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)|
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.
Answer
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 BHz, and sup|f(t)|=A), then |df(t)dt|≤2ABπ.
Note that the original result by Bernstein established a bound of 4ABπ; later, that bound was tightened to 2ABπ.
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)=∞∑−∞ckejkx (with real x), then max 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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