Characterization of Euler's Number by Inequality

Theorem
Let $a$ be a (strictly) positive real number.

Then:


 * $a = e \iff \forall x \in \R: a^x \ge x + 1$

where $e$ denotes Euler's number.

Forward Implication
Proved in Exponential Function Inequality.

Reverse Implication
Consider $\map {f_a} x = a^x - x - 1$.

Then we need to prove:


 * $a = e \impliedby \forall x \in \R: \map {f_a} x \ge 0$

By Linear Combination of Derivatives, Derivative of Power of Constant, Derivative of Identity Function, and Derivative of Constant:


 * $\map {f'_a} x = a^x \ln a - 1$

By Linear Combination of Derivatives, Derivative of Power of Constant, Derivative of Constant Multiple, and Derivative of Constant:
 * $\map {f''_a} x = a^x \paren {\ln a}^2$

Now, we divide the theorem into two cases:

Case 1: $0 < a \le 1$
Consider $x = 1$.

Then:

Thus the is not true.

Case 2: $a > 1$
We have:

Hence, by Twice Differentiable Real Function with Positive Second Derivative is Strictly Convex, we have that $f_a$ is strictly convex.

Trying to find its minimum, we solve:
 * $\map {f'_a} x = 0$

Therefore:

At this point:

It is to be noted that when $a = e$, the minimum is:
 * $\dfrac {1 + \map \ln {\ln e} } {\ln e} - 1 = 0$

meaning that the only solution to $\map {f_e} x = 0$ is $x = 0$.

Also, from Exponential Function Inequality, we have:
 * $\forall x \in \R: e^x \ge x + 1$

Substituting $x = \map \ln {\ln a}$:
 * $\ln a \ge \map \ln {\ln a} + 1$

From Logarithm is Strictly Increasing and Logarithm of 1 is 0:
 * $\ln a > 0$

Thus:
 * $\dfrac {\ln a} {\ln a} \ge \dfrac {\map \ln {\ln a} + 1} {\ln a}$

Therefore:
 * $\dfrac {1 + \map \ln {\ln a} } {\ln a} - 1 \le 0$

Then, we solve the case of equality:

From earlier, we have that the only solution is $\map \ln {\ln a} = 0$, when $a = e$.

Therefore, for other values of $a$, the minimum is negative.