# Bounds of the derivative of a bounded band-limited function

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.

• Have you checked this? – Tendero Aug 30 '18 at 13:24
• @Tendero thanks. There, the signal energy is known, rather than the peak absolute value as in my question. – Olli Niemitalo Aug 30 '18 at 13:39
• 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 $$\max \left| \frac{\mathrm dx}{\mathrm dt}\right| \leq 2\pi f_0.$$ – Dilip Sarwate Aug 30 '18 at 14:50
• 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. – Olli Niemitalo Aug 31 '18 at 7:02

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 $$\sum \prime$$.)

• @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. – Dilip Sarwate Aug 30 '18 at 15:11
• @OlliNiemitalo I found the proof given by Pinksy here (I hope that link works!). He definitely uses $4AB\pi$ as the bound, not $2AB\pi$. – MBaz Aug 30 '18 at 15:46
• @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. – Olli Niemitalo Aug 30 '18 at 19:14
• 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. – Olli Niemitalo Aug 30 '18 at 20:26
• @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. – MBaz Aug 30 '18 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.

• @Olli Niemitalo , I had derived the sinusoid case I think this is the general case we were looking at. Thanks Matt L. – MimSaad Aug 30 '18 at 18:22