Is $\{f_n \}$ uniformly convergent on $[0,1]\ $?

Question

Let $\{f_n \}$ be a sequence of functions in $[0,1]$ defined by $$f_n (x) = x^n(1-x),\ x \in [0,1].$$ Does $f_n \to 0$ as $n \to \infty$ uniformly on $[0,1]$?

How to proceed? Any help will be highly appreciated.

Answer 1

We have

$$ f_n'(x) = -x^n + nx^{n-1}(1-x)=[n-(n+1)x]x^{n-1}. $$ By inspection, $f_n'(x) = 0$ when $x=\frac{n}{n+1}$ or $x=0$. The value $\frac{n}{n+1}$ is the local maximum with $$ f_n(\tfrac{n}{n+1})= (\tfrac{n}{n+1})^n(1-\tfrac{n}{n+1}) = \frac{n^n}{(n+1)^{n+1}}. $$ Since $$ \lim_{n \to \infty} \frac{n^n}{(n+1)^{n+1}}=\lim_{n \to \infty} \frac{n^n}{(n+1)^n} \cdot \lim_{n \to \infty} \frac{1}{n+1} = 0, $$ it follows that $\sup_{x \in [0,1]} f_n(x) \to 0$, so $f_n \to 0$ uniformly.

Answer 2

Define $M_n := \max_{0 \leq x \leq 1} f_n(x)$.

Since $0 \leq f_n(x) \leq M_n$ on $[0, 1]$, $f_n(x) \rightarrow 0$ uniformly on $[0, 1]$ iff $M_n \rightarrow 0$ as $n \rightarrow \infty$.

As a differentiable function, $f_n(x) = x^n - x^{n+1}$ must attain its max on $[0, 1]$ when $x = 0, 1$ (at an endpoint) or when $$f'_n(x) = nx^{n-1} - (n+1)x^n = x^{n-1}(n - (n+1)x) = 0,$$ which means $x = 0$ or $x = n/(n+1)$. Clearly if $x = 0, 1,$ $f_n(x) = 0$, so $$M_n = f_n(n/(n+1)) = \frac{n^n}{(n+1)^n}*\frac{1}{(n+1)} \approx \frac{1}{e}*\frac{1}{(n+1)} \rightarrow 0 \text{ as } n \rightarrow \infty.$$

This shows $f_n(x) \rightarrow 0$ uniformly on $[0, 1]$.