Derivatives of radial functions can be bounded by derivatives in terms of radial distance?

Question

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 $r=|x|.$

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

Instead of a general smooth $f$, can we prove the assertion for polynomials(or an uniformly and absolutely converging power series) with only even powers, namely, $$f(r) = \sum_{j=0}^m c_j r^{2j} \quad(m\text{ can be}+\infty)\quad?$$

PS: I asked this question on Math.SE but no one answered so it is posted here. This is a quite simple/straightforwad question (that a Freshman in math can fully understand) but it is surprising that till now, no one(me included) could answer it or at least give some idea.

Answer 1

To avoid confusion, I shall denote $f=f(r)$ and $F(x)=f(|x|)$ the corresponding radial function.

Because of Van der Corput-Schaake Inequality, it is enough to prove the following more general inequality: for every point $x\ne0$, integer $k\ge1$ and unit vector $e$, $$|D^kF_xe^{\otimes k}|\le\sup_{0\le s\le r}|f^{(k)}(s)|\quad ?$$ This is obvious for $k=1$. Let me first prove that it is true for $k=2,3$. Then I'll give a general strategy for a proof. In the following, I denote $$\nu:=\frac{x\cdot e}r\in[-1,1].$$

For $k=2$, one has $$D^2F_xe^{\otimes 2}=(1-\nu^2)\frac1rf'+\nu^2f''.$$ Because $F$ is smooth, we know that the derivatives of $f$ of odd order vanish at $0$. Using $f'(0)=0$, we obtain $$D^2F_xe^{\otimes 2}=(1-\nu^2)\frac1r\int_0^rf''(s)ds+\nu^2f''=\langle\mu_2,f''\rangle.$$ There remains to check that the measure $\mu_2$ has total mass $1$ (i.e. is a probability). This is obvious.

If $k=3$, we have $$D^3F_xe^{\otimes 3}=3(\nu-\nu^3)\left(\frac1rf''-\frac1{r^2}f'\right)+\nu^3f'''.$$ Using again $f'(0)=0$, we obtain $$D^3F_xe^{\otimes 3}=\langle\mu_3,f'''\rangle$$ where $$\mu_3:=\nu^3\delta_{s=r}+3(\nu-\nu^3)\frac{s}{r^2}\chi_{[0,r]}(s).$$ Again, one checks that $$|\mu_3|=|\nu^3+\frac32(\nu-\nu^3)|\le1.$$

The general strategy is to prove that if $r=|x|$, then $$D^kF_xe^{\otimes k}=\langle\mu_k,f^{(k)}\rangle$$ for some measure $\mu_k$ over $[0,r]$. Mind that the coefficients of $\mu_k$ involve only $\nu$ and $r$. The measure $\mu_k$ can be calculated from the Faa di Bruno Formula (I suspect that $\mu_k$ has a constant sign). To conclude, one has to check that its total mass is less than $1$. This total mass involves only $\nu$ ; it is likely to be the absolute value of a polynomial $P_k(\nu)$.

More precisely, $\mu_k$ is the sum of the Dirac mass $\nu^k\delta_{x=r}$ and a continuous density over $s\in(0,r)$ : $$\frac1rN_k(\nu,t),\qquad t:=\frac sr.$$ The density is determined by induction with the following rules $$\partial_t(N_{k+1}-\nu tN_k) = -(1-\nu^2)\partial_\nu N_k, $$ $$N_{k+1}(\nu,1) = \nu N_k(\nu,1)+k\nu^{k-1}(1-\nu^2),$$ and we have $N_k(\nu,0)$ if $k$ is odd. The second line gives explicitly $$N_k(\nu,1)=\frac{k(k-1}2\nu^{k-2}(1-\nu^2).$$ The first few $N_k$'s are $$N_1\equiv0,\quad N_2=1-\nu^2,\quad N_3=3\nu(1-\nu^2)t,$$ $$N_4=\frac32(1-\nu^2)(5\nu^2-1)(t^2-1)+6\nu^2(1-\nu^2),$$ $$N_5=\frac32\nu(1-\nu^2)(5\nu^2-1)t(t^2-1)+6\nu^3(1-\nu^2)(t-1)-2(3\nu-5\nu^3)(1-\nu^2)(t^3-3t+2)-12\nu(1-\nu^2)(1-2\nu^2)(t-1)+10\nu^3(1-\nu^2).$$ One verifies easily that $\mu_4$ is a probability (Question : is $\mu_{2\ell}$ always a probability ?).

I checked that for every value of the angle $\nu$, $$0\le\langle\mu_5,{\bf1}\rangle\le1.$$ However, I gave up about the sign of $N_5$, and therefore I cannot claim that the total mass $|\mu_5|\le1$.

Answer 2

Here is an argument to show that $\nabla^k F_x\cdot e^{\otimes^k}=f^{(k)}(r)$ if $x=re$. This shows that the inequality Denis is seeking to prove is sharp, if true; I can verify this inequality in the particular case $f(r)=e^{-r^2}$. Here is the argument:

Let $g$ be a function such that $g(r^2)=f(r)$, so that in Denis' notation, $F(x)=g(|x|^2)$. The derivatives of $f$ and the expressions $\nabla^k F_x\cdot e^{\otimes^k}$ can be expressed in terms of the derivatives of $g$ as follows:

For every $k$, there exist polynomials $P_{i,k}(t)$ such that $$ f^{(k)}(r)=\sum_{i=1}^k P_{i,k}(r)g^{(i)}(r^2) $$ and $$ \nabla^k F_x\cdot e^{\otimes^k}=\sum_{i=1}^k P_{i,k}(x\cdot e)g^{(i)}(|x|^2). $$ Taking $x=re$, we see that the RHS of both equations are equal.

To prove these equations, use induction and the following formulas, valid for any $h, P$: $$ \nabla(h(|x|^2))\cdot e = 2(x\cdot e)h'(|x|^2)\qquad \nabla(P(x\cdot e))\cdot e = P'(x\cdot e) $$ and $$ \frac{d}{dr}h(r^2)=2r h'(r^2)\qquad \frac{d}{dr}P(r)=P'(r). $$ In addition, these polynomials satisfy the recurrence $P_{i,k+1}(t)=P_{i,k}'(t)+2tP_{i-1,k}(t)$, so that, for $k$ even, the $P_{i,k}$ are even, and for $k$ odd, the $P_{i,k}$ are odd. Thus if $x=-re$, we have equality up to $\pm1$.

If $g(t)=e^{-t}$ (i.e. $f(r)=e^{-r^2}$), then $$ \nabla^k F_x\cdot e^{\otimes^k}=H_k(x\cdot e)g(r^2), $$ where $H_n(t)$ is the $n$th Hermite polynomial (which satisfy a similar recurrence to the $P_{i,k}$). Since $H_k(t)g(r^2)$ is decreasing in $r$, the sup of $\nabla^k F_x\cdot e^{\otimes^k}$ over $|x|\leq r$ (i.e. $|t|\leq r$) is attained when $x=re$.

Answer 3

There are many estimates of Sobolev type of the form $$ ||D^{\alpha}f||_1 \le ||\Delta f||_2 $$ for different pairs of norms. For your case the r.h.s. will be one-dimensional in r $||(D^2-\frac{n-1}{r}-\frac{1}{r^2})f||_2$. Now the question is reduced to one-dimensional estimates of this norm, say from the embedding theorem bounding derivatives uniformly.

Of course it is just a plan to attack the problem, no pretence for being an answer complete, bounty cake and so on.

E.g. in case of the Lebesgue norms with powers of Bessel operators instead standard derivatives-we have Kipriyanov spaces, like Sobolev ones. I proved some years ago that these norms are equivalent to Sobolev with power weights. For such spaces there are embedding in $C$---spaces. So estimates with Bessel powers lead to uniform estimates.