Let $f(t)$ be a function with properties:

$$\begin{array}{ll} t\in\mathbf{R}&t\text{ is in reals}\\ f(t)\in\mathbf{R}\text{ for all } t&f(t)\text{ is in reals}\\ |f(t)|<A\text{ for all }t&\text{absolute value of }f(t)\text{ is bounded above by }A\\ \int_{-\infty}^{\infty} f(t) \ e^{- i \omega t} \ {\rm d}t = 0\text{ for all }|\omega|\ge B&f(t)\text{ is band-limited by frequency B in radians} \end{array}$$

Given $A$ and $B,$ what is the tight upper bound for $|f'(t)|,$ the absolute value of the derivative of the function?

Nothing else shall be assumed about $f(t)$ than what has been stated above. The bound should accommodate for this uncertainty.

For a sinusoid of amplitude $A$ and frequency $B,$ the maximum absolute value of the derivative is $AB.$ I wonder if this is an upper bound, and in that case also the tight upper bound. Or maybe a non-sinusoidal function has a steeper slope.

  • $\begingroup$ Have you checked this? $\endgroup$
    – Tendero
    Commented Aug 30, 2018 at 13:24
  • $\begingroup$ @Tendero thanks. There, the signal energy is known, rather than the peak absolute value as in my question. $\endgroup$ Commented Aug 30, 2018 at 13:39
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    $\begingroup$ See my answer for the bound you seek. It says More generally, a result due to Bernstein says that if the maximum frequency in a generic $x(t)$ bounded within $[-1,1]$ is $f_0$, that is, $X(f) = 0$ for $|f| > f_0$, then $$\max \left| \frac{\mathrm dx}{\mathrm dt}\right| \leq 2\pi f_0.$$ $\endgroup$ Commented Aug 30, 2018 at 14:50
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    $\begingroup$ Based on the sharp version of Bernstein's inequality, from Dilip's linked answers, MBaz's edited answer and the literature cited, $AB$ is indeed the sharp (I called it tight meaning the same) upper bound for the maximum absolute value of the derivative, a full-scale sinusoid at exactly the band limit (not strictly allowed by the constraints I give) making the inequality an equality. $\endgroup$ Commented Aug 31, 2018 at 7:02

2 Answers 2


You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92).

With a well-behaved signal $f(t)$ as you defined it above (in particular, $f(t)$ is integrable and bandlimited to $B\,\text{Hz}$, and $\text{sup}\,|f(t)| = A$), then $$\left|\frac{\text{d}f(t)}{\text{d}t}\right| \leq 2AB\pi. $$

Note that the original result by Bernstein established a bound of $4AB\pi$; later, that bound was tightened to $2AB\pi$.

I have spent some time reading Zygmund's "Trigonometric Series"; all I'll say is that it is the perfect remedy for those under the impression that they know trigonometry. A full understanding of the proof is beyond my mathematical skill, but I think I can highlight the main points.

First, what Zygmund calls Bernstein's inequality is a more limited result. Given the trigonometric polynomial $$T(x) = \sum_{-\infty}^\infty c_k e^{jkx}$$ (with real $x$), then $$\max_x |T'(x)| \leq n \max_x |T(x)|$$ with strict inequality unless $T$ is a monomial $A \cos(nx+\alpha)$.

To generalize this we need a preliminay result. Consider a function $F$ that is in $\text{E}^\pi$ and in $\text{L}^2$. ($\text{E}^\sigma$ is the class of integral functions of type at most $\sigma$ -- this is one of the places where my math starts to fray at the edges. My understanding is that this is a mathematically rigorous way of stating that $f=\text{IFT}\lbrace F \rbrace$ has bandwidth $\sigma$.)

For any such $F$ we have the interpolation formula $$F(z) = \frac{\sin(\pi z)}{\pi}F_1(z),$$ where $z$ is complex and $$F_1(z) = F'(0) + \frac{F(0)}{\pi} + \sum_{n=-\infty}^\infty {^\prime} (-1)^nF(n) \left( \frac{1}{z-n}+\frac{1}{n} \right).$$ (This is theorem 7.19.)

Now we can state the main theorem. If:

  • $F$ is in $\text{E}^\sigma$ with $\sigma>0$
  • $F$ is bounded on the real axis
  • $M=\sup |F(x)|$ for real $x$

then $$|F'(x)| \leq \sigma M$$ with equality possible iff $F(z) = a e^{j\sigma z} + b e{-j\sigma x}$ for arbitrary $a,b$. We suppose that $\sigma=\pi$ (otherwise we take $F(z\pi/\sigma)$ instead of $F(z)$.)

To prove this, we write the derivative of $F$ using the interpolation formula above: $$F'(x) = F_1(x)\cos(\pi x)+\frac{\sin(\pi x)}{\pi} \sum_{n=-\infty}^\infty \frac{(-1)^nF(n)}{(x-n)^2}.$$ Setting $x=1/2$ we get $$F'(1/2) = \frac{4}{\pi} \sum_{n=-\infty}^\infty \frac{(-1)^nF(n)}{(2n-1)^2}$$ which implies $$|F'(1/2)| \leq \frac{4}{\pi} \sum_{n=-\infty}^\infty \frac{1}{(2n-1)^2} = \frac{4M\pi^2}{4\pi} = M\pi.$$

Now we need a nice little trick: Take an arbitrary $x_0$ and define $G(z) = F(x_0+z-1/2)$. Then, $$|F'(x_0)| = |G'(1/2)| \leq M\pi.$$

(TODO: Show the proof for the case of equality. Define $\sum \prime$.)

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    $\begingroup$ @OlliNiemitalo As pointed out in MattL's answer, the sinusoid $\sin(2\pi Bt)$ has maximum derivative $2\pi B$. This meets Bernstein's bound, as stated in my answer here on dsp.SE (cited in a comment on your question) and in my answer on math.SE that you found, with equality. $\endgroup$ Commented Aug 30, 2018 at 15:11
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    $\begingroup$ @OlliNiemitalo I found the proof given by Pinksy here (I hope that link works!). He definitely uses $4AB\pi$ as the bound, not $2AB\pi$. $\endgroup$
    – MBaz
    Commented Aug 30, 2018 at 15:46
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    $\begingroup$ @MBaz Your link works indeed! At the end of the section 2.3.8 they say that the best known version of Bernstein's inequality has the factor 2 instead of 4, which is sharp, and that for details consult Zygmund (1959) Vol. 2, p. 276. I think that's Zygmund, A. Trigonometric series. 2nd ed. Vol. II. Cambridge University Press, New York 1959. $\endgroup$ Commented Aug 30, 2018 at 19:14
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    $\begingroup$ RP Boas, Some theorems on Fourier transforms and conjugate trigonometric integrals, Transactions of the American Mathematical Society 40 (2), 287-308, 1936 cites the relevant articles by Bernstein, Szegö, and Zygmund, already with the sharp bound, as far as I can tell. $\endgroup$ Commented Aug 30, 2018 at 20:26
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    $\begingroup$ @OlliNiemitalo Excellent! I had missed that note at the end of section 2.3.8. I'll update my answer. Also: that book by Zygmund is in my university's library, but it's not online. I'll take it out tomorrow and see what it says. $\endgroup$
    – MBaz
    Commented Aug 30, 2018 at 22:15

In general you would get something like this, but it might not be tight:

$$\begin{align}|f'(t)|&=\left|\frac{1}{2\pi}\int_{-\infty}^{\infty}j\omega F(j\omega)e^{-j\omega t}d\omega\right|\\&\le \frac{1}{2\pi}\int_{-\infty}^{\infty}|\omega||F(j\omega)|d\omega\\&=\frac{1}{2\pi}\int_{-\omega_c}^{\omega_c}|\omega||F(j\omega)|d\omega\\&\le \frac{|\omega_c|}{2\pi}\int_{-\omega_c}^{\omega_c}|F(j\omega)|d\omega\\&=\frac{\omega_c}{\pi}\int_{0}^{\omega_c}|F(j\omega)|d\omega\tag{1}\end{align}$$

The upper bound on $|f(t)|$ is of course implicit in $|F(j\omega)|$.

For a sinusoid $A\sin(\omega_ct)$, $(1)$ gives $A\omega_c$ as an upper bound, as expected.

  • $\begingroup$ @Olli Niemitalo , I had derived the sinusoid case I think this is the general case we were looking at. Thanks Matt L. $\endgroup$
    – MimSaad
    Commented Aug 30, 2018 at 18:22
  • $\begingroup$ Hi Matt, I really like your answer but I have a question, How you make sure that $\int\limits_0^{\omega_c} |F(j\omega)|d\omega < \infty$ is always absolute integrable? $\endgroup$
    – Joako
    Commented Apr 20, 2022 at 3:18
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    $\begingroup$ @Joako: In general you don't know if that integral converges. If it doesn't, the bound is of course useless. I would think that in most practical cases the integral does converge. $\endgroup$
    – Matt L.
    Commented Apr 21, 2022 at 8:10

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