# Pick out the sequences which are uniformly convergent:

[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?

• Related Aug 1, 2017 at 5:21
• (a) is not uniformly convergent since $||f_n-f||\not \to0$ Aug 1, 2017 at 5:27
• @Naive:Sorry,it was the typo. Aug 1, 2017 at 5:41

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)

• :$f'_n(x)=\frac{-2(x-n)}{[1+(x-n)^2]^2}$.So,the point of extrema is $x$ if $f'_n(x)=0$ i.e.,$\frac{-2(x-n)}{[1+(x-n)^2]^2}=0 \implies x=n$.So,$f_n(n)=1$.Hence,$\left\|f_n-f\right\|=1$.Thus $\left\|f_n-f\right\|\rightarrow1$ as $n\rightarrow \infty$. Aug 1, 2017 at 6:32
• This argument is perfect for (d). But for (c) the domain is $(-\infty,0)$ and the critical point $x=n$ is not in the domain. If you can show that for each $n\in \Bbb N$, $f_n^{'}(x)$ is increasing for $x\in (-\infty,n)$ then what can you conclude from this? Aug 1, 2017 at 6:38
• :$f_n(x)$ is increasing if $f'_n(x)>0 \implies \frac{-2(x-n)}{[1+(x-n)^2]^2}>0 \implies x<n$.Now,{$f_n(x)$} is monotonically increasing sequence on ($-\infty,n$) which has upper bound zero.Hence,for (c),{$f_n(x)$} is a Uniformly convergent sequence. Aug 1, 2017 at 6:51
• :Is my argument correct? Aug 1, 2017 at 6:52
• Avoid using implication '$\implies$', other than that your argument for $\{f_n\}$ is increasing is correct. But the latter part of it does not make much sense. Aug 1, 2017 at 7:08