I'm working with the derivative of a radial function $f(r)\in C^\infty(\mathbb R),$ so I need to calculate $\partial^\alpha f(|x|)$ for $x$ in $\mathbb{R}^n$ not equal to the origin. $\alpha\in \mathbb{N}^n$ is a multi-index. By chain rule, I have to calculate $\partial^\beta|x|,$ $\forall \beta\le\alpha.$ But I find it quite hard to give an explicit expression.

I try to estimate $|\partial^\alpha f(|x|)|$ then, which means I have to estimate $|\partial^\alpha|x||$, $\forall \alpha\in \mathbb{N}^n.$ I guess we have $|\partial^\alpha|x||\le 1$ but fail to prove it. I tried induction but it's too complex. I have to consider many cases of $\alpha$ be like.

Is there an elegant way to derive or at least estimate $\partial^\alpha|x|$?


1 Answer 1


There is a good way to induct. For $k\in\mathbb R$, let's call a function $f:\mathbb R^n\setminus \{0\} \to \mathbb R$ $k$-homogeneous if $f(\lambda x) =\lambda^k f(x)$ for all $x\in \mathbb R^n\setminus \{0\}$ and $\lambda > 0$. For example, $|x|$ is $1$-homogeneous. $0$ happens to be $42$-homogenous. With this concept in mind, its not hard to prove

Prop. Let $f:\mathbb R^n\setminus \{0\} \to \mathbb R$ be smooth and $k$-homogeneous. Then any partial $\partial_i f$ is $(k-1)$-homogeneous.

Proof. \begin{align} (\partial_i f)(\lambda x) &= \lim_{h\to 0}\frac{f(\lambda x+he_i)-f(\lambda x)}h \\&=\lambda^k\lim_{h\to 0}\frac{f( x+he_i/\lambda)-f( x)}h \\&=\lambda^{k-1}\lim_{h\to 0}\frac{f( x+(h/\lambda)e_i)-f( x)}{h/\lambda} \\&=\lambda^{k-1}\lim_{t\to 0}\frac{f( x+te_i)-f( x)}t \\&= \lambda^{k-1} \partial_i f(x). \text{QED} \end{align}

Now induction immediately gives

Corollary. $$|\partial^\beta|x|| \le C_{\beta} |x|^{1-|\beta|}. $$ ($C_{\beta}$ is the maximum of the smooth function $\partial^\beta|x|$ on the unit sphere.)

  • $\begingroup$ Thanks a lot. It also solve my another question: What if taking $\ell^p$ norm of $x$ instead $p=2$? Now taking $\ell^p$ norm is also $1$-homogeneous so the answer is same. $\endgroup$
    – DreamAR
    Nov 11, 2021 at 15:09
  • $\begingroup$ As long as they are smooth, yes. Glad to help :) $\endgroup$ Nov 11, 2021 at 15:25

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