How can I show uniform convergence?

Question

Let

$f_n(x)=\frac{x^n}{1+x^n},~x\geq 0, ~n\in\mathbb{N}$.

1.1.: Determine the pointwise limi of $(f_n)$, $x\geq 0$.

1.2.: Show that the sequence $(f_n)$ is uniformly convergent on the intervals

$[0,c]$ for $0<c<1$.

$[b,+\infty[$ for $b>1$.

1.3.: Show that there is no uniform convergence on $[1,+\infty[$.

Learning for an upcoming test and I found this one in some problem sheets.

I never dealt with pointwise convergence so I'm not sure how to proceed here. Is it similar to finding the limit of a sequence which is just a point?

About b: I never heard of uniform convergence so I looked up the wiki entry on it. 'If the speed of the convergence does not depend on x.'

How do I approach this type of problem?

So basically I can't deal with this exercise at all. I'm sorry for the lack of work shown here, but I'm really struggling with metric spaces in general.

Answer 1

Hints:

Check first that

$$\lim_{n\to\infty}\frac{x^n}{1+x^n}=f(x)=\begin{cases}0&,\;\;0\le x<1\\{}\\\dfrac12&,\;\;x=1\\{}\\1&,\;\;x>1\end{cases}$$

and for example, for $\;0\le x\le c<1\;$ :

$$\frac{x^n}{1+x^n}\le c^n$$

and you can apply Wierstrass's M-test with the geometric series.

The case $\;x>1\;$ is similar and I'll let it to you.

Answer 2

1.1 So,here we go..First you will need the definition of Pointwise Convergence.That is:

For any $x\in D$, where $D$ is a domain,and for any $ε>0$,there exists an $N$ such that,for every $n>N$:$∣f_n(x)-f(x)∣<ε$

Also,bear in mind that if $∣$x$∣<1$,then $lim_{n\rightarrow∞}∣$x$∣^t=0$,for any real number $t$ and if $∣x∣>1$,then$∣x∣^t\rightarrow∞$ for any real $t$.

We will show that the given function $f_n(x)$ converges to the function $f(x)=0$ if $0≤∣x∣<1$.

Indeed, notice that $lim_{n\rightarrow∞}∣\frac{x^n}{1+x^n}∣=0$,since $x^n\rightarrow0$.

If $∣x∣>1$ , notice that: $lim_{n\rightarrow∞}∣\frac{x^n}{1+x^n}-1∣=∣\frac{1}{1+x^n}∣=0$. So $f_n$ converges to $f(x)=1$ for $∣x∣>1$

Finally,we check the case for $x=1$. Then $f_n(x)=1/2$ as you immediately verify.