Upper Bound of Natural Logarithm

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Theorem

Let $\ln y$ be the natural logarithm of $y$ where $y \in \R_{>0}$.


Then:

$(1): \quad \ln y \le y - 1$
$(2): \quad \forall s \in \R_{>0}: \ln x \le \dfrac {x^s} s$


Proof

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

From Logarithm is Strictly Increasing and Strictly Concave, $\ln$ is (strictly) concave.

From Mean Value of Concave Real Function:

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

From Derivative of Natural Logarithm:

$D \ln 1 = \dfrac 1 1 = 1$

So:

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

But from Logarithm of 1 is 0:

$\ln 1 = 0$

Hence the result.

$\Box$


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

\(\displaystyle s \ln x\) \(=\) \(\displaystyle \ln {x^s}\)          Logarithm of Power          
\(\displaystyle \) \(\le\) \(\displaystyle x^s - 1\)          from above          
\(\displaystyle \) \(\le\) \(\displaystyle x^s\)                    

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

$\blacksquare$


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