# If $f$ is a smooth real valued function on real line such that $f'(0)=1$ and $|f^{(n)} (x)|$ is uniformly bounded by $1$ , then $f(x)=\sin x$?

Let $f : \mathbb R \to \mathbb R$ be a smooth ( infinitely differentiable everywhere ) function such that $f '(0)=1$

and $|f^{(n)} (x)| \le 1 , \forall x \in \mathbb R , \forall n \ge 0$ ( as usual denoting $f^{(0)}(x):=f(x)$) ; then is it true that

$f(x)=\sin x , \forall x \in \mathbb R$ ?

• Multiply $f$ by some bump function. This shows that the answer is no for just smooth functions. I don't know the answer for analytic functions, though.
– Dirk
Commented Jul 18, 2016 at 15:31
• @Dirk : But there are other conditions imposed besides being smooth ... and I don't really understand what you are saying ...
– user228168
Commented Jul 18, 2016 at 15:32
• @Dirk Bump function does not satisfies $|f^{(n)}(x)|\le 1$. Indeed this condition implies that $f$ is analytic.
– user99914
Commented Jul 18, 2016 at 15:44
• @ArcticChar I was just gonna say that. It's an entire function. And it's clear that $|f(iy)|\le e^y$. And that argument, using power series centered at $x$, shows that in fact $|f(x+iy)|\le e^{|y|}$. And in fact $|f^{(k)}(x+iy)|\le e^{|y|}$. Commented Jul 18, 2016 at 15:55
• Might be useful for various reasons to note that $e^{it}/i$ satisfies all the given conditions except for being real-valued... Commented Jul 18, 2016 at 16:09

Yes, $$f(x)=\sin(x)$$. The proof below is not entirely elementary. There's probably a more elementary proof, although nobody's come up with one yet; the proof below does tell us a few interesting things about the class of functions satisfying the given condition.

Say $$f\in X$$ if $$f$$ is smooth and $$\Vert f \Vert = \Vert f \Vert _X=\sup_{k\ge0}\sup_{x\in\Bbb R}|f^{(k)}(x)|<\infty.$$ (An example of one of the interesting things mentioned above: We will see below that if $$f\in X$$ then $$\Vert f \Vert _X= \Vert f \Vert _\infty$$.)

As has been noted, if $$f\in X$$ then Taylor's Theorem with a suitable form of the remainder shows that $$f$$ is equal to its Maclaurin series on all of $$\Bbb R$$; thus $$f$$ is (the restriction to $$\Bbb R$$ of) an entire function. Simply noting $$|\sum a_nz^n|\le\sum|a_n|\,|z|^n$$ shows that $$|f(z)|\le \Vert f \Vert e^{|z|}$$. In particular $$|f(iy)|\le \Vert f \Vert e^{|y|}$$; now since $$X$$ is translation-invariant it follows that $$|f(x+iy)|\le \Vert f \Vert e^{|y|}.$$

Functions in $$X$$ enjoy a certain magical property:

Magical Property If $$f\in X$$ then $$f'(x)=\sum_{k=-\infty}^\infty \frac{4(-1)^k}{(2k+1)^2\pi^2}f(x+\pi(k+1/2)).$$

(Hence $$\Vert f' \Vert _\infty\le \Vert f \Vert _\infty$$ as noted above.)

The proof of the Magical Property is the not quite so elementary part of the argument (maybe it follows by some elementary complex analysis, I don't see how). First we show how the result follows:

Suppose as in the OP that $$\Vert f \Vert _X\le 1$$, $$f$$ is real-valued on $$\Bbb R$$, and $$f'(0)=1$$. Then $$1=f'(0) =\sum_{k=-\infty}^\infty \frac{4(-1)^k}{(2k+1)^2\pi^2}f(\pi(k+1/2)).$$Now since $$\sum_{k=-\infty}^\infty \frac{4}{(2k+1)^2\pi^2}=1$$and $$-1\le f\le 1$$ we cannot have any cancellation here; we must have $$f(\pi(k+1/2))=(-1)^k=\sin(\pi(k+1/2)).$$

So if we let $$g(z)=\frac{f(z)-\sin(z)}{\cos(z)}$$then $$g$$ is an entire function. Since $$|f(x+iy)|\le \Vert f \Vert e^{|y|}$$ it is clear that $$g$$ is bounded in the complement of the union of small disks centered at $$\pi(k+1/2)$$. Hence $$g$$ is bounded and so $$g$$ is constant. So $$f(z)=\sin(z)+c\cos(z).$$Since $$f$$ is real-valued on $$\Bbb R$$ it follows that $$c\in\Bbb R$$, and now it's easy to show that $$|f|\le1$$ implies $$c=0$$.

Proof of the Magical Property

Here we assume the reader is familiar with the basic properties of tempered distributions and their Fourier transforms. If you don't know that stuff probably you shouldn't bother asking about the stuff in this section, it would take a lot of space to explain. Of course if you do know that stuff and something below seems problematic please say so.

Suppose $$f\in X$$.

Since $$f\in L^\infty(\Bbb R)$$, $$f$$ is a tempered distribution, and as such it has a Fourier transform, that being another tempered distribution. A version of the Paley-Wiener Theorem for tempered distributions, in for example Rudin Functional Analysis, shows that $$supp(\hat f)\subset[-1,1].$$(This special case is easier than the theorem in Rudin; I may add a few words about that if it looks like anyone's reading this.)

The Magical Property is really just an explicit version of Bernstein's inequality. Recall: If $$g\in L^p(\Bbb R)$$ and $$supp(\hat g)\subset[-1,1]$$ then $$\Vert g' \Vert _p\le \Vert g \Vert _p$$. The proof proceeds by showing that there exists a complex measure $$\mu$$ with $$\Vert \mu \Vert =1$$ and $$\hat\mu(\xi)=i\xi$$ for $$-1\le\xi\le 1$$; then considering the Fourier transform shows that $$g'=g*\mu$$, hence $$\Vert g' \Vert _p\le \Vert g \Vert _p \Vert \mu \Vert = \Vert g \Vert _p$$. Modulo technicalities the Magic Property for $$f\in X$$ follows by the same argument, except we need to say exactly what $$\mu$$ is.

It's straightforward, if tedious, to verify that $$\sum_{k=-\infty}^\infty\frac{4(-1)^k}{(2k+1)^2\pi^2}e^{i\pi(k+1/2)t}= \begin{cases}it,&(-1\le t\le 1), \\i(2-t),&(1\le t\le 3).\end{cases}$$(Simply extend the function on the right to a function of period $$4$$ and calculate the Fourier coefficients.) Hence if $$\mu=\sum_{k=-\infty}^\infty\frac{4(-1)^k}{(2k+1)^2\pi^2}\delta_{-\pi(k+1/2)}$$then $$\hat\mu(\xi)=i\xi\quad(-1\le\xi\le 1).$$Since $$supp(\hat f)\subset[-1,1]$$ this shows that $$f'=f*\mu$$, which is exactly what the Magic Property asserts.

A Technicality Asserting that $$f'=f*\mu$$ because $$\hat\mu(\xi)=i\xi$$ on the support of $$\hat f$$ is a little glib. The problem is that $$\hat f$$ is just a distribution, and $$\hat\mu$$ is not smooth on a neighborhood of $$[-1,1]$$. (This really is a problem; for example if $$f$$ is an arbitrary tempered distibution with $$\hat f$$ supported in $$[-1,1]$$ the convolution $$f*\mu$$ need not even exist.)

This is not hard to fix. For $$0<\lambda<1$$ let $$f_\lambda(x)=f(\lambda x).$$ It's enough to show that $$f_\lambda'=f_\lambda*\mu$$. This is straightforward, since $$\hat\mu(\xi)=i\xi$$ on a neighborhood of the support of $$\hat f_\lambda$$. Again, I may add details if it seems worthwhile.

Amuusing Note We've actually shown this:

Theorem Suppose $$f\in L^\infty(\Bbb R)$$. TFAE:

1. $$f\in X$$.

2. $$f$$ extends to an entire function with $$|f(x+iy)|\le ce^{|y|}$$.

3. $$\hat f$$ is supported in $$[-1,1]$$.

(We've shown explicitly that (1) implies (2) and (2) implies (3). To show (3) implies (1), note that (3) implies that $$\Vert f' \Vert _\infty\le \Vert f \Vert _\infty$$.)

• Could you please explain why "since $|f(x+iy)|\le||f||e^{|y|}$ it is clear that $g$ is bounded in the complement of the union of small disks centered at $\pi(k+1/2)$."? $\cos(z), \sin(z)$ and $f(z)$ all have the same asymptotic growth rate. How is it clear that the denominator $\cos(z)$ suppresses the numerator?
– Hans
Commented Jul 31, 2016 at 8:54
• Why is "$X$ translation-invariant" which I suppose you mean $f(x+iy)=f(iy), \forall x$? The last equation does not hold for $\sin(z)$. I have to conclude that is not what you mean. So what do you mean by that?
– Hans
Commented Jul 31, 2016 at 9:34
• @Hans Say $S$ is the set of all $z=x+iy$ such that $|y|\le 1$ and $|z-(k+1/2)\pi|\ge1/2$ for every $k$. There exists $\delta>0$ such that $|\cos(z)|\ge\delta$ for $z\in S$. So $g$ is bounded in $S$ since the numerator is bounded in $S$; if the numerator is less than $M$ then $|g|<M/\delta$. On the other hand, there exists $c$ so $|\cos(z)|\ge ce^{|y|}$ for $|y|>1$, so when $|y|>1$ we have $|g|\le c_1e^{|y|}/(c_2e^{|y|})$. Commented Jul 31, 2016 at 13:07
• @Hans When I say $X$ is translation-invariant I mean $X$ is translation-invariant. This is a standard term, meaning that if $f\in X$, $a\in \Bbb R$ and $g(x)=f(x+a)$ then $g\in X$. Commented Jul 31, 2016 at 13:09
• Thank you, David. Regarding the boundedness of $g$, I was obtuse and missed the exponential lower bound of $\cos(z)$. Regarding $X$, I understand your definition now. However, my real question is how you deduce "since $X$ is translation-invariant it follows that $|f(x+iy)|\le||f||e^{|y|}$". Could you please explain that? You have never defined $X$ explicitly. I assume you implicitly define it as the space of complex valued functions satisfying OP's conditions. Is that correct?
– Hans
Commented Jul 31, 2016 at 21:32

Here's my attempt, some of the details are missing. It is likely that an easier solution can be found.

Clarification: As David pointed out, there is no such Paley-Wiener type theorem. I will sketch what I think can be done to fix it in the end.

As David mentioned in the comments, your assumptions imply $f$ is an entire function (of exponential type $1$), and also give the following bound: $$| f(z) | \le e^{|\Im z |}.$$

Suppose that a Paley-Wiener type theorem implies we can write $f$ as a Fourier transform:

$$$f(z) = \int_{-1}^{1} e^{i t z} \, \rm{d} \nu(z), \tag{a}\label{a}$$$

where $\nu$ is a complex measure on $[-1,1]$. In particular,

$$$f^{(n)}(z) = \int_{-1}^{1} (it)^n e^{i t z} \, \rm{d} \nu(z). \tag{b}\label{b}$$$

Consider odd $n>0$. Notice that the function $(it)^n$ is not continuous on $[-1,1]$, if we identify $-1$ and $1$. To fix it, note that $(it)^n e^{-\frac12 i\pi t}$ is continuous, and we have an absolutely convergent Fourier series, $$(it)^n e^{-\frac12 i\pi t} = \sum_{k \in \mathbb{Z}} C_n(k) e^{i \pi k t}, \quad t\in [-1,1]. \tag{c}\label{c}$$ For example, $$it e^{-\frac12 i\pi t} = \frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{(-1)^k}{(2k + 1)^2} e^{i \pi k t}, \quad t\in [-1,1]. \tag{d}\label{d}$$

Combining $\eqref{a}$, $\eqref{b}$, and $\eqref{d}$, we find \begin{align*} f^\prime (z) & = & \int_{-1}^1 i t e^{izt} \, \rm{d} \nu(t) = \int_{-1}^1 i t e^{-\frac12 i \pi t} e^{i(z + \frac12 \pi)t} \, \rm{d} \nu(t)\\ &=& \frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{(-1)^k}{(2k+1)^2} \int_{-1}^1 e^{i t (z + \pi k + \frac12 \pi)} \, \rm{d} \nu(t) \\ &=& \frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{(-1)^k}{(2k+1)^2} \, f\left(z + \pi (k+\frac12)\right). \end{align*} In particular, $$1 = f^\prime(0) = \frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{(-1)^k}{(2k+1)^2} \, f\left(\pi (k+\frac12)\right).$$ Since $\frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{1}{(2k+1)^2} = 1$, and $|f|\le 1$, we find that $f\left(\pi (k+\frac12)\right) = (-1)^k$. Notice also that $f^{\prime\prime}(0) = 0$, since otherwise this would contradict $|f^\prime(z)|\le 1$, for $z$ close to $0$.

Now (for odd $n>0$), using $\eqref{b}$, and $\eqref{c}$, we get in a similar way to the above $$f^{(n)}(0) = \sum_{k\in\mathbb{Z}} C_n(k) \, f\left(\pi (k+\frac12)\right) = \sum_{k\in\mathbb{Z}} C_n(k) \, (-1)^k = i^n \cdot e^{-\frac12 \pi i} = (-1)^{\frac12 (n-1)}.$$ So for example, $f^{(3)}(0) = -1$, and this implies $f^{(4)}(0)=0$, and so on. Since $f$ is analytic and has the same derivatives as $\sin(z)$ at $z=0$, we find that $\sin(z) + c$ is the only function satisfying the requirements, for some real constant $c$. But $c=0$ is the only possibility, since $|f| \le 1$.

One can prove that $f$ can be approximated uniformly on compact sets by a sequence $$$$g_j(z) = \int_{-1}^{1} e^{i t z} \, \rm{d} \nu_j(z),$$$$ where $\nu_j$ are complex measures on $[-1,1]$. See Boas, Entire Functions, Theorem 6.8.14. We have to prove by induction that the first $n$ derivatives of $f$ and $\sin(z)$ at $z=0$ are the same. Then we need to repeat the above arguments, using $\varepsilon, \delta$ methods whenever we had equalities before. Since we have uniform convergence of entire functions, we can approximate (a fixed number of) the derivatives of $f$ on any compact set using the functions $g_j$.

• I see that David just posted essentially the same solution. Commented Jul 19, 2016 at 15:26
• How do you know that $\hat f$ is a measure? That simplifies some technicalities. (Very funny, takes me a day to work out and a few hours to type up, then the two posts appear within 15 minutes of each other...) Commented Jul 19, 2016 at 15:33
• Pretty sure that no, your $\nu$ need not be a measure. Say $M$ is the space of complex measures on $[-1,1]$. The FT certainly maps $M$ into what I called $X$. If this map were surjective then, recalling that the norm in $X$ is equal to the sup norm, we'd have $||\hat\mu||_\infty\ge c||\mu||$ for all $\mu$ in $M$. Something like a Dirichlet kernel shows this is not so. (Makes a difference because if $\nu$ is not measure then things ike (a) and especially (b) are not so clear...) Commented Jul 19, 2016 at 16:43
• This was the reason I used the word attempt. We can write $f$ as the limit of Fourier integrals with respect to a sequence of measures $\mu_n$. The question is whether using the additional assumptions on $f$, we can choose a convergent subsequence. Commented Jul 19, 2016 at 16:51
• I guess. You might clarify things by indicating which parts are false, or are things you don't know how to prove - the words "A Paley-Wiener type theorem then implies..." seem to indicate that a Paley-Wiener type theorem actually implies what follows. Commented Jul 19, 2016 at 17:05