X to the x is not of Exponential Order/Lemma

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Lemma

Let $f: \R_{>0} \to \R$ be defined as:

$\forall x \in \R_{>0}: \map f x = x^x$

Let there exist strictly positive real constants $M, K, a \in \R_{> 0}$ such that:

$\forall t \ge M: \size {\map f t} < K e^{a t}$


Then there exists a constant $C$ such that:

$\forall t > C: \size {\map f t} > K e^{a t}$


Proof

By the definition of power:

$\map f t = \map \exp {t \ln t}$

By Exponential of Real Number is Strictly Positive, we can reduce the lemma into the existence of $C$ such that:

$\forall t > C: \map f t > K e^{a t}$


We will divide into two cases.

Case 1: $K > 1$

Assume that $t > K e^a$.

\(\ds a\) \(>\) \(\ds 0\)
\(\ds e^a\) \(>\) \(\ds e^0\) Exponential is Strictly Increasing
\(\ds e^a\) \(>\) \(\ds 1\) Exponential of Zero
\(\ds K e^a\) \(>\) \(\ds 1\) As $K > 1$
\(\ds t\) \(>\) \(\ds 1\) As $t > K e^a$
\(\text {(1)}: \quad\) \(\ds t \ln K\) \(>\) \(\ds \ln K\)
\(\ds t\) \(>\) \(\ds K e^a\) Assumption
\(\ds \ln t\) \(>\) \(\ds \map \ln {K e^a}\) Logarithm is Strictly Increasing
\(\ds \ln t\) \(>\) \(\ds a + \ln K\)
\(\ds t \ln t\) \(>\) \(\ds a t + t \ln K\)
\(\ds t \ln t\) \(>\) \(\ds a t + \ln K\) from $(1)$
\(\ds \map \exp {t \ln t}\) \(>\) \(\ds \map \exp {a t + \ln K}\) Exponential is Strictly Increasing
\(\ds \map f t\) \(>\) \(\ds K e^{a t}\) Exponential is Strictly Increasing

Here, $C = K e^a$.


Case 2: $K \le 1$

Assume that $t > e^a$.

\(\ds \ln K\) \(\le\) \(\ds \ln 1\) Logarithm is Strictly Increasing
\(\text {(1)}: \quad\) \(\ds \ln K\) \(\le\) \(\ds 0\) Logarithm of 1 is 0
\(\ds t\) \(>\) \(\ds e^a\) by assumption
\(\ds \ln t\) \(>\) \(\ds a\) Logarithm is Strictly Increasing
\(\ds t \ln t\) \(>\) \(\ds a t\)
\(\ds t \ln t\) \(>\) \(\ds a t + \ln k\) from $(1)$
\(\ds \map \exp {t \ln t}\) \(>\) \(\ds \map \exp {a t + \ln K}\) Exponential is Strictly Increasing
\(\ds \map f t\) \(>\) \(\ds K e^{a t}\) Exponential is Strictly Increasing

Here:

$C = e^a$

$\blacksquare$