# Inequality in Schwartz space

I am trying to prove theorem 9.2 from book "Lectures on Linear Partial Differential Equations" wtitten by G. Eskin. In proof of this theorem is inequality which makes problems for me. Firstly I remind designation.

Multi-index: $\alpha=(\alpha_{1}, \ldots , \alpha_{n}) \in \mathbb{Z}_{+}^{n}$, length of multi-index: $| \alpha | = \sum_{i=1}^{n} \alpha_{i}$, partial differential operator: $$D^{\alpha} = \frac{\partial ^{|\alpha|}}{\partial x_{1}^{\alpha_{1}} \ldots \partial x_{n}^{\alpha_{n}}}.$$ Moreover $|x|=\sqrt{x_1^2+ \ldots + x_n^2}$, Schwartz space: $$\mathcal{S}(\mathbb{R}^n) = \{f \in C^{\infty}(\mathbb{R}^n) \colon \sup_{x \in \mathbb{R}^{n}} (1+|x|)^{m}|D^{\beta}f(x)| < \infty, \: m \in \mathbb{N}, \: \beta \in \mathbb{Z}_{+}^{n} \}.$$ Seminorm in Schwartz space: $$\|f\|_{m,\mathcal{S}} = \sup_{x \in \mathbb{R}^{n}} (1+|x|)^{m} \sum_{|\beta| = 0}^{m} |D^{\beta}f(x)|.$$ For every function $f \in \mathcal{S}(\mathbb{R}^n)$, for every multi-index $\beta \in \mathbb{Z}_{+}^{n}$ and for every $m \in \mathbb{N}$ we have \begin{equation*} (1+|x|)^{m}D^{\beta}f(x) = \int \limits_{-\infty}^{x_1} \cdots \int \limits_{-\infty}^{x_n} D^{(1,\ldots,1)}[(1+|y|)^{m}D^{\beta}f(y)] \; dy_n \ldots dy_1. \end{equation*}

We need to show inequality. $$\| f \|_{m,\mathcal{S}} \leq C_{1} \sum_{|\beta|=0}^{n+m} \int _{\mathbb{R}^n} (1+|y|)^m |D^{\beta} f(y)| \; dy, \quad f \in \mathcal{S}(\mathbb{R}^n).$$

I used above properties, used Liebniz and I've got this:

\begin{eqnarray*} \|f\|_{m,\mathcal{S}} &=& \sup_{x \in \mathbb{R}^{n}} (1+|x|)^{m} \sum_{|\beta| = 0}^{m} |D^{\beta}f(x)| = \sup_{x \in \mathbb{R}^{n}} \sum_{|\beta| = 0}^{m} \left| (1+|x|)^{m} D^{\beta}f(x) \right|\\ &=& \sup_{x \in \mathbb{R}^{n}} \sum_{|\beta| = 0}^{m} \left| \int \limits_{-\infty}^{x_1} \cdots \int \limits_{-\infty}^{x_n} D^{(1,\ldots,1)}[(1+|y|)^{m}D^{\beta}f(y)] \; dy_n \ldots dy_1 \right|\\ &\leq & \sup_{x \in \mathbb{R}^{n}} \sum_{|\beta| = 0}^{m} \int \limits_{-\infty}^{x_1} \cdots \int \limits_{-\infty}^{x_n} \left| D^{(1,\ldots,1)}[(1+|y|)^{m}D^{\beta}f(y)] \right| \; dy_n \ldots dy_1 \\ & \leq & \sum_{|\beta| = 0}^{m} \int _{\mathbb{R}^n} \left| D^{(1,\ldots,1)}[(1+|y|)^{m}D^{\beta}f(y)] \right| \; dy = \\ & = & \sum_{|\beta| = 0}^{m} \int _{\mathbb{R}^n} \left| \sum _{\alpha + \gamma = (1,\ldots ,1)} \binom{n}{|\gamma|} D^{\alpha}\left( (1+|y|)^{m} \right) \cdot D^{\gamma} \left( D^{\beta}f(y) \right) \right| \; dy \\ &=& \sum_{|\beta| = 0}^{m} \int _{\mathbb{R}^n} \left| \sum _{\alpha + \gamma = (1,\ldots ,1)} \binom{n}{|\gamma|} c_0 \cdot y^{\alpha} |y|^{-2|\alpha|} (1+|y|)^{m} \cdot D^{\gamma} \left( D^{\beta}f(y) \right) \right| \; dy \\ & \leq & \sum_{|\beta| = 0}^{m} \int _{\mathbb{R}^n} \sum _{\alpha + \gamma = (1,\ldots ,1)} \binom{n}{|\gamma|} \left| c_0 \cdot y^{\alpha} |y|^{-2|\alpha|} (1+|y|)^{m} \cdot D^{\gamma} \left( D^{\beta}f(y) \right) \right| \; dy \\ & \leq & \sum_{|\beta| = 0}^{m} \int _{\mathbb{R}^n} \sum _{\alpha + \gamma = (1,\ldots ,1)} \binom{n}{|\gamma|} c_0 \cdot |y|^{|\alpha|} |y|^{-2|\alpha|} (1+|y|)^{m} \cdot \left| D^{\gamma} \left( D^{\beta}f(y) \right) \right| \; dy \\ &=&\sum_{|\beta| = 0}^{m} \int _{\mathbb{R}^n} \sum _{\alpha + \gamma = (1,\ldots ,1)} \binom{n}{|\gamma|} c_0 \cdot |y|^{-|\alpha|} (1+|y|)^{m} \cdot \left| D^{\gamma} \left( D^{\beta}f(y) \right) \right| \; dy \\ &=&\sum_{|\beta| = 0}^{m}\sum _{\alpha + \gamma = (1,\ldots ,1)} \binom{n}{|\gamma|} c_0 \cdot \int _{\mathbb{R}^n} |y|^{-|\alpha|} (1+|y|)^{m} \cdot \left| D^{\gamma} \left( D^{\beta}f(y) \right) \right| \; dy \\ \end{eqnarray*}

It is almost what I need - problem is with factor $|y|^{-|\alpha|}$. Has someone idea how to prove it? Thanks in advance.

• When you're differentiating $(1+\lvert y\rvert)^m$, you reduce the exponent. But you still have a problem near $0$. Instead of $(1+\lvert y\rvert)^m$, I suggest you use $(1 + \lvert y\rvert^2)^{m/2}$. That's much nicer for differentiating. – Daniel Fischer Sep 5 '13 at 21:30
• Yeah, you are right about diferentiation in $0$, but we ignore this. I have to use this factor, because even if we can find constant $c$ such that $$(1+|y|)^m \leq c \cdot (1+|y|^2)^{m/2}$$ then there is no relation between $D^{\alpha}(1+|y|)^m$ and $D^{\alpha}(1+|y|^2)^{m/2}$. We know that $\alpha \in \{0,1\}^n$ so $$D^{\alpha}(1+|y|)^m= c \cdot y^{\alpha} |y|^{-2|\alpha|}\cdot (1+|y|)^m$$. Moreover we know that $$|y^{\alpha}| \leq |y|^{|\alpha|}.$$ I use this in calculation above – Marcin Sep 5 '13 at 22:15
• First, I want to ask why you got $|y|^{-|\alpha|}$? I think it should be $|y|^{-|\alpha|+1}$, then the integral converges at $0$. Also, could you post a copy of Theorem 9.2, so I may see more clearly what's going on? – L. Xu Sep 7 '13 at 20:57
• @L.Xu I took a screenshots from google books, you can find them here -> link. How you got $|y|^{1-|\alpha|}$?? – Marcin Sep 14 '13 at 19:35
• @L.Xu From Newton formula $$D^{\alpha}\left( (1+|y|)^{m} \right) = \sum _{j=0}^{m} \binom{m}{j}D^{\alpha}\left( |y|^{j} \right).$$ We know that $\alpha \in \{0,1\}^n$, so $$D^{\alpha}\left( |y|^{j} \right) = c_{j} \cdot y^{\alpha} \cdot |y|^{m-2|\alpha|}.$$ We know also $$|y^{\alpha}| \leq |y|^{|alpha|}$$ so when $c = \max_{j}|c_j|$ then we have $$\left| D^{\alpha}\left( (1+|y|)^{m} \right) \right| \leq c |y|^{-\alpha} (1+|y|)^{m}$$ – Marcin Sep 14 '13 at 21:18

First for $j$ is odd, \begin{align} D^\alpha\left(|y|^j\right)=c_j\cdot y^\alpha\cdot |y|^{j-2|\alpha|}. \end{align} For $j$ is even, $D^\alpha\left(|y|^j\right)=0$ if $j<2|\alpha|$. Then let's check \begin{align} \left|D^\alpha\left((1+|y|)^m\right)\right|\le\sum_{j=0}^m \left(\begin{array}{cc}m\\j \end{array}\right)\left|D^\alpha\left(|y|^j\right)\right|=\sum_{j=1}^m c_j\cdot \left|y^\alpha\right|\cdot |y|^{j-2|\alpha|}\le \sum_{j=1}^m c_j\cdot |y|^{j-|\alpha|}. \end{align} In the above sum, the first term vanishes since it's a constant, the rest of them have similar form with the exponent of $|y|$ bigger than or equal to $1-|\alpha|$. Then it's clear that \begin{align} \left|D^\alpha\left((1+|y|)^m\right)\right|\le C|y|^{1-|\alpha|}(1+|y|)^m, \end{align} so the integral (cited from your question) \begin{align} \int _{\mathbb{R}^n} \left|D^\alpha\left((1+|y|)^m\right)\right|\cdot \left| D^{\gamma} \left( D^{\beta}f(y) \right) \right| \; dy,\quad |\alpha|\le n, \end{align} converges, so the whole integral converges.
In the following I'll answer why $$\| f \|_{m,\mathcal{S}} \leq C \sum_{|\beta|=0}^{n+m} \int _{\mathbb{R}^n} (1+|y|)^m |D^{\beta} f(y)| \; dy, \quad f \in \mathcal{S}(\mathbb{R}^n).$$ From the link of the textbook you provided, it seems to me that the author wants to say this is obvious and easy to derive. I considered about it for long, but I'm not able to give any easy way to it (just as you said, the term $|y|^{1-|\alpha|}$ is hard to get rid of). First, as you deduced in the question, we have \begin{align} \| f \|_{m,\mathcal{S}} \le& \sum_{|\beta| = 0}^{m} \int _{\mathbb{R}^n} \left| \sum _{\alpha + \gamma = (1,\ldots ,1)} \binom{n}{|\gamma|} D^{\alpha}\left( (1+|y|)^{m} \right) \cdot D^{\gamma} \left( D^{\beta}f(y) \right) \right| \; dy. \end{align} Then from the calculating result I give above, we know \begin{align} \| f \|_{m,\mathcal{S}} \le& \sum_{|\beta| = 0}^{m} \sum _{\alpha + \gamma = (1,\ldots ,1)} \binom{n}{|\gamma|} \int _{\mathbb{R}^n} \left| D^{\alpha}\left( (1+|y|)^{m} \right)\right| \cdot \left| D^{\beta+\gamma}f(y) \right| \; dy \\ \le& \sum_{|\beta| = 0}^{m} \sum _{\alpha + \gamma = (1,\ldots ,1)} \binom{n}{|\gamma|} \sum_{j=1}^m c_j\cdot \int _{\mathbb{R}^n} \left|y^\alpha\right|\cdot |y|^{j-2|\alpha|} \cdot \left| D^{\beta+\gamma}f(y) \right| \; dy. \end{align} Now let's consider to simplify the integral \begin{align} \int _{\mathbb{R}^n} \left|y^\alpha\right|\cdot |y|^{j-2|\alpha|} \cdot \left| D^{\beta+\gamma}f(y) \right| \; dy. \end{align} I'll just show for $\alpha=(1,1,\ldots,1)$ and $j=1$, other cases are similar and easier. In this case, the above integral becomes \begin{align} \int _{\mathbb{R}^n} \left|y_1y_2\ldots y_n\right|\cdot |y|^{1-2n} \cdot \left| D^{\beta}f(y) \right| \; dy. \end{align} Rewrite it as \begin{align} \int _{\mathbb{R}^n} y_1|y|^{1-2n} \cdot F_1(y) \; dy, \end{align} where $F_1(y)=\textrm{sign}(y_1)\left|y_2\ldots y_n\right|\cdot \left| D^{\beta}f(y) \right|$. Note that $F_1$ is non-differentiable only on zero measure set (because $D^{\beta}f(y)$ is smooth), then integrating by part (this integration by part is not easy to prove since the derivative has jumps somewhere) we have \begin{align} \int _{\mathbb{R}^n} \left|y_1y_2\ldots y_n\right|\cdot |y|^{1-2n} \cdot \left| D^{\beta}f(y) \right| \; dy&=\int _{\mathbb{R}^{n-1}} \left(\int _{\mathbb{R}} F_1(y) \cdot |y|^{1-2n}y_1 \; dy_1\right)dy_2\ldots dy_n \\ &=-C\int _{\mathbb{R}^{n-1}} \left(\int _{\mathbb{R}} \frac{\partial F_1}{\partial y_1}(y) \cdot |y|^{3-2n} \; dy_1\right)dy_2\ldots dy_n \\ &\le C\int _{\mathbb{R}^{n}} \left|\frac{\partial F_1}{\partial y_1}(y)\right| \cdot |y|^{3-2n} \; dy \\ &\le C\int _{\mathbb{R}^{n}} \left|y_2\ldots y_n\right|\cdot \left| D_1D^{\beta}f(y) \right| \cdot |y|^{3-2n} \; dy. \end{align} Repeating the above procedure (set $F_2(y)=\textrm{sign}(y_2)\left|y_3\ldots y_n\right|\cdot \left| D_1D^{\beta}f(y) \right|\ldots$ and so on), we can finally get \begin{align} \int _{\mathbb{R}^n} \left|y_1y_2\ldots y_n\right|\cdot |y|^{1-2n} \cdot \left| D^{\beta}f(y) \right| \; dy\le& C\int _{\mathbb{R}^{n}} \left| D^{\alpha}D^{\beta}f(y) \right| \cdot |y| \; dy \\ \le& C\int _{\mathbb{R}^{n}} (1+|y|)^m \cdot \left| D^{\alpha}D^{\beta}f(y) \right| \; dy. \end{align} So we have for all $\alpha+\gamma=(1,1,\ldots,1)$, $|\beta|\le m$, \begin{align} \int _{\mathbb{R}^n} \left|y^\alpha\right|\cdot |y|^{j-2|\alpha|} \cdot \left| D^{\beta+\gamma}f(y) \right| \; dy\le C\int _{\mathbb{R}^{n}} (1+|y|)^m \cdot \left| D^{(1,1,\ldots,1)}D^{\beta}f(y) \right| \; dy. \end{align} Summing up all the above, we can conclude $$\| f \|_{m,\mathcal{S}} \leq C \sum_{|\beta|=0}^{n+m} \int _{\mathbb{R}^n} (1+|y|)^m |D^{\beta} f(y)| \; dy, \quad f \in \mathcal{S}(\mathbb{R}^n).$$
Moreover, I realized that in the textbook by this result the author just wants to show $$\| f \|_{m,\mathcal{S}}^2 \leq C \sum_{|\beta|=0}^{n+m} \int _{\mathbb{R}^n} (1+|y|)^{2m+n+1} |D^{\beta} f(y)|^2 \; dy, \quad f \in \mathcal{S}(\mathbb{R}^n).$$ If we show this directly, things can be less complicated. First by C-S inequality and Newton's formula, \begin{align} \left(\int _{\mathbb{R}^n} \left|y^\alpha\right|\cdot |y|^{j-2|\alpha|} \cdot \left| D^{\beta+\gamma}f(y) \right| \; dy\right)^2\le& C\int _{\mathbb{R}^n} \left|y^\alpha\right|\cdot |y|^{j-2|\alpha|} (1+|y|)^{n+1} \cdot \left| D^{\beta+\gamma}f(y) \right|^2 \; dy \\ =& C\sum_{k=0}^{n+1}\left(\begin{array}{cc}n+1\\k \end{array}\right)\int _{\mathbb{R}^n} \left|y^\alpha\right|\cdot |y|^{j+k-2|\alpha|} \cdot \left| D^{\beta+\gamma}f(y) \right|^2 \; dy. \end{align} Then we set $$F_1(y)=\textrm{sign}(y_1)\left|y_2\ldots y_n\right|\cdot \left| D^{\beta+\gamma}f(y) \right|^2.$$ In this case, $F_1(y)$ is everywhere smooth with respect to $y_1$ except at $0$, and the integration by part is easy to prove (split the integral by $\int_{-\infty}^{-\epsilon}+\int^{\infty}_{\epsilon}$ and let $\epsilon\to0$ to approximate it) and we have
\begin{align} \int _{\mathbb{R}^n} |y|^{j+k-2|\alpha|}y_1 \cdot F_1(y) \; dy&=-C\int _{\mathbb{R}^{n-1}} \left(\int _{\mathbb{R}} \frac{\partial F_1}{\partial y_1}(y) \cdot |y|^{j+k-2|\alpha|+2} \; dy_1\right)dy_2\ldots dy_n \\ &\le C\int _{\mathbb{R}^{n}} \left|\frac{\partial F_1}{\partial y_1}(y)\right| \cdot |y|^{j+k-2|\alpha|+2} \; dy \\ &\le C\int _{\mathbb{R}^{n}} \left|y_2\ldots y_n\right|\left(\left| D^{\beta+\gamma}f(y) \right|^2+\left| D_1D^{\beta+\gamma}f(y) \right|^2\right) \cdot |y|^{j+k-2|\alpha|+2} \; dy. \end{align} Repeating this procedure we can finally get \begin{align} \left(\int _{\mathbb{R}^n} \left|y^\alpha\right|\cdot |y|^{j-2|\alpha|} \cdot \left| D^{\beta+\gamma}f(y) \right| \; dy\right)^2\le C \sum_{|\beta|=0}^{n+m} \int _{\mathbb{R}^n} (1+|y|)^{m+n+1} |D^{\beta} f(y)|^2 \; dy. \end{align} Therefore $$\| f \|_{m,\mathcal{S}}^2 \leq C \sum_{|\beta|=0}^{n+m} \int _{\mathbb{R}^n} (1+|y|)^{m+n+1} |D^{\beta} f(y)|^2 \; dy, \quad f \in \mathcal{S}(\mathbb{R}^n).$$ This is stronger than what we want.
• Thanks for reply. I got it 3 days before you ;) a I used similar explanation. I add only that if $\alpha = 0$ and $j=0$ then $D^{\alpha}(|x|^j)= 1$, but this is not a problem, because integral $\int_{\mathbb{R}^n}|D^{\beta+\gamma}f(y)| \; dy$ is finite as Schwartz space is subset of $L^{p}(\mathbb{R}^n)$ and $p \geq 1$. – Marcin Sep 21 '13 at 10:48
• What do you mean by "similar explanation"? You mean you had got all my above argument similarly, or just for the "$|y|^{1-|\alpha|}$" argument? – L. Xu Sep 21 '13 at 16:06