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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.

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  • $\begingroup$ @ChristianRemling Despite the first question, what about the second statement in my question ? $\endgroup$ – booksee May 20 '15 at 20:30
  • $\begingroup$ This example does not quite seem to work. Does not $d^2f/dr^2$ get large when you actually do the modification near 0? $\endgroup$ – Michael Renardy May 21 '15 at 0:16
  • $\begingroup$ May be use known estimates via the Laplace operator, so you will derive estimates via the radial Bessel in r.h.s. $\endgroup$ – Sergei May 26 '15 at 17:44
  • $\begingroup$ @Sergei Thanks for your comment. Could you provide some details ? I did not see its relation to Laplacian. $\endgroup$ – booksee May 26 '15 at 21:26
  • $\begingroup$ @booksee. See my edits today. I answer to some of your worries. $\endgroup$ – Denis Serre Jun 1 '15 at 7:16
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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 $|\mu_5|\le1$.

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    $\begingroup$ It looks like this reduction is in fact equivalent to the original problem, since one could select a smooth $f$ by requiring $f^{(k)}$ to be close in $L^1$ to the sign of $\mu_k$. In the event that $\mu_k$ is unsigned, then the extremal comes when $f^{(k)}$ is constant, thus $F(x)$ is a constant multiple of $|x|^k$ (this is not smooth, but can be mollified). The question then boils down to whether the $k^{th}$ derivative of $(1 + |x|^2)^{k/2}$ is always bounded above by $k!$, which I believe to be the case (I think a contour integration argument should do this, though I haven't checked.) $\endgroup$ – Terry Tao May 29 '15 at 18:58
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    $\begingroup$ @DenisSerre Thanks for your answer but I don't think it is general to answer my original question. For example, the tensor you choose is $e\otimes\dots\otimes e$ where each unit vector is the same($=e$) and the derivative you compute for $k=2$ cannot represent , say, $\dfrac{\partial^2 f}{\partial x_1\partial x_2}$ in general. In my question, the sup-norm on the left is taken over any multi-index $\alpha$ s.t. $|\alpha|=k$, so we may need to use the tensor like $e_1\otimes \dots \otimes e_k$ where each $e_i$ is an arbitrary unit vector. $\endgroup$ – booksee May 30 '15 at 4:36
  • $\begingroup$ @DenisSerre But if you can prove that the magnitude of derivative w.r.t. $e_1\otimes\dots\otimes e_k$ is bounded by the one w.r.t. $e\otimes\dots\otimes e$, that is nice. $\endgroup$ – booksee May 30 '15 at 4:39
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    $\begingroup$ If it helps, one can write $N_k(\nu,t) = \frac{1}{(k-1)!} \frac{d^k}{dx^k} ( \sqrt{x^2 + 1-\nu^2} - t )^{k-1}|_{x = \nu}$ (this corresponds to the situation where $f^{(k)}$ is a Dirac mass at $tr$). One also has $\mu_k([0,r]) = \frac{1}{k!} \frac{d^k}{dx^k} (x^2 + (1-\nu^2) r^2)^{k/2}|_{x=\nu r}$, which is non-negative thanks to the identity at terrytao.wordpress.com/2015/05/30/a-differentiation-identity and the fundamental theorem of calculus. So it all boils down to checking that $N_k$ is nonnegative. $\endgroup$ – Terry Tao Jun 3 '15 at 18:36
  • $\begingroup$ @Terry. Nice contribution. I'll think about it. By the way, I remember your definitive answer to my question mathoverflow.net/q/51848/8799 . $\endgroup$ – Denis Serre Jun 3 '15 at 18:53
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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 $x=re$.

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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.

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  • $\begingroup$ Good point. I've not proceeded in this way before. However, honestly speaking, I don't think it can be proved via $L^p$ inequalities. Note that the constant on the r.h.s. is actually $1$, which means that it is independent of the domain(and dimension $n$). Hence, in my opinion, any inequality involving integrals is "dangerous". Moreover, I think the inequality is a pointwise one in nature. $\endgroup$ – booksee May 28 '15 at 0:45

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