Does $f_n$ uniformly converge to $f$?

Question

$f_n :[0, 1]\rightarrow\mathbb{R} \qquad x \mapsto x^n - x^{n+1}$

The sequence converges pointwise to the zero function. It converges uniformly if

$$\sup_{x \in [0, 1]} \; \big| \, f_{n}(x) - f(x) \, \big|$$

tends to zero. But I am not sure if it does or how to prove.

Answer 1

Note that $f_n'(x)=nx^{n-1}-(n+1)x^n=x^{n-1}\bigl(n-(n+1)x\bigr)$. Therefore, the maximum of $f_n$ is $f_n\left(\frac n{n+1}\right)$. But$$f_n\left(\frac n{n+1}\right)=\left(\frac n{n+1}\right)^n\frac1{n+1}.$$Since$$\lim_{n\to\infty}\left(\frac n{n+1}\right)^n\frac1{n+1}=0,$$your sequence converges uniformly.

Answer 2

First note that we can write $x^n-x^{n+1}=x^n(1-x)$.

Given $0<\varepsilon<1$, if $1-\varepsilon<x\leq 1$ then $$ x^n(1-x)\leq1-x<\varepsilon,$$ while if $0\leq x\leq 1-\varepsilon$ then $$ x^n(1-x)<x^n<\varepsilon$$ for all sufficiently large $n$. Hence $f_n\to 0$ uniformly on $[0,1]$.

Answer 3

Notice that $f_{n+1} = x f_n$ as $x\in[0,1]$ you have that $f_n$ is monotone. Use Dini's theorem Link to complete the proof.

Answer 4

The maximum of $x^n-x^{n+1}$ for $x\in [0,1]$ occurs when $x=\frac{n}{n+1}$ such that

$$x^n-x^{n+1}=\frac{1}{n+1}\left(\frac1{1+\frac1n}\right)^n\le \frac1{n+1}\to 0\,\,\text{as}\,\,n\to \infty$$