I'm reading Functional Analysis by Rudin about distribution theory. I have a problem of derivation of the inequality (10) in theorem 6.25, page 166.

It first proves an inequality $$(5)\quad\quad\quad\quad |D^{\alpha}\phi(x)|\le\eta n^{N-|\alpha|}|x|^{N-|\alpha|}\quad\quad\quad(x\in K,|x|\le N)$$ where $\phi\in\mathscr{D}(\Omega)$ ,$\eta\gt 0$,$n$ is that in $R^{n}$ and $N$ is a natural number. I have no problem with this inequality.

Then choose an auxiliary function $\psi\in\mathscr{D}(R^{n})$, which is $1$ in some neighborhood of $0$ and whose support is in the unit ball B of $R^{n}$. Define $$\psi_{r}(x)=\psi(\frac{x}{r})\quad\quad(r\gt 0)$$ I think the precise definition of $\psi$ has nothing to do with the following derivation. I put it here just for completeness.

By Leibniz formula, $$(9)\quad\quad\quad D^{\alpha}(\psi_{r}\phi)=\sum_{\beta\le\alpha}c_{\alpha\beta}(D^{\alpha-\beta}\psi)(\frac{x}{r})(D^{\beta}\phi)(x)r^{|\beta|-|\alpha|}$$

It now follows from (5) that $$(10)\quad\quad\quad\|\psi_{r}\phi\|_{N}\le\eta C\|\psi\|_{N}$$ where C is a constant depending on $n$ and $N$. The norm $\|\cdot\|_{N}$ is defined as $$\|f\|_{N}=\sup\{D^{\alpha}f(x),x\in X,\alpha\le N\}$$

My question is how to derive (10) from (5) and what $C$ is. I tried myself but I got a constant depending on $r$ as well.


1 Answer 1


Look at $(9)$. If $\lvert x\rvert > r$, then $(D^{\alpha-\beta}\psi)(x/r) = 0$ since the support of $\psi_r$ is contained in $rB$. If $\lvert x\rvert \leqslant r$, then $(5)$ shows

$$\Bigl\lvert (D^{\beta}\phi)(x) r^{\lvert\beta\rvert -\lvert\alpha\rvert}\Bigr\rvert \leqslant \eta n^{N-\lvert\beta\rvert} \lvert x\rvert^{N-\lvert\beta\rvert} r^{\lvert\beta\rvert - \lvert\alpha\rvert} \leqslant\eta n^{N-\lvert\beta\rvert} r^{N-\lvert\alpha\rvert}.\tag{$\ast$}$$

Since we are interested in small $r > 0$, we may without loss of generality assume $r \leqslant 1$, and then from $(\ast)$ we obtain

$$\Bigl\lvert (D^{\beta}\phi)(x) r^{\lvert\beta\rvert -\lvert\alpha\rvert}\Bigr\rvert \leqslant \eta n^{N-\lvert\beta\rvert} \tag{${\ast}{\ast}$}$$

as $r^{N-\lvert\alpha\rvert} \leqslant 1$ because $\lvert\alpha\rvert \leqslant N$. Inserting $({\ast}{\ast})$ into $(9)$ yields

$$\lvert D^{\alpha}(\psi_r\phi)(x)\rvert \leqslant \sum_{\beta\leqslant \alpha} c_{\alpha\beta} \bigl\lvert(D^{\alpha-\beta}\psi)(x/r)\bigr\rvert\cdot \eta n^{N-\lvert\beta\rvert} \leqslant \eta \lVert\psi\rVert_N\cdot \underbrace{\sum_{\beta\leqslant \alpha} c_{\alpha\beta} n^{N-\lvert\beta\rvert}}_{C_{\alpha}}$$

for $\lvert x\rvert \leqslant r$. Since the inequality is trivial for $\lvert x\rvert > r$,

$$\sup \{ \lvert D^{\alpha}(\psi_r\phi)(x)\rvert : x \in \mathbb{R}^n\} \leqslant \eta \cdot C_{\alpha} \cdot \lVert \psi\rVert_N.$$

Take $C = \max \{ C_{\alpha} : \lvert \alpha \rvert \leqslant N\}$ to obtain $(10)$.

  • $\begingroup$ Thanks for this detailed answer. I think I get through your derivation. But I still need some clarification on a point that the proof in text says we can choose a compact ball $K$ centered at $0$ such that $|D^{N}\phi|\le\eta$ in $K$. I think this is because $(D^{N}\phi)(0)=0$ and $D^{N}\phi$ is continuous around $0$. Then we can choose a small neighborhood such that the values of $D^{N}\phi$ in this neighborhood is close to $0$. In this case I think the compactness of $K$ is actually of no use to this proof and actually any open set $K$ should work, right? $\endgroup$
    – Hua
    Mar 9, 2016 at 11:34
  • $\begingroup$ Yes, the compactness of $K$ isn't used. But the convexity - actually, only that $K$ is star-shaped with respect to $0$ - is used to prove $(5)$, in the invocation of the mean value theorem. So one could use any star-shaped neighbourhood of $0$ on which $\lvert D^{\alpha}\phi\rvert$ is small for $\lvert\alpha\rvert = N$. $\endgroup$ Mar 9, 2016 at 12:13
  • $\begingroup$ Thanks for point the convexity thing out! $\endgroup$
    – Hua
    Mar 9, 2016 at 12:27

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