Derivative of Natural Logarithm Function/Proof 4/Lemma
Theorem
Let $\sequence {f_n}_n$ be the sequence of real functions $f_n: \R_{>0} \to \R$ defined as:
- $\map {f_n} x = n \paren {\sqrt [n] x - 1}$
Let $k \in \N$.
Let $J = \closedint {\dfrac 1 k} k$.
Then the sequence of derivatives $\sequence { {f_n}'}_n$ converges uniformly to some real function $g: J \to \R$.
Proof
From Derivative of $n$th Root and Combination Theorem for Sequences:
- $\forall n \in \N: \forall x \in J : D_x \map {f_n} x = \dfrac {\sqrt [n] x} x$
From Closed Bounded Subset of Real Numbers is Compact, $J$ is compact.
Thus:
| \(\ds \lim_{n \mathop \to \infty} D_x \map {f_n} x\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \dfrac {\sqrt [n] x} x\) | from above | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 x \lim_{n \mathop \to \infty} \sqrt [n] x\) | Multiple Rule for Real Sequences | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 x \lim_{n \mathop \to \infty} x^{1/n}\) | Definition of $n$th Root | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 x\) | Power Function tends to One as Power tends to Zero: Rational Number |
In particular, $\sequence { {f_n}'}_n$ is pointwise convergent to a continuous function on $J$.
For $x \in \closedint {\frac 1 k} 1$:
| \(\ds x^{1/n}\) | \(\le\) | \(\ds x^{1 / \paren {n + 1} }\) | Exponential is Monotonic: Rational Number | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqrt [n] x\) | \(\le\) | \(\ds \sqrt [n + 1] x\) | Definition of $n$th Root | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\sqrt [n] x} x\) | \(\le\) | \(\ds \frac {\sqrt [n + 1] x} x\) | Real Number Ordering is Compatible with Multiplication |
So $\sequence {\map { {f_n}'} x}_n$ is increasing when $x \in \closedint {\dfrac 1 k} 1$.
For $x \in \closedint 1 k$:
| \(\ds x^{1/n}\) | \(\ge\) | \(\ds x^{1 / \paren {n + 1} }\) | Exponential is Monotonic: Rational Number | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqrt [n] x\) | \(\ge\) | \(\ds \sqrt [n + 1] x\) | Definition of $n$th Root | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\sqrt [n] x} x\) | \(\ge\) | \(\ds \frac {\sqrt [n + 1] x} x\) | Real Number Ordering is Compatible with Multiplication |
So $\sequence {\map { {f_n}'} x}_n$ is decreasing when $x \in \closedint {\dfrac 1 k} 1$.
Thus $\sequence {\map { {f_n}'} x}_n$ is monotone for all $x \in \closedint {\frac 1 k} 1 \cup \closedint 1 k = J$.
From Dini's Theorem, $\sequence { {f_n}'}_n$ converges uniformly to $\dfrac 1 x$ on $J$.
Hence the result.
$\blacksquare$