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.
 A: 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 $1$.
A: Hint: Since $\ln$ is monotonic, he inequality is equivalent to
$$\frac{\ln x - \ln y}{x - y} \geq C;$$
the l.h.s. is the slope of the secant line to the graph through $(x, \ln x)$ and $(y, \ln y)$, so the best $C$ is precisely the infimum those slopes for $(x, y) \in (0, 1]$.
On the other hand, $(\ln x)'' = -\frac{1}{x^2} < 0$, so given any secant line, one can produce a secant line with smaller slope by moving either endpoint to the right.

The last observation means we may as well one of the endpoints, say, $y$, to be $1$, and so we are looking for a constant $C$ such that $$\frac{\ln x - \ln 1}{x - 1} \geq C$$ for all $y \in (0, 1]$. By the same observation, the quantity on the l.h.s. decreases as $x$ increases, so we can take $$C = \lim_{x \to 1} \frac{\ln x - \ln 1}{x - 1} = \left.\frac{d}{dx} \ln x \right\vert_{x = 1} = \left.\frac{1}{x}\right\vert_{x = 1} = 1.$$

A: This may not be the most rigorous approach, but I would start by rearranging the equation as $$|\frac{\ln(x)-\ln(y)}{x-y}|\ge C.$$
Since we want this to work for any $x,y$ in our interval we want to find the minimum value of $|\frac{\ln(x)-\ln(y)}{x-y}|$ on the interval $(0,1]$ and let that equal $C$.  This is going to happen as both $x$ and $y$ draw towards 1.  Then let's set $y=1$ and take the limit as $x$ approaches 1:
$$\lim\limits_{x\rightarrow1}\frac{\ln(x)-0}{x-1}.$$
applying l'Hopital's Rule, we get $$\lim\limits_{x\rightarrow1}\frac{1/x}{1}=1$$
A: $f(b)- f(a) = f'(\xi) (b - a),\, \xi \in (a, b)$. When $x, y$ near $0$, the constant $C$ will become very large.
