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Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of smooth functions $f_n:\mathbb{R}\rightarrow\mathbb{R}$ which are bounded together with all their derivatives $f^{(j)}_n$, $j,n\in\mathbb{N}$, and $(\|f_n\|_\infty)_{n\in\mathbb{N}}$ the corresponding numerical sequence of their supremum norms: $$\|f_n\|_\infty=\sup\{|f_n(t)|\ |\ t\in\mathbb{R}\}\ ,\quad n\in\mathbb{N}\ .$$

Question: Can we choose the $f_n$'s so that:

  1. $(\|f_n\|_\infty)_{n\in\mathbb{N}}$ has rapid decay: $(n^k\|f_n\|_\infty)_{n\in\mathbb{N}}$ is bounded for all $k\in\mathbb{N}$; and
  2. For each $j\in\mathbb{N}$ we have that $(\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ does not have rapid decay but instead only polynomial growth: $(n^{-k}\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ is bounded for some $k\in\mathbb{N}$?

Property 2. means that for all (positive) $j\in\mathbb{N}$ there are (positive) $k_+,k_-\in\mathbb{N}$ such that $(n^{-k_-}\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ is bounded but $(n^{k_+}\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ is not. By replacing $k_+,k_-$ with $\max\{k_+,k_-\}$, one can rephrase Property 2. as follows: for all (positive) $j\in\mathbb{N}$ there is a (positive) $k\in\mathbb{N}$ such that $(n^{-k}\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ is bounded but $(n^{k}\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ is not.

I am particularly interested in specific examples, but abstract (non-)existence arguments are also fine.

Edit: in view of the negative answer I have found below, the following weaker situation should prove more interesting:

Follow-up question: Can we choose the $f_n$'s so that:

  1. $(|f_n(t)|)_{n\in\mathbb{N}}$ has rapid decay for each $t\in\mathbb{R}$ (i.e. we only require bounds on $(n^k|f_n(t)|)_{n\in\mathbb{N}}$ for each $k\in\mathbb{N}$ which are pointwise in $t$, and no longer uniform as before);
  2. For each $j\in\mathbb{N}\cup\{0\}$ we have that $(\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ has polynomial growth: $(n^{-k}\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ is bounded for some $k\in\mathbb{N}$; and
  3. For some $j\in\mathbb{N}\cup\{0\}$ we have that $(\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ does not have rapid decay: $(n^k\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ is unbounded for some $k\in\mathbb{N}$?

The answer to the follow-up question would be negative if it were somehow possible to use 2) to upgrade the pointwise bounds in 1) to uniform bounds, for if so we are be back to the original question.

Pedro Lauridsen Ribeiro
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  • For 1), can't you just take $f_n \equiv 0$ for each $n$ – mathworker21 Dec 19 '18 at 23:23
  • It's not difficult to come up with examples satisfying either 1) or 2). A non-zero example of 1) would be e.g. the sequence of constant functions $f_n(t)\equiv 2^{-n}$, whereas an example of 2) (which must be non-zero) would be e.g. the sequence of Gaussians $f_n(t)=\exp(-nt^2)$. The first example (as well as yours) violates 2), whereas the second example violates 1). The challenge is to get both 1) and 2) - it may be the case that both requirements cannot even be met simultaneously, I don't know (and would love to)... – Pedro Lauridsen Ribeiro Dec 20 '18 at 02:42
  • Another example of 1), which actually violates 2) in the opposite direction as the previous ones, would be $f_n(t)=2^{-n}\sin(2^n t)$, whose derivatives have supremum norms which grow exponentially with $n$ from the second derivative onwards. – Pedro Lauridsen Ribeiro Dec 20 '18 at 02:49
  • Locally, the answer to the follow-up question is positive, as suggested in https://math.stackexchange.com/a/478056/52543 - take e.g. $f_n(t)=n^2 t^n(1−t)$, $t\in[0,1]$. Since $f_n(0)=f_n(1)=0$ for all $n$ and $\lim_{n\rightarrow\infty}n^ k f_n(t)=0$ for all $t\in(0,1)$, $k\in\mathbb{N}$, 1) holds. 2) is obvious, and the maximum of $f_n(t)$ in $[0,1]$ takes place at the only positive zero $t_n=\frac{n}{n+1}$ of $f'_n(t)=n^3 t^{n−1}(1−\frac{n+1}{n}t)$, with $f_n(t_n)=\frac{n^2}{n+1}(1−\frac{1}{n+1})^n$. Asymptotically we have $f_n(t_n)\sim n/e$ for large $n$ and therefore 3) holds for $j=0$. – Pedro Lauridsen Ribeiro Dec 20 '18 at 20:25

1 Answers1

1

It turns out there is no such sequence, due to the following special case of the Gagliardo-Nirenberg interpolation inequality: given any smooth function $f$ on $\mathbb{R}$ which is bounded together with all its derivatives, then for all $j,m\in\mathbb{N}$ with $j<m$ there is a constant $C=C_{j,m}>0$ such that $$\|f^{(j)}\|_\infty\leq C\|f^{(m)}\|^{\frac{j}{m}}_\infty\|f\|^{1-\frac{j}{m}}_\infty\ .$$ This bound also has a local version: given any $f\in\mathscr{C}^\infty(I)$, $I=[a,b]$ with $a<b\in\mathbb{R}$ and $j,m$ as above, there are constants $C_1=C_{1,j,k}>0$, $C_2=C_{2,j,k}>0$ such that $$\|f^{(j)}\|_{\infty,I}\leq C_1\|f^{(m)}\|^{\frac{j}{m}}_{\infty,I}\|f\|^{1-\frac{j}{m}}_{\infty,I}+C_2\|f\|_{\infty,I}\ ,$$ where $\|f\|_{\infty,I}=\sup\{|f(t)|\ |\ t\in I\}$.

Suppose we are given $(f_n)_{n\in\mathbb{N}}$ such that $(\|f_n\|_\infty)_{n\in\mathbb{N}}$ has rapid decay and $(\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ has polynomial growth, that is:

  • $(n^k\|f\|_\infty)_{n\in\mathbb{N}}$ is bounded for all $k\in\mathbb{N}$;
  • For all $j\in\mathbb{N}$ there is a $k\in\mathbb{N}$ such that $(n^{-k}\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ is bounded.

The first condition is just 1), whereas the second condition is weaker than 2) in that we do not forbid $(\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ to have rapid decay. We will show now that the latter in fact must happen under these two conditions. Given $j<m\in\mathbb{N}$ and $k'\in\mathbb{N}$ such that $(n^{-k'}\|f^{(m)}_n\|_\infty)_{n\in\mathbb{N}}$ is bounded, it follows from the first GN inequality that for all $n,k\in\mathbb{N}$ we have $$n^k\|f^{(j)}_n\|_\infty\leq C (n^{-k'}\|f^{(m)}_n\|_\infty)^{\frac{j}{m}}(n^{\frac{mk+jk'}{m-j}}\|f_n\|_\infty)^{1-\frac{j}{m}}$$ and therefore $(n^k\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ is bounded for all $k\in\mathbb{N}$. Since $j<m\in\mathbb{N}$ were arbitrary, we conclude that $(n^k\|f^{(j)}\|_\infty)_{n\in\mathbb{N}}$ is bounded for all $j,k\in\mathbb{N}$, as asserted.

It is clear that a local version of this result must also hold: if $(f_n)_{n\in\mathbb{N}}$ is a sequence of smooth functions on $I=[a,b]$, $a<b\in\mathbb{R}$ such that $(\|f_n\|_{\infty,I})_{n\in\mathbb{N}}$ has rapid decay and $(\|f^{(j)}_n\|_{\infty,I})_{n\in\mathbb{N}}$ has polynomial growth, that is:

  • $(n^k\|f\|_{\infty,I})_{n\in\mathbb{N}}$ is bounded for all $k\in\mathbb{N}$;
  • For all $j\in\mathbb{N}$ there is a $k\in\mathbb{N}$ such that $(n^{-k}\|f^{(j)}_n\|_{\infty,I})_{n\in\mathbb{N}}$ is bounded,

then $(\|f^{(j)}_n\|_{\infty,I})_{n\in\mathbb{N}}$ must also have rapid decay for all $j\in\mathbb{N}$, for given $j<m\in\mathbb{N}$ and $k'\in\mathbb{N}$ such that $(n^{-k'}\|f^{(m)}_n\|_{\infty,I})_{n\in\mathbb{N}}$ is bounded, it follows from the second GN inequality that for all $n,k\in\mathbb{N}$ we have $$n^k\|f^{(j)}_n\|_{\infty,I}\leq C_1 (n^{-k'}\|f^{(m)}_n\|_{\infty,I})^{\frac{j}{m}}(n^{\frac{mk+jk'}{m-j}}\|f_n\|_{\infty,I})^{1-\frac{j}{m}}+C_2 n^k\|f_n\|_{\infty,I}$$ and therefore $(n^k\|f^{(j)}_n\|_\infty)_{n\in\mathbb{N}}$ is bounded for all $j,k\in\mathbb{N}$, as before.

Pedro Lauridsen Ribeiro
  • 1,337