Uniform Convergence of a sequence on [0,1]

Question

For each integer $n \geq 2$, define the function $f_n : [0, 1] \to \mathbb{R}$ by

$$f_n(x)=\begin{cases} xn^2, & \text{if $0 \leq x\leq 1/n$} \\[2ex] -n^2(x-2/n), & \text{if}{ ~ 1/n < x\leq 2/n} \\[3ex] 0, & \text{if}{~ 2/n <x \leq 1 }\end{cases}$$

I'm pretty sure the sequence $f_n$ doesn't converge uniformly on [0,1] but i'm unsure how to show this, but i think you have to use $sup_x[0,1]$|$f_n$(x) - $f(x)$|<0 as n tends to infinity?

Answer 1

Note that $f_n(1/n)=n$ and the pointwise limit function is $f=0$ in $[0,1]$. Hence $$\sup_{x\in [0,1]}\left|f_n(x)-f(x)\right|\geq |f_n(1/n)|=n\to +\infty$$ which means that $f_n$ does not converge uniformly on $[0,1]$.

On the other hand, for $a\in(0,1]$ we have that $f_n=0$ in $[a,1]$ for $n>2/a$. Therefore $$\sup_{x\in [a,1]}\left|f_n(x)-f(x)\right|\to 0$$ which means that $f_n$ converges uniformly on $[a,1]$.

Answer 2

First note that for all $x\in[0,1]$, $\lim_{n\to \infty}f_n(x)=0$.

Now let $\epsilon=2$. Then, for all $N> 1$, there exists $n=N+1>N$ and there exists $x=1/(N+2)\in[0,1]$, such that we have

$$\left|f_n(x)-0\right|=\frac{(N+1)^2}{N+2}>\epsilon = 2$$

This is precisely the negation of uniform convergence.

Answer 3

let $x$ be in $[0,1]$. if $x=0$ then for all $n$ $f_n(0)=0$ and $ f(0)=0$.

assume now that $x>0$, and let $n_0>\frac{2}{x}$

for $n>n_0$, $\frac{2}{n}<x$ and

$f_n(x)=0$ which gives $f(x)=0$.

the pointwise limit of the sequence is $f=0$ at $[0,1]$.

for the uniform convergence, a good solution has been given just before.