I'm studying a Stein-Shakarchi Fourier Analysis textbook and I'm stuck on the proof on page 116.
Help me understand why differentiability at one point allows us to establish such an inequality. It seems to use the mean value theorem, but it requires that the function be differentiable over the interval. And it’s still not clear how exactly to apply it here
One way to see this is as follows, if $g$ is assumed to be continuous on the compact interval $[x_0 - \pi, x_0 + \pi]$:
Since $g$ is differentiable at $x_0$, we have: $$\frac{g(x_0 - t) - g(x_0)}{-t} \xrightarrow[t \to 0]{} g'(x_0) \in \mathbb{R}$$ In particular, there exists $\delta > 0$ such that: $$\forall t \in (-\delta, \delta),\quad \left|\frac{g(x_0 - t) - g(x_0)}{-t} - g'(x_0)\right| \leq 1$$ Hence, since $|x - y| \geq \big||x| - |y|\big|$ for any reals $x$ and $y$: $$\forall t \in (-\delta, \delta),\quad \left|\left|\frac{g(x_0 - t) - g(x_0)}{-t}\right| - \left|g'(x_0)\right|\right| \leq 1$$ Which then provides: $$\forall t \in (-\delta, \delta),\quad \left|\frac{g(x_0 - t) - g(x_0)}{-t}\right| \leq \left|g'(x_0)\right| + 1$$ On the other hand, the function $t \mapsto \frac{g(x_0 - t) - g(x_0)}{-t}$ is well-defined and continuous on $[-\pi, \pi] \setminus (-\delta, \delta)$ which is compact, thus is bounded by some $M > 0$, which means that: $$\forall t \in [-\pi, \pi], \quad \left|\frac{g(x_0 - t) - g(x_0)}{-t}\right| = \frac{|g(x_0 - t) - g(x_0)|}{|t|} \leq C := \max(|g'(x_0)| + 1, M)$$ From there we can derive the desired result since: $$\begin{split}|\sigma_N(g)'(x_0)| &= \left|\int_{-\pi}^\pi F_N'(t)(g(x_0 - t) - g(x_0))\mathrm{d}t\right|\\ &\leq \int_{-\pi}^\pi \big|F_N'(t)(g(x_0 - t) - g(x_0))\big|\mathrm{d}t\\ &\leq \int_{-\pi}^\pi \left|F_N'(t)\right| \cdot C|t| \mathrm{d}t\end{split}$$