Bounds of the difference of a bounded band-limited function
As shown in the answers to this question, for continuous-time signals, the bound predicted by Bernstein's inequality is achieved with equality by a sinusoidal signal with a frequency equal to the upper frequency limit. 

In analogy to this, in this answer I'll show a bound on $\big|x[n]-x[n-1]\big|$ for a discrete-time sinusoidal signal $x[n]$ at angular frequency $W$:

$$x[n]=A\sin(nW+\phi)\tag{1}$$

For the signal $x[n]$ given by $(1)$, the largest value of $\big|x[n]-x[n-1]\big|$ for all $n\in\mathbb{Z}$ occurs for two values $x[k]$ and $x[k-1]$ symmetrical to the point of the largest derivative of a (continuous) sinusoid, i.e., for $x[k]=A\sin(W/2)$ and $x[k-1]=A\sin(-W/2)$. Consequently, the bound for a sinusoid with frequency $W$ is given by

$$\big|x[n]-x[n-1]\big|\le 2A\sin\left(\frac{W}{2}\right)\tag{2}$$

This bound is achieved with equality for the signals

$$x[n]=A\sin\left(nW+\frac{(2l+1)W}{2}\right),\qquad l\in\mathbb{Z}\tag{3}$$

Note that the constant $A$ does not generally equal the maximum amplitude $B=\max |x[n]|$ of $x[n]$. Depending on the sampling phase, the maximum amplitude $B$ lies in the following interval:

$$A\cos\left(\frac{W}{2}\right)\le B\le A\tag{4}$$

Consequently, we have

$$A\le\frac{B}{\cos\left(\frac{W}{2}\right)},\qquad 0&lt;W&lt;\pi\tag{5}$$

Combining $(2)$ and $(5)$ we get

$$\big|x[n]-x[n-1]\big|\le 2B\tan\left(\frac{W}{2}\right),\qquad B=\max_n\big|x[n]\big|\tag{6}$$

Note, however, that $(6)$ is only useful for $\tan(W/2)&lt;1$, i.e., for $W&lt;\pi/2$ because for any $x[n]$ with $\max|x[n]|=B$ we must have

$$\big|x[n]-x[n-1]\big|\le 2B\tag{7}$$

I believe that the bound $(6)$ holds for all discrete-time band-limited signals with a maximum frequency $W$ (i.e., with $X(e^{j\omega})=0$ for $|\omega|\in (W,\pi]$), but I don't know how to show it.