Possible Duplicate:
uniform convergence of few sequence of functions
Pick out the sequences {$f_n$} which are uniformly convergent.
(a)$f_n(x)$= $1/(1+(x-n)^2)$ on ($-∞,0$)
(b) $f_n(x)$= $1/(1+(x-n)^2)$ on ($0, ∞$)
I am completely helpless. how to find the solution. Thanks for your help.
First you should verify that the pointwise limit is $f(x)=0$ in both cases (though with different domains- prove this). For uniform convergence in (b) we must see whether or not $$\left\|f_n-f\right\|=\sup_{x\in [0,\infty)}\left|f_n(x)-f(x)\right|\to 0$$ But $$\sup_{x\in [0,\infty)}\left|f_n(x)-f(x)\right|=\sup_{x\in [0,\infty)}\left|\frac{1}{1+(x-n)^2}\right|\ge^{*} \left|\frac{1}{1+(n-n)^2}\right|=1$$ What does this imply? Why is $\ge^*$ valid in the above reasoning?
(a) unfortunately can't be treated the same way. Indeed, $$\sup_{x\in (-\infty,0]}\left|f_n(x)-f(x)\right|=\sup_{x\in (-\infty,0]}\left|\frac{1}{1+(x-n)^2}\right|=\sup_{x\in (-\infty,0]}\frac{1}{1+(x-n)^2}$$ Observe that as we are confined in $(-\infty,0]$ we can't let $x=n$. Even if we let $x=-n$ then we won't arrive at any results. Instead, it seems as if the convergence is uniform. Let $$g(x)=\frac{1}{1+(x-n)^2}$$ in $(-\infty,0]$ and observe $g$ is increasing (why?). Thus, $$\sup_{x\in (-\infty,0]}g(x)=g(0)=\frac{1}{1+n^2}$$ I think you can finish this argument
Observe that $\forall$ $x\neq 0,\lim\limits_{n\to\infty}f_n(x)=\lim\limits_{n\to\infty}\frac{1}{1+(x-n)^2}=\lim\limits_{n\to\infty}\frac{\frac{1}{n^2}}{\frac{1}{n^2}+(\frac{x}{n}-1)^2}=0.$
a) $\sup\limits_{-\infty<x<0}|f_n(x)-f(x)|=\sup\limits_{-\infty<x<0}|\frac{1}{1+(x-n)^2}|=0\to0$ (Since $|\frac{1}{1+(x-n)^2}|> 0$ & for $\epsilon>0$ $\exists $ $x<0$ such that $|\frac{1}{1+(x-n)^2}|<\epsilon$) $\implies f_n$ is uniformly convergent on $(-\infty,0).$
b) $\sup\limits_{0<x<\infty}|f_n(x)-f(x)|=\sup\limits_{0<x<\infty}|\frac{1}{1+(x-n)^2}|=1\to 1\neq0$ (Since $|\frac{1}{1+(x-n)^2}|\leq 1$ where '=' holds for $x=n$) $\implies f_n$ is not uniformly convergent on $(0,\infty).$
