofs of the fact that the Fourier transform maps Schwartz space to itself. There are other books whose proofs of this I can understand. I can't see how Folland does it, and this is what is bothering me. – user912312 Nov 3 '11 at 2:43
Fair Enough. Sorry for that. – JavaMan Nov 3 '11 at 2:44

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 $\displaystyle C_{N,\beta}:=B_{N,\beta}\max_{|\gamma'|\leqslant N}D(\gamma')$.