Let $(f_n)_{n\in \mathbb{N}}$ be the sequence of functions defined by \begin{align*}f_n:[0, \infty)&\rightarrow \mathbb{R}\\ x&\mapsto \min \{n,x\}\end{align*} and let $f$ be the function defined by \begin{align*}f:[0, \infty)&\rightarrow \mathbb{R}\\ x&\mapsto x\end{align*}
(a) Let $a>0$. Show that sequence of functions $f_n\mid_{[0,a]}:[0,a]\rightarrow \mathbb{R}$ converges uniformly to the function $f\mid_{[0,a]}:[0,a]\rightarrow \mathbb{R}$.
(b) Show that sequence of functions $f_n:[0,\infty)\rightarrow \mathbb{R}$ doesn't converge uniformly to the function $f:[0,\infty)\rightarrow \mathbb{R}$.
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I have done the following:
(a) We have to show that, given $\epsilon >0$, there exists a positive integer $N$ such that $n \ge N$ implies $|f_n(x)-f(x)|<\epsilon$ for every $x\in [0,a]$.
We have that $a>0$ is fixed. For $N\geq a$ we have then that $$\left |f_n(x)-f(x)\right |=\left |x-x\right |=0<\epsilon$$ which means that $f_n$ converges uniformly to $f$.
Is that correct and complete?
At (b) we have that $$\left |f_n(x)-f(x)\right |=\left |\min\{n,x\}-x\right |=\begin{cases}n-x & \text{ if } n<x \\ x-x & \text{ if } n\geq x\end{cases}=\begin{cases}n-x & \text{ if } n<x \\ 0 & \text{ if } n\geq x\end{cases}$$ In the first case we cannot bound it by an $\epsilon$, can we?
Best Answer
There is a wrong sign in part $(b)$ $$|f_n(x)-f(x)|=\left |\min\{n,x\}-x\right |=\begin{cases}x - n & \text{ if } n<x \\ 0 & \text{ if } n\geq x\end{cases}$$ $$\sup_{x \in [0,\infty)} |f_n(x)-f(x)| \geq |f_n(n + 1) - f(n + 1)| = n +1 -n = 1$$ for every $n \in \mathbb{N}$, hence the convergence is not uniform on $[0,\infty)$. The proof of $(a)$ looks correct