I am examining the uniform convergence of the sequence of functions $\{f_n\}$ defined by $$ f_n(x) = n(x^{1/n} - 1) $$ on intervals of the form $(1/k, k)$ for $k \in \mathbb{N}$. I am particularly interested in understanding whether this sequence converges uniformly to $f(x) = \ln(x)$ on these intervals.
Problem Statement:
The sequence $\{f_n\}$ is given by: $$ f_n(x) = n(x^{1/n} - 1) $$ We aim to show that $\{f_n\}$ converges uniformly to $f(x) = \ln(x)$ on intervals of the form $(1/k, k)$.
Analysis of Pointwise Convergence
It is known that $\{f_n\}$ converges pointwise to $f(x) = \ln(x)$ on $(0, \infty)$. This is established through the limit: $$ \lim_{n \to \infty} f_n(x) = \ln(x) $$
Analysis of Uniform Convergence on $(1/k, k)$
The confusion arises when examining the uniform convergence on the intervals $(1/k, k)$. Specifically, the points of contention are:
For $x = 1/k$ and $x = k$, the limits of the differences $|f_n(x) - \ln(x)|$ as $n \to \infty$ do not tend to $0$, but instead, they approach $|\ln(k)|$ and $|\ln(1/k)|$, respectively.
Supremum of the Difference: Despite the behavior at the endpoints, there is a claim that the supremum of the difference $|f_n(x) - \ln(x)|$ over the interval $(1/k, k)$ goes to $0$ as $n \to \infty$. This would imply uniform convergence, but it's unclear how this aligns with the non-zero limits at the endpoints.
How does the behavior of $|f_n(x) - \ln(x)|$ at the endpoints $x = 1/k$ and $x = k$ influence the argument for uniform convergence on $(1/k, k)$? If the supremum of the difference on the interval $(1/k, k)$ indeed goes to $0$, how does this reconcile with the endpoint behavior where the differences approach non-zero constants?
Any insights or clarifications on this matter would be greatly appreciated.
Just look at the function $g_{n}(x)=n(x^{1/n}-1)-\ln(x)$ in the interval $(\frac{1}{k},k)$ for a fixed $k$.
(It does not make a difference if you include the endpoints $k$ and $1/k$. It will only strengthen the uniform convergence to a larger set)
$g_{n}'(x)=\frac{1}{x}(x^{\frac{1}{n}}-1)$ and notice that $g_{n}(x)$ is decreasing in $(\frac{1}{k},1)$ and increasing in $[1,k)$ and $g'_{n}(x)$ is $0$ at $1$.
Thus, you have that $g_{n}(x)$ has a minimum at $1$. Here's a graphical illustration which should show you that in this interval, the functions should behave like a skewed parabola.
It should now be apparent to you that $|f_{n}(x)-\ln(x)|=|g_{n}(x)|\leq \max\bigg(|g_{n}(\frac{1}{k})|,|g_{n}(1)|,|g_{n}(k)|\bigg)$ for all $x\in(\frac{1}{k},k)$ due to the Intermediate Value Property.
And now as $g_{n}(\frac{1}{k})\to 0$, $g_{n}(1)=0$ and $g_{n}(k)\to 0$, you can now find for a given $\epsilon>0$, an $N$ such that $\max\bigg(|g_{n}(1/k)|,|g_{n}(1)|,|g_{n}(k)|\bigg)<\epsilon$ for all $n\geq N$
Which means that $\sup_{x\in (\frac{1}{k},k)}|f_{n}(x)-\ln(x)|<\epsilon$ for all $n\geq N$ and this $N$ is only dependent on $\epsilon$.