Pick out the sequences which are uniformly convergent:

Question

[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?

Answer 1

We make use the following result.

$f_n\to f$ uniformly on (say) $E$ $\iff$ $||f_n-f||=\sup_{x\in E}|f_n(x)-f(x)|\to0$ as $n\to\infty$

For (a) you've already shown $$||f_n-f|| \to 1 $$

Thus (a) is not UC.

For (b) we have $$||f_n-f||=1/n\to 0$$

Hence (b) is UC.

For (c) I suggest you to find the supremum by using First Derrivative Test for Maxima. You should get the following $$||f_n-f||=\frac{1}{1+n^2}\to 0$$

and hence (c) is UC.

Finally for (d) $$||f_n-f||=1\to1$$

Hence it is not UC. To obtain $||f_n-f|| $ you can follow the same technique as the one i've mentioned in (c)