[NBHM-2005-PhD Screening Test, Analysis]
Pick out the sequences which are uniformly convergent:
$(a)f_n(x)=sin^n(x)$ on [$0,\pi/2$[.
$(b)f_n(x)=\frac{x^n}{n}+1$ on [$0,1$[.
$(c)f_n(x)=\frac{1}{1+(x-n)^2}$ on ]$-\infty,0$[.
$(d)f_n(x)=\frac{1}{1+(x-n)^2}$ on ]$0,\infty$[.
SOLUTION
(a)Limit function of {$f_n(x)$} on [$0,\pi/2$[ is $$f(x)= \lim_{n\to \infty}sin^n(x)=0$$because on $x\in$[$0,\pi/2$[ $$0\leq sin(x)<1\implies0\leq sin^n(x)<1\implies f(x)= \lim_{n\to \infty}sin^n(x)=0 $$ Now, $$\left\|f_n-f\right\|=\sup_{x\in [0,pi/2)}\left|sin^n(x)-0\right|= 1.$$Hence,$\left\|f_n-f\right\|\rightarrow1 $ as $n\rightarrow \infty$.Thus,{$f_n$} is NOT UNIFORMLY convergent.
$(b)$Limit function of {$f_n(x)$} on [$0,1$[ is $$f(x)=\lim_{n\to \infty}f_n(x)$$
Now,$$0\leq x <1\implies 0\leq x^n 1<\implies 0\leq \frac{x^n}{n}<\frac{1}{n}\implies 1\leq \frac{x^n}{n}+1<\frac{1}{n}+1$$ $$\implies 1\leq f(x)=\lim_{n\to \infty}f_n(x)<\lim_{n\to\infty}(1+\frac{1}{n})$$.Hence,by Sandwich theorem $f(x)=1$.
Now,
$$\left\|f_n-f\right\|=\sup_{x\in [0,1[}\left|\frac{x^n}{n}+1-1\right|= \sup_{x\in [0,1[}\left|\frac{x^n}{n}\right|=\frac{1}{n}$$.So,$\left\|f_n-f\right\|\rightarrow 0 $ as $n\rightarrow \infty$.Hence,{$f_n$} is UNIFORMLY convergent on [$0,1$[
$(c)$Limit function of {$f_n(x)$} on ]$-\infty,0$[,$$f(x)=\lim_{n\to\infty}\frac{1}{1+(x-n)^2}=0$$
Now,$(x-n)^2\geq 0 \forall x\in \mathbb R\implies 1+(x-n)^2\geq 1 \forall x\in \mathbb R\implies \frac{1}{1+(x-n)^2}\leq 1 \forall x\in \mathbb R $.Hence,$\left\|f_n-f\right\|=\sup_{x\in [-\infty,0[}\left|\frac{1}{1+(x-n)^2}-0\right|=1$ So,$\left\|f_n-f\right\|\rightarrow 1 $ as $n\rightarrow \infty$.Hence,{$f_n$} is NOT UNIFORMLY convergent on [$0,1$[.
$(d)$for the same reason as in (c),{$f_n$} is NOT UNIFORMLY cconvergent on ]$0,\infty$[.
In the answer key it is given that among the above sequences only (b) & (c) are uniformly convergent.
Please check my solutions.Also,please point out where i've made mistake in part $c$.
Is there any other way(LIKE SOME GEOMETRIC METHOD)of showing the given sequence to be uniform convergence apart from the above NORM method and Cauchy criterion for uniform convergence?