16
$\begingroup$

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

$\endgroup$
10
$\begingroup$

Looking at the form of $f_n(x) = {x^2+nx\over n} = {x^2\over n}+x$, it is reasonable to propose $f(x)=x$ as the pointwise limit. To verify this, note that $f_n(x)\to f(x)$ pointwise on an interval $I$ means $$ |f_n(x)-f(x)|\to 0\text{ as }n\to \infty \text{ for each fixed }x\in I. $$

With that in mind, consider $$ |f_n(x)-f(x)|=\left|\left({x^2\over n}+x\right)-x\right|=|x^2/n|\to 0\text{ as }n\to\infty\text{ for each fixed }x\in\mathbb R. $$ We conclude that indeed $f_n(x)\to x$ pointwise on $\mathbb R$.

On the other hand, $f_n(x)\to f(x)$ uniformly on $I$ means $$ \sup_{x\in I}|f_n(x)-f(x)|\to 0\text{ as }n\to \infty. $$ Examining this (stronger!) condition in our case, we see $$ \sup_{x\in\mathbb R}|f_n(x)-f(x)|=\sup_{x\in\mathbb R}\left|\left({x^2\over n}+x\right)-x\right|=\sup_{x\in \mathbb R}|x^2/n|\geq |n^2/n|=n\not\to 0\text{ as }n\to\infty. $$ Thus, $f_n(x)$ does not not converge uniformly to $f(x)$ on $\mathbb R$.

$\endgroup$
  • $\begingroup$ I may be wrong but I think sup is not defined, so the final series of equalities and inequalities is badly phrased. I think you need to show that the sup doesn't exist by showing that the content is unbounded. $\endgroup$ – Suzu Hirose Dec 1 '14 at 0:56
  • 1
    $\begingroup$ But if it doesn't exist why do you state that it does exist in the first equation? Why not just show that |fn-f| is unbounded, and then argue that the sup doesn't exist? $\endgroup$ – Suzu Hirose Dec 1 '14 at 1:10
  • 2
    $\begingroup$ Nowhere in there says the sup exists. It is a statement examining what the the sup is, in effort to establish whether it is finite or not. You can't establish that the sup infinite by starting out saying the sup doesn't exist. That is question begging. $\endgroup$ – JohnD Dec 1 '14 at 1:12
  • 1
    $\begingroup$ I'm going to differ with you here, your method of argument is sloppy. We should show |fn-f| is unbounded and say that the sup doesn't exist rather than argue from an equation featuring the non-existent sup. Anyway I have no more to say. $\endgroup$ – Suzu Hirose Dec 1 '14 at 1:14
  • 1
    $\begingroup$ Obviously I disagree. You should think about it more. $\endgroup$ – JohnD Dec 1 '14 at 1:30
3
$\begingroup$

What is $\sup\{|f_n(x)-f(x)|\}\to?$ as $n\to\infty$?

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\epsilon_{0}<1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.