Finding the largest constant $C$ such that $|\ln x−\ln y| \geq C|x−y|$ for all $x, y \in (0, 1]$ Find the greatest value of C such that
$|\ln x−\ln y|≥C|x−y|$
for any $x,y∈(0,1]$. What should my approach be? I can't think of much options.
To find the largest $C$ so that $|\log(x)-\log(y)|\ge C|x-y|$, note that, by the Mean Value Theorem, for some $t$ strictly between $x$ and $y$
$$
\frac{\log(x)-\log(y)}{x-y}=\frac1t
$$
The minimum of $\frac1t$ on $(0,1]$ is $1$. Therefore, the largest $C$ can be is $1$.
$f(b)- f(a) = f'(\xi) (b - a),\, \xi \in (a, b)$. When $x, y$ near $0$, the constant $C$ will become very large.