2
$\begingroup$

Fix a non-empty closed subset $\Omega\subset\mathbb{R}$.

Let $f:\mathbb{R}\to\mathbb{R}$ be smooth and such that $\sup_{x\in \Omega}\,|\partial^k f(x)|\lesssim k^r$ for some $r\ge 0$ for all $k\in \mathbb{N}$ (including $0$).

If $\Omega= \mathbb{R}$ then, from this nice MSE post, we know that if $r=0$ then $f$ is a multiple/translation of the $\sin$ function; i.e.\ $f = c\sin(\cdot +b)$ for some $c,b\in \mathbb{R}$.

Is there a characterization of such functions when $r>0$ for $\Omega=\mathbb{R}$?

More importantly, what if $\Omega =[0,1]$?

Improve this question
$\endgroup$
8
  • $\begingroup$ sorry I mean $s->k$ (I sometimes use "s" for "smoothness"). So I'm wondering what classes of functions are included when we fix a "polynomial-esque" growth rate for the maximal $k^{th}$ order of its derivatives. $\endgroup$
    – Math_Newbie
    Commented May 15 at 19:07
  • $\begingroup$ What do you mean by $\lesssim k^r$? $\endgroup$
    – Iosif Pinelis
    Commented May 15 at 19:40
  • 1
    $\begingroup$ most likely $\exists C>0$, $\forall k\ge 0$, the sup is $\le Ck^r$. $\endgroup$
    – Abdelmalek Abdesselam
    Commented May 15 at 19:57
  • 1
    $\begingroup$ Of course, we can also write $\le k^r$ instead of $\lesssim k^r$ and it's still the same question (since we can replace $f$ by $cf$). $\endgroup$
    – Christian Remling
    Commented May 15 at 20:21
  • 1
    $\begingroup$ I think the argument from the linked MSE post still works (without having checked all the fine details): if $\|f^{(k)}\|_{\infty}\le k^N$, then $f$ is entire of exponential type $\le 1$. By Paley-Wiener, $\widehat{f}$ is supported by $[-1,1]$, and this should give $\|f^{(k)}\|_{\infty}\le \|f\|_{\infty}$, so that there are no additional functions. $\endgroup$
    – Christian Remling
    Commented May 15 at 21:33

1 Answer 1

Reset to default
3
$\begingroup$

For the bounded case on $[0,1]$ there's not much to say.

First, let $T(x)$ be the power series $$ \sum \frac{a_k}{k!} x^k $$ with the sequence $a_k$ satisfying $|a_k| \leq C (1+k)^r$ for some $C > 0$. This implies that $T$ has an infinite radius of convergence and converges to a real analytic function on $\mathbb{R}$. In particular, we can compute $$ T^{(\ell)}(x) = \sum \frac{a_{k+\ell}}{k!} x^k $$ We have that for $\ell > 0$ $$ |a_{k+\ell}| \leq C(1 + k + \ell)^r \leq C\ell^r (2+k)^r $$ so for $x\in [0,1]$ we find $$ |T^{(\ell)}(x) | \leq C \left( \sum \frac{(2+k)^r}{k!} \right) \ell^r =: \tilde{C} \ell^r \tag{*}\label{eq:T:der:bn}$$ and so $T$ defines a function with the requisite property.

Going backwards, let $T(x)$ be the power series $$ T(x) = \sum \frac{a_k}{k!} x^k, \qquad a_k = f^{(k)}(0) $$ Our hypothesis on $f$ implies that $|a_k| \leq C(1+k)^r$ for some $C$, and hence the previous paragraph can be used to show that $T$ converges to a real analytic function on $\mathbb{R}$ with \eqref{eq:T:der:bn} satisfied for every $\ell > 0$ and $x\in [0,1]$. For any $m > 0$, the order $m$ Taylor remainder theorem states that $$ \sup_{x\in [0,1]} |T(x) - f(x)| \leq \frac{1}{(m+1)!} \sup_{x\in [0,1]} | T^{(m+1)}(x) - f^{(m+1)}(x)| \leq \frac{(C + \tilde{C}) (m+1)^r}{(m+1)!} $$ and hence $T = f$ on $[0,1]$.

Conclusion If $f$ is a smooth function on $\mathbb{R}$, then $\sup_{x\in [0,1], k\in \mathbb{N}}k^{-r} |f^{(k)}(x)| < \infty$ if and only if $f|_{[0,1]}$ is the restriction of some real analytic function defined by a power series $\sum \frac{a_k}{k!} x^k$ with $\sup_{k\in \mathbb{N}} (1+k)^{-r} |a_k| < \infty$.

(Which is admittedly rather underwhelming.)

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .