Why doesn't $f_n(x) = \frac{n^2x + e^{nx} - 1}{3^{2nx}}$ converge uniformly on $[0, 1]$?

Question

It converges uniformly on the interval $[a, \infty)$ to its limit function $f(x) = 0$ for all $x \ge 0$. The limit as $x \rightarrow 0^+$ is equal to $0$, and I can't really find a reason as to why it's not uniformly convergent on $[0, 1]$. What am I missing?

Answer 1

A sequence of functions $\{f_n\}$ converges uniformly to $f$ if and only if $$ a_n := \sup_x |f_n(x)-f(x)|\stackrel{\mathrm{i.i.d.}}\longrightarrow 0. $$ Here $f_n$ is continuous on $[0,1]$, so it attains a maximum value at some $x_n\in[0,1]$. Since $f_n$ is differentiable on $(0,1)$, we must have that $x_n\in\partial[0,1]=\{0,1\}$ or $f_n'(x_n) = 0$. We compute $f_n$ at the boundary points: \begin{align} f_n(0) &= 0\\ f_n(1) &= 3^{-2 n} \left(n^2+e^n-1\right)\stackrel{n\to\infty}\longrightarrow 0, \end{align} and then compute the derivative: $$ f_n'(x) = n 9^{-n x} \left(n^2 (-x) \log (9)-(\log (9)-1) e^{n x}+n+\log (9)\right), $$ from which we see that $f_n'(x)=0$ if and only if $$ x_n:= \frac{n (-\log (9)) W\left(\frac{(\log (9)-1) e^{\frac{1}{n}+\frac{1}{\log (9)}}}{n \log (9)}\right)+n+\log (9)}{n^2 \log (9)}., $$ where $W$ denotes the Lambert $W$ function. Now, $f_n$ converges pointwise to zero, , and so $$ a_n \leqslant|f_n(x_n)| = \frac{\left.\left| n \left(W\left(\frac{e^{\frac{1}{\log (9)}+\frac{1}{n}} (\log (9)-1)}{n \log (9)}\right) \log (9)+\log (9)-1\right)\right.\right| 3^{-2 \left(-\Re\left(W\left(\frac{(\log (9)-1) e^{\frac{1}{n}+\frac{1}{\log (9)}}}{n \log (9)}\right)\right)+\Re\left(\frac{1}{n}\right)+\frac{1}{\log (9)}\right)}}{(\log (9)-1) \log (9)}. $$ Since $\lim_{n\to\infty}|f_n(x_n)|=\infty$, it follows that $\{f_n\}$ doe not converge uniformly.