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

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It's a long answer since the computations are long as often in the study of Schwartz spaces, and the main difficulty is the name of the variables. What we showed is that the Fourier transform is continuous with respect to the semi-norms. – Davide Giraudo Nov 3 '11 at 21:04

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