The Question: Prove that the sequence of functions $f_n(x)=\frac{x^2+nx}{n}$ converges pointwise on $\mathbb{R}$, but does not converge uniformly on $\mathbb{R}$.
My Work: Prove Pointwise: First, $\lim\limits_{n\to\infty} \frac{x^2+nx}{n}=\lim\limits_{n\to\infty} \frac{x^2}{n}+x=x$.
My Problem: I am not sure where this fails to be uniformly convergent. Any help is appreciated. Thanks
On
Choose $\epsilon > 0$. If the convergence is uniform, then you can find an $n$ such that $|\frac{x^2}{n} + x -x|_{\infty} = |\frac{x^2}{n}|_{\infty}$ is smaller than $\epsilon$. That is, there exists an $n$ such that for ALL $x$, $x^2/n$ is smaller than $\epsilon$. Is this possible?
On
I remember that converging sequences are always bounded, that is, there is a metric space structure on which convergence is defined. In this time the elements of the given sequence don't have distances. Hence there cannot be convergence.
On
A sequence of functions $f_{n}:\mathbb{R}\rightarrow\mathbb{R}$ is said to non-uniformly contionous on $\mathbb{R}$ , if $\exists \epsilon_{0}>0:\exists$ subsequences $n_{k},x_{k}$,such that
$|f_{n_{k}}(x_{k})-f(x_{k})|\ge\epsilon_{o} \forall k\in \mathbb{N}$
From a simple calculation,we have that $f_{n}(x)\rightarrow x,\forall x\in\mathbb{R}$
Now,define $n_{k}:=k,x_{k}:=\sqrt{k}, \forall k\in \mathbb{N}$,so that, $|f_{n_{k}}(x_{k})-f(x_{k})|=|\frac{k}{k}+k-k|=1>\epsilon_{0},$ for a certain $\epsilon_{0}<1$
What is $\sup\{|f_n(x)-f(x)|\}\to?$ as $n\to\infty$?