Upper Bound of Natural Logarithm

Theorem
Let $$\ln y$$ be the natural logarithm of $$y$$ where $$y \in \R, y > 0$$.

Then:
 * $$\ln y \le y - 1$$.
 * $$\forall s \in \R: s > 0: \ln x \le \frac {x^s} s$$

Proof

 * First, to show that $$\ln y \le y - 1$$:

We have that the natural logarithm function is concave.

From Mean Value of Convex and Concave Functions, we have $$\ln y - \ln 1 \le \left({D \ln 1}\right) \left({y - 1}\right)$$.

From Derivative of Natural Logarithm we have $$D \ln 1 = \frac 1 1 = 1$$.

So $$\ln y - \ln 1 \le \left({y - 1}\right)$$.

But from the Basic Properties of Natural Logarithm, $$\ln 1 = 0$$.

Hence the result.


 * Next, to show that$$\ln x \le \frac {x^s} s$$:

$$ $$ $$

The result follows by dividing both sides by $$s$$.