Sequences $\{f_n\}$ which are uniformly convergent

Question

Pick out the sequences $\{f_n\}$ which are uniformly convergent.

(a) $f_n(x)$=$\sin^n x$ on [$0,\pi/2$)
(b) $f_n(x)$=$\dfrac{x^n}{n}+1$ on [$0,1$)

I am totally stuck on this problem. How can I able to solve this problem. Can anybody help me please

First figure out what they converge to pointwise. Then use that as a candidate to see if you can pick your $\delta$ independent of $x$. Likely, both will break near the right hand side of the domain. — toypajme, Jan 02 '13 at 16:44

score 3 · Answer 1 · answered Jan 02 '13 at 17:10

a) $\lim_{n\to \infty}f_n(x)=0$ for $0\leq x<\frac{\pi}{2}\implies f_n\to f=0.$Now, $\operatorname{Sup}_{x\in[0,\frac{\pi}{2})}|f_n(x)-f(x)|=\operatorname{Sup}_{x\in[0,\frac{\pi}{2})}|\sin^nx|=1\to 1\neq0\implies \{f_n\}$ is not uniformly convergent to $1$ on $[0,\frac{\pi}{2}).$

b) $x=0\implies \lim_{n\to \infty}f_n(x)=1$ & $0<x<1\implies\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty}[\frac{x^n}{n}+1]=1.$ So, $f_n\to f=1.$ Now, $\operatorname{Sup}_{x\in[0,1)}|f_n(x)-f(x)|=\operatorname{Sup}_{x\in[0,1)}|\frac{x^n}{n}|=\frac{1}{n}\to0\implies\{f_n\}$ is uniformly convergent to $1$ on $[0,1).$

Nameless · Answer 2 · 2013-01-02T17:11:26.257

2

For (b) prove $f_n\to f$ pointwise where $f(x)=1$, $x\in [0,1)$. Observe: $$\left\|f_n-f\right\|=\sup_{x\in [0,1)}\left|\frac{x^n}{n}+1-1\right|=\sup_{x\in [0,1)}\frac{x^n}{n}= \frac{1}{n}$$ What does this tell you?

(a) can be treated similarly by noticing $\sin (x)<1$ in that domain

EDIT: For (a) the limit is $f(x)=0$. $$\left\|f_n-f\right\|=\sup_{x\in [0,\frac{\pi}2)}\left|\sin^nx\right|=\sup_{x\in [0, \frac{\pi}2)}\sin^n x=1$$

edited Jan 02 '13 at 17:11

answered Jan 02 '13 at 16:51

Nameless

13,456

then both the option are correct.am I correct? – ramu Jan 02 '13 at 17:03
The sequence in a) does not converge uniformly on $[0,\pi/2)$. The sequence converges pointwise to $0$ on $[0,\pi/2)$; but for any $n$, one can find an $x_n\in[0,\pi/2)$ so that $\sin^n (x_n)>1/2$. – David Mitra Jan 02 '13 at 17:06
@DavidMitra By saying treated similarly, I didn't mean the convergence is uniform as well. I will edit my post to clarify this – Nameless Jan 02 '13 at 17:08
@ramu Look at the edited answer – Nameless Jan 02 '13 at 17:12

score 2 · Answer 3 · answered Jan 02 '13 at 17:13

Recall the definition of uniform convergence: a sequence of functions $f_n$ converges uniformly to $f$ if, for every $\epsilon \gt 0$, there exists some $N(\epsilon) \in \mathbb{N}$ independent of $x$ such that $\forall n \gt N$, and $\forall x$ in the domain, $|f_n(x) - f(x)| \lt \epsilon$. Intuitively: how "far" you have to go "out" in the sequence in order to get within a given error $\epsilon$ is the same for all $x$.

So, with that in mind, here's a hint:

For (a), what happens as $x$ gets closer to $\pi/2$? Can you show that you have to take larger and larger $n$ for the sequence to get within a given $\epsilon$ of its limit function there? Remember, this is what it means to have pointwise convergence only, and for uniform convergence to fail.

You might consider the similar but perhaps easier situation: $f_n(x) = x^n$ on $[0,1)$.

For (b), do you run into the same problem?

Sequences $\{f_n\}$ which are uniformly convergent

3 Answers3