Suppose $f$ is a radial function, i.e., $f(x)=f(|x|)$, and $f \in C^\infty(\bar{B})$, where $\bar{B}$ is the closure of the unit ball in $\mathbb{R}^n$. Prove or disprove the following.

Given any positive integer $k$, $$\sup_{|\alpha|=k,x\in B} |D^\alpha f(x)| \leq \sup_{r < 1} \lvert f^{(k)}(r) \rvert,$$ where $\alpha$ is a multi-index and $D^\alpha f$ is the corresponding derivative of $f$. By $f^{(k)}(r)$, we mean the $k^{th}$ derivative of $f$ as a function of $r=|x|.$ The norm on the left is in fact $|\cdot|_{W^{k,\infty}(B)}.$

I try some functions and the inequality hold for all of them. The case where $k=1$ is easy to prove but I can't prove for a general $k$.

Update: Instead of a general smooth $f$, can we prove the assertion for polynomials in which $f(r) = \sum_{j=0}^m c_j r^{2j}$ ?

  • $\begingroup$ If we drop the assumption $f \in C^\infty (B)$, the desired estimate fails, as can be seen by taking $f(r)=r$, and $k=2$, since the right-hand side vanishes with this choice, but the left-hand side does not. Thus, the estimate is probably false. $\endgroup$
    – PhoemueX
    May 17, 2015 at 7:12
  • $\begingroup$ @Yes Just taking the maximum. For smooth $f$ , $\lvert f \rvert_{W^{k,\infty}(B)}=\sup_{|\alpha|=k} \sup_{x\in B}|D^\alpha f(x)|.$ $\endgroup$
    – booksee
    May 17, 2015 at 20:37
  • $\begingroup$ @PhoemueX Not so simple. The assumption $f\in C^\infty$ implies $f'(0)=0$, hence $|f'(r)/r| \le \sup|f''|$. And $|f'(r)/r|$ is the second derivative in the tangential direction. This (and symmetry considerations) imply the inequality is true for $k=2$. $\endgroup$
    – user147263
    May 18, 2015 at 4:20


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