Bounds of the derivative of a bounded band-limited function
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 $\sigma&gt;0$
$F$ 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 $\sum \prime$.)