Let $\Phi : \{0,1\}^{N} \to \mathbb{R}$ be a function of a Markov chain on state space $\{0,1\}^N$ where $$\Phi(x) = \sum\limits_{i=1}^N \cos\left(\frac{\pi(2i-1)}{2N}\right)x(i).$$ Starting from any given vector $x \in \{0,1\}^{N}$ the Markov chain can change in a single step by, at most, transposing two adjacent values in the vector. (For those familiar this is the exclusion process on a path of length $N$).
In the book I am reading (Markov Chains and Mixing Times Page 328) the authors say:
We have $\left|\Phi\left(X_1\right)-\Phi(x)\right| \leq \pi / N$, since the derivative of $u \mapsto \cos \left(\frac{\pi u}{ 2 N}\right)$ is bounded by $\frac{\pi}{2 N}$.
Where $x$ is the initial state of the Markov chain and $X_1$ is the state after taking a single step.
Originally, my thought was that this was achieved by applying the fundamental theorem of calculus:
$$f(a)-f(b)=\int_b^a f^{\prime}(t) \,\Bbb d t$$
But I’m starting to thing I’m wrong and I’m not sure how it would work even if I was right. The bounds of integration would then be vectors, and the authors don’t use the derivative of $\Phi$ it self but only the cosine part to get an upper bound. I’m quite confused about what is going on here, I would be grateful if someone could please explain how they were able to get such a bound using the derivative.
Let $j$ and $k$ the two positions on which $X_1$ and $x$ differ which satisfies $|j-k| = 1$ and $X_1(j)-x(j) = x(k)-X_1(k)$ by the interchange condition. Then $$ \big|\Phi(X_1)-\Phi(x)\big| = \left|\cos\left(\frac{\pi(2j-1)}{2N}\right)\Big(X_1(j)-x(j)\Big) + \cos\left(\frac{\pi(2k-1)}{2N}\right)\Big(X_1(k)-x(k)\Big)\right| \le \left|\cos\left(\frac{\pi(2j-1)}{2N}\right) - \cos\left(\frac{\pi(2k-1)}{2N}\right)\right| \le \frac{\pi}{2N}|2(j-k)| $$ where in the last step we used that the $u \mapsto \cos\left(\frac{\pi u}{2N}\right)$ function is $\big(\pi/(2N)\big)$-Lipschitz because its derivative is bounded by $\pi/(2N)$. From this the claim follows by using $|j-k|=1$.