I was explaining to my students the other day why $\cos(2x)$ is not a linear combination of $\sin(x)$ and $\cos(x)$ over $\mathbb{R}$. Besides the canonical method of using special values of sine and cosine, I noticed something interesting. In the following, all vector spaces are over $\mathbb{R}$.

Consider the linear space $C^\infty_b(\mathbb{R})$ of real-valued bounded smooth functions on $\mathbb{R}$, and take any $c > 0$. We say a function $f \in C^\infty_b(\mathbb{R})$ has property $P(c)$, if for all $k \in \mathbb{N}$ (including $0$), we have $$\sup f^{(k+1)} = c \sup f^{(k)} = -\inf f^{(k+1)} = -c \inf f^{(k)}.$$ Here, the supremum and infimum are of course taken over $\mathbb{R}$, and $f^{(k)}$ is the $k$-th derivative of $f$, with the convention that $f^{(0)}=f$.

Define $$S(c) = \{f \in C^\infty_b(\mathbb{R}) \,\vert\, f \text{ has property } P(c)\}.$$ Since for fixed $a,b$ and all $x$, we have $a \sin(x) + b \cos(x) = \sqrt{a^2 + b^2} \sin(x + \theta) $ for some fixed $\theta$, it is clear that all linear combinations of $\sin(cx)$ and $\cos(cx)$ belong to $S(c)$. In particular, linear combinations of $\sin(x)$ and $\cos(x)$ are all in $S(1)$, while $\cos(2x)$ is not.

Question: is it true that $S(c) = \operatorname{Vect}\bigl(\sin(cx), \cos(cx)\bigr)$?

If $f \in S(1)$ and $f$ is periodic, then using Fourier series, I can prove that $f$ is indeed $2 \pi$-periodic and $f \in \operatorname{Vect}\bigl(\sin(x), \cos(x)\bigr)$ with some work. Although I haven't checked this yet, I also believe that the periodic case for $S(c)$ where $c>0$ is arbitrary could be established by a more elaborate Fourier series argument (of course, I could be wrong). So the real interest lies in treating the non-periodic case, i.e., answering the following

Special case: does $f \in S(c)$ imply $f$ is periodic?

At first, I suspect the answer to the above special case is negative. But after some experiment, I am not so sure. Note that the radius of convergence of the Taylor series (say around $0$) for all $f \in S(c)$ is infinite, so the Taylor series of $f$ converges to $f$ itself. In particular, all functions in $S(c)$ are automatically analytic, so one does not have much freedom when trying to construct a (counter-)example.

If the answer turns out to be negative, then can one at least assert that $S(c)$ is a linear subspace? What if we only consider periodic functions for some fixed period in case $S(c)$ is not a linear subspace? Of course, these probably depend on the explicit form of the answer which is not yet known to me, and all of these are just some (perhaps stupid and naive) speculation on an old exercise of a first-year undergraduate. But it seems interesting, and any thought is appreciated.

Edit: I was a bit careless in formulating the question since the questions for all different $c$ are equivalent merely by rescaling, so one can simply assume $c = 1$, in which case we still have much work to do.

Edit 2: Proof of the periodic case can be found here in case anyone is interested.

  • $\begingroup$ Could you modulate $\sin(x)$ by some sufficiently smooth bump function $\varphi(x)$ with $\varphi(x)\to 1$ as $x\to\infty$? At first glance that seems like it should work, and since you're in $C^\infty$, I would think that the usual bumps are sufficient for this... $\endgroup$ Feb 18, 2020 at 17:23
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    $\begingroup$ All such functions are analytic, so once it vanishes on some interval, it is zero everywhere. There's unfortunately no analytic bump functions! $\endgroup$ Feb 18, 2020 at 17:32
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    $\begingroup$ @Zero Just consider the remainder of the Taylor expansion to the order $n$, and you will see immediately that due to the restrictions of the higher derivatives, the remainder tends to zero as $n$ goes to infinity, which means exactly that the Taylor series converges pointwise to the function itself and the function is indeed given by a power series (its Taylor series). $\endgroup$
    – Hua Wang
    Feb 18, 2020 at 21:30
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    $\begingroup$ @HuaWang : You wrote in your post: "If $f \in S(1)$ and $f$ is periodic, then using Fourier series, I can prove that $f$ is indeed $2 \pi$-periodic and $f \in \operatorname{Vect}\bigl(\sin(x), \cos(x)\bigr)$ with some work." If can indeed prove this, I think it could well be useful for some of us to see that proof, which you could write elsewhere, and here provide a link to that proof. $\endgroup$ Feb 20, 2020 at 22:13
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    $\begingroup$ @IosifPinelis OK. Following your advice, I wrote a detailed version of the periodic case, and put it on my google drive. Here is the link drive.google.com/file/d/1Vyp5bcaGVXHvdxuvksNaaXQcrNlG4pFT/… $\endgroup$
    – Hua Wang
    Feb 21, 2020 at 2:20

1 Answer 1


As noted by the OP we can replace $f$ by $af(bx)$ for suitable $a,b\in\mathbb{R}$ so that wlog we can take $c=1$ and ensure that $\sup f=-\inf f=1$.

Firstly we note that $f(z)$ is infinitely differentiable on $\mathbb{R}$ so we can form the taylor series at 0, $f(z)=\sum_{i=0}^{\infty}\frac{f^{(i)}(0)}{i!}z^i$. Since $|f^{(i)}(0)|\leq 1$ for all $i\geq 0$ we have $\sum_{i=0}^{\infty}|\frac{f^{(i)}(0)}{i!}z^i|=\sum_{i=0}^{\infty}\frac{|f^{(i)}(0)|}{i!}|z|^i\leq \sum_{i=0}^{\infty}\frac{|z|^i}{i!}$ which converges for all $z$ to $e^{|z|}$.

Hence $F(z)=\sum_{i=0}^{\infty}\frac{a_i}{i!}z^i$ is an absolutely convergent series defining an entire function on $\mathbb{C}$ agreeing with $f$ on $\mathbb{R}$ s.t. $\sup f^{(k)}=-\inf f^{(k)}=1$ for all $k\in \mathbb{Z}_{\geq0}$ and $|F(z)|\leq e^{|z|}$ for all $z\in \mathbb{C}$.

We now determine the form of $F$ given the preceding conditions.

First note that Bernstein proved the following (see Rahman and Tariq$^1$) as an extension of his related inequality for polynomials:

Theorem Let $g$ be an entire function of exponential type $\tau>0$ such that $|g(x)|\leq M$ on the real axis. Then $$\sup_{-\infty<x<+\infty}|g^{'}(x)|\leq M\tau.$$

Now $|F(z)|\leq e^{|z|}$ for all $z\in \mathbb{C}$ and therefore we know that $F$ is of exponential type 1. In addition $|F(x)|=|f(x)|\leq 1$ on the real axis.

We are therefore interested in the conditions of equality in the above.

Fortunately in their book "Analytic Theory of Polynomials"$^2$ Rahman and Schmeisser prove (Theorem 14.1.7) that equality holds in the above if and only if $g(z)=ae^{i\tau z}+be^{-i\tau z}$ where $|a|+|b|=M$. (You can access the relevant pages 513-514 on google books)

$|F(x)|\leq 1$ for $x\in\mathbb{R}$ and hence in the above theorem we may take $M=1$, $\tau=1$ and $g=F$ since $|F|$ is of exponential type 1. Also $\sup_{x\in\mathbb{R}}|F^{'}(x)|=1$ which means we have the case of equality.

Thus $F(z)=ae^{iz}+be^{-iz}$ where $|a|+|b|=M=1$.

On the real axis $F$ is real valued and agrees with $f$ so we must have $f(x)=ae^{ix}+be^{-ix}$, $x\in \mathbb{R}$ with $a=\bar{b}$. Setting $a=c+id$ we obtain $f(x)=2c\cos x-2d\sin x$ with $|a|=|b|=2\sqrt{c^2+d^2}=1$. Rewriting with $C=2c$, $D=-2d$, we obtain

$$f(x)=C\cos x+D\sin x$$ for some constants $C,D\in \mathbb{R}$, $C^2+D^2=1$.

This proves the OP's conjecture. Note that the condition $C^2+D^2=1$ is due to the bound we imposed on $f$ due to our normalisation which is not part of the definition of $S(c)$.

1 Rahman, Q. I.; Tariq, Q. M., On Bernstein’s inequality for entire functions of exponential type, J. Math. Anal. Appl. 359, No. 1, 168-180 (2009). ZBL1168.30002.

2 Rahman, Q. I.; Schmeisser, G., Analytic theory of polynomials, London Mathematical Society Monographs. New Series 26. Oxford: Oxford University Press (ISBN 0-19-853493-0/hbk). xiv, 742 p. (2002). ZBL1072.30006.

  • $\begingroup$ Reading your solution, it looks as if the condition $\inf = -\sup$ is actually superfluous and the result follows only from the conditions $\sup \vert f^{(k)} \vert = 1$ for all $k \ge 0$. So, you did prove a stronger result (unless I have overlooked something). $\endgroup$
    – Luc Guyot
    Feb 27, 2020 at 20:36
  • $\begingroup$ @LucGuyot I think you can lose all the sup and inf conditions except one of them say $\sup |f^{(i)}|=1$ or $\inf|f^{(i)}|=-1$, $i>0$ as long as you have $|f^{(k)}|\leq1$ for all $k\geq 1$. You need the bounds on the derivatives to get the $|F(z)|\leq e^{|z|}$ so that $F$ has exponential type 1, essential for this solution. $\endgroup$
    – Ivan Meir
    Feb 27, 2020 at 21:29
  • $\begingroup$ @LucGuyot Note that if you have $\sup |f^{(i)}|=1$ or $\inf|f^{(i)}|=-1$, $i>0$ then you just apply the same method to $f^{(i-1)}$ to get the result. $\endgroup$
    – Ivan Meir
    Feb 27, 2020 at 21:31

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