Sometime ago, I asked the following question Prove or disprove that there exists a constant $K$ such that $|f(x)-f(y)|\leq K |x-y|,\;\;\forall\;\;x,y\in[0,1]$ and was answered. However, I did not really understand this line
"However, for each $x$ there is a $K_x$. Indeed, $K_0=1$ works, and for $x\in(0,1]$, let $x_0\leq x$ be such that $f'$ has a local maximum or minimum at $x_0$. Then $K_x=|f'(x_0)|$ works."
Please, can anyone please give me a detailed explanation?
Suppose $f$ is bounded and differentiable on $[a,b].$ Then for every $x\in [a,b]$ there exists such a constant $K_x.$
Proof: Let $M=\sup_{[a,b]}|f|.$ Fix $x\in [a,b].$ Then there is a neighborhood $I_x$ of $x$ such that
$$\tag 1 \left |\frac{f(y)-f(x)}{y-x}\right | \le |f'(x)| + 1\,\, \text { for } y\in I_x\setminus\{x\}.$$
This follows from the definition of $f'(x).$ Now there there is $\delta > 0$ such that $|y-x|\ge \delta$ for $y\in [a,b]\setminus I_x.$ For such $y$ we have
$$\tag 2 \left |\frac{f(y)-f(x)}{y-x}\right | \le \frac{2M}{\delta} <\infty.$$
From $(1),(2)$ the existence of $K_x$ follows.