# Prove $\varphi\in\mathcal{S}(\mathbb R^n)$ if and only if the following inequality holds..

I need some help for showing the following result: Let $\varphi\in C^\infty(\mathbb R^n)$. Then $\varphi\in \mathcal{S}(\mathbb R^n)$ if and only if for all $\alpha\in\mathbb N^n$ and $N\geq 0$ there is a constant $C_{\alpha, N}$ such that $$|\partial^\alpha \varphi(x)|\leq \frac{C_{\alpha, N}}{(1+|x|)^N}.$$

For me:

$C^\infty_c(\mathbb R^n)$ is the set of all $\varphi:\mathbb R^n\rightarrow \mathbb C$ of class $C^\infty$.

$\mathcal{S}(\mathbb R^n)$ is the Schwartz space defined as the class of all $C^\infty$ functions $\varphi:\mathbb R^n\rightarrow \mathbb C$ such that $$\sup_{x\in\mathbb R^n}|x^\beta \partial^\alpha \varphi(x)|<\infty,$$ for all multi-indices $\alpha, \beta\in\mathbb N^n$.

I proved the implication $(\Rightarrow)$ as follows: Given $\alpha, \beta\in\mathbb N^n$, $$|x^\beta \partial^\alpha \varphi(x)|=|x^\beta||\partial^\alpha \varphi(x)|\leq |x|^{|\beta|}|\partial^\alpha \varphi(x)|\leq (1+|x|)^{|\beta|}\frac{C_{\alpha, |\beta|}}{(1+|x|)^{|\beta|}}=C_{\alpha, |\beta|},$$ hence $\displaystyle\sup_{x\in\mathbb R^n}|x^\beta \partial^\alpha \varphi(x)|<\infty$, i.e., $\varphi\in\mathcal{S}(\mathbb R^n)$.

The main problem is the converse. I conjecture the proof must be made by contradiction but I wasn't managed to do it..

Any help with be great..Thanks

• The title is quite uninformative, don't you think? – Mariano Suárez-Álvarez Jul 19 '13 at 20:03
• Should that be "Let $\varphi\in C^\infty(\mathbb R^n)$"? – Rasmus Jul 19 '13 at 20:09
• @Rasmus might be $C_0^\infty = C_{c(ompact)}^\infty$ too, idk about what exactly is the requirement here. However, the definition doesnt make sense here. $\mathcal{S}$ is defined as the set of all fast-falling functions. The functions that have compact support fulfill this property trivially? – CBenni Jul 19 '13 at 20:37
• @Rasmus I agree that must be $C^\infty(\mathbb R^n)$.. – PtF Jul 19 '13 at 20:50

If $\varphi\in\mathcal S(\mathbb R^n)$, then $x^{\alpha}\partial^{\beta}\varphi$ and $P(x)Q(\partial)\varphi$ are in $\mathcal S(\mathbb R^n)$, for all polynomials $P$, $Q$ and multi-indices $\alpha$, $\beta$. Then, selecting any polynomial $P=P(x)$ of degree $k$ and $Q(\partial)=\partial^{\beta}$ we can write

$$|P(x)\partial^{\beta}\varphi(x)|\leq constant_{k,\beta};$$

taking $P(x)=(1+x^2_1+\dots+x^2_n)^k$ we arrive at the thesis.

• thanks for the help I'll try to write that.. – PtF Jul 19 '13 at 20:52
• how can I bound the term $(1+x^2_1+\ldots+x_n^2)^N$? The problem is that $N$ is not a natural number hence I can't use the multinomial theorem as I did below.. – PtF Jul 20 '13 at 17:46
• You do not bound it; you divide both sides of the inequality by the polynomial P(x) because it is non zero (this is the reason of the "+1" and the squares) and non negative (so the inequality does not change). Little trick ;) – Avitus Jul 20 '13 at 17:54
• It is a good trick @Avitus but how will you make the constant depend on $N$ without bounding $P(x)$? – PtF Jul 20 '13 at 19:03
• In Folland's book on real analysis he says this $N$ is natural, now I don't know.. – PtF Jul 20 '13 at 21:56

Using @Avitus idea I wrote the following proof for the converse:

On the other hand, given $\alpha=(\alpha_1, \ldots, \alpha_n)\in\mathbb N^n$, $$|(1+\sum_{i=1}^nx_i^2)^N\partial^\alpha \varphi(x)|=|(1+\sum_{i=1}^nx_i^2)^N||\partial^\alpha\varphi(x)|.$$ But by the multinomial theorem, $$(1+\sum_{i=1}^nx_i^2)^N=\sum_{|(\beta_0, \beta)|=N}\frac{N!}{\beta_0!\beta!}x^{2\beta}, \beta_0\in\mathbb N.$$ Notice both $\beta_0$ and $\beta$ depend on $N$. Hence, $$|(1+\sum_{i=1}^nx_i^2)^N\partial^\alpha \varphi(x)|\leq \sum_{|(\beta_0, \beta)|=N}\frac{N!}{\beta_0!\beta!}|x^{2\beta}\partial^\alpha \varphi(x)|\leq \sum_{|(\beta_0, \beta)|=N}\frac{N!}{\beta_0!\beta!}\sup_{x\in\mathbb R^n}|x^{2\beta}\partial^\alpha \varphi(x)|=C_{\alpha, N}.$$ The result follows noting that $$(1+|x|)^N\leq (1+\sum_{i=1}^nx_i^2)^N.$$

The above argument works well if $N\in\mathbb Z^+$ otherwise we can't use the multinomial theorem..

• Well I made an awful mistake, I can't use the multinomial theorem because $N$ is not natural..I will try to fix that.. – PtF Jul 20 '13 at 17:31
• are you sure that $(1+|x|)^N \leq (1+|x|^2)^N$? – Jorge.Squared May 31 '17 at 12:07
• That holds for $|x|≥1$ the other case must be treated separately. – PtF Jun 1 '17 at 11:08
• For $0 \leq |x| < 1$ the inequality $(1+|x|)^{N} \leq (1 + |x|^2)^{N}$ simply isn't true. But in the proof of your proposition, why do you work with $(1+ |x|^2)^N$ instead of taking $(1+|x|)^{N}$ right at the beginning? This should be enough, shouldN't it – Jorge.Squared Jun 1 '17 at 11:50
• Yes, it is not true but it won't be a problem. Because the definition I had was using $(1+|x|)^N$ but it is equivalent after all since $(1+|x|)^N$ and $(1+|x|^2)^N$ are comparable.. – PtF Jun 1 '17 at 13:38