Pointwise but not Uniformly Convergent
The Question: Prove that the sequence of functions $f_n(x)=\frac$ converges pointwise on $\mathbb$, but does not converge uniformly on $\mathbb$. My Work: Prove Pointwise: First, $\lim\limits_ \frac=\lim\limits_ \frac+x=x$. My Problem: I am not sure where this fails to be uniformly convergent. Any help is appreciated. Thanks
- real-analysis
- convergence-divergence
- uniform-convergence
asked Jun 4, 2013 at 17:25
Mathstudent Mathstudent
751 2 2 gold badges 6 6 silver badges 19 19 bronze badges
5 Answers 5
$\begingroup$
Looking at the form of $f_n(x) = = +x$, 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 $$ |f_n(x)-f(x)|=\left|\left(+x\right)-x\right|=|x^2/n|\to 0\text< as >n\to\infty\text< for each fixed >x\in\mathbb R. $$ 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 $$ \sup_|f_n(x)-f(x)|\to 0\text< as >n\to \infty. $$ Examining this (stronger!) condition in our case, we see $$ \sup_|f_n(x)-f(x)|=\sup_\left|\left(+x\right)-x\right|=\sup_|x^2/n|\geq |n^2/n|=n\not\to 0\text< as >n\to\infty. $$ Thus, $f_n(x)$ does not not converge uniformly to $f(x)$ on $\mathbb R$.
