1. Uniform convergence

Can someone give me some help on this problem? Let $\displaystyle f_n(x)=\frac{4}{4+x^n}$ on $\displaystyle [0,\infty)$
I think $\displaystyle \{f_n(x)\}$ converges to $\displaystyle f(x)=0$.
Now I want to determine if this convergence is uniform or not. I tried to take the derivative to use the Weierstrass-M test, but I am stuck on finding the sup.
Also, does this sequence converge uniformly if we change the domain to $\displaystyle (1,\infty)$

2. Originally Posted by jackie
Can someone give me some help on this problem? Let $\displaystyle f_n(x)=\frac{4}{4+x^n}$ on $\displaystyle [0,\infty)$
I think $\displaystyle \{f_n(x)\}$ converges to $\displaystyle f(x)=0$.
Now I want to determine if this convergence is uniform or not. I tried to take the derivative to use the Weierstrass-M test, but I am stuck on finding the sup.
Also, does this sequence converge uniformly if we change the domain to $\displaystyle (1,\infty)$
The function it converges to is not the zero function.

If $\displaystyle 0\leq x < 1$ then $\displaystyle x^n \to 0$ so $\displaystyle \frac{4}{4+x^n} \to 1$.
If $\displaystyle x=1$ then $\displaystyle x^n \to 1$ so $\displaystyle \frac{4}{4+x^n} \to \frac{4}{5}$.
If $\displaystyle x>1$ then $\displaystyle x^n \to \text{me}$ so $\displaystyle \frac{4}{4+x^n} \to 0$.

The converges is not uniform because the function it coverges to is not continous. Remember, uniform convergence of continous functions is continous, this is not the case here. Thus, it cannot be uniform.

3. Originally Posted by ThePerfectHacker
The function it converges to is not the zero function.

If $\displaystyle 0\leq x < 1$ then $\displaystyle x^n \to 0$ so $\displaystyle \frac{4}{4+x^n} \to 1$.
If $\displaystyle x=1$ then $\displaystyle x^n \to 1$ so $\displaystyle \frac{4}{4+x^n} \to \frac{4}{5}$.
If $\displaystyle x>1$ then $\displaystyle x^n \to \text{me}$ so $\displaystyle \frac{4}{4+x^n} \to 0$.

The converges is not uniform because the function it coverges to is not continous. Remember, uniform convergence of continous functions is continous, this is not the case here. Thus, it cannot be uniform.
Thank you so much, TPH. I forgot to look at cases of x on the given domain. So, if the domain is $\displaystyle (1,\infty)$, then it will be the third case, and the function $\displaystyle f(x)=0$ is a continuous function. Is the converse of the theorem you said true?

4. Originally Posted by jackie
Thank you so much, TPH. I forgot to look at cases of x on the given domain. So, if the domain is $\displaystyle (0,\infty)$, then it will be the third case, and the function $\displaystyle f(x)=0$ is a continuous function. Is the converse of the theorem you said true?
Think you meant:

"if the domain is $\displaystyle (1,\infty)$, ..."

5. Originally Posted by jackie
Thank you so much, TPH. I forgot to look at cases of x on the given domain. So, if the domain is $\displaystyle (1,\infty)$, then it will be the third case, and the function $\displaystyle f(x)=0$ is a continuous function. Is the converse of the theorem you said true?
The converse of the theorem is not true. Also, even on $\displaystyle (1,\infty)$ the convergence is not uniform. Before we explain why remember that $\displaystyle (1+\tfrac{1}{n})^n < e$ for all $\displaystyle n\geq 1$. Assume that $\displaystyle f_n(x) = \tfrac{4}{4+x^n}$ converged uniformly to the zero function, then it means for any $\displaystyle \varepsilon>0$ there is integer $\displaystyle N>0$ such that if $\displaystyle n\geq N$ then $\displaystyle |f_n(x)| < \varepsilon$ for all $\displaystyle x\in (1,\infty)$. Let us assume that this is true. So for a specific $\displaystyle \varepsilon > 0$ we have $\displaystyle \tfrac{4}{4+x^n} < \varepsilon$ for all $\displaystyle x\in (1,\infty)$ where $\displaystyle n\geq N$. If this inequality is true for all $\displaystyle x$ then it is true for $\displaystyle x=1+\tfrac{1}{n}$. Thus, we must have $\displaystyle \tfrac{4}{4+\left(1+\tfrac{1}{n} \right)^n} < \varepsilon$. However, $\displaystyle \tfrac{4}{4+\left(1+\tfrac{1}{n} \right)^n} \geq \tfrac{4}{4+e}$. This means if we choose $\displaystyle \varepsilon < \tfrac{4}{4+e}$ then we would have a contradition for $\displaystyle x=1+\tfrac{1}{n}$. Therefore, the sequence cannot converge uniformly.

6. Originally Posted by ThePerfectHacker
The function it converges to is not the zero function.

If $\displaystyle 0\leq x < 1$ then $\displaystyle x^n \to 0$ so $\displaystyle \frac{4}{4+x^n} \to 1$.
If $\displaystyle x=1$ then $\displaystyle x^n \to 1$ so $\displaystyle \frac{4}{4+x^n} \to \frac{4}{5}$.
If $\displaystyle x>1$ then $\displaystyle x^n \to \text{me}$ so $\displaystyle \frac{4}{4+x^n} \to 0$.

The converges is not uniform because the function it coverges to is not continous. Remember, uniform convergence of continous functions is continous, this is not the case here. Thus, it cannot be uniform.
$\displaystyle x^n$ goes to you? Because you are infinite?

7. Originally Posted by Sampras
$\displaystyle x^n$ goes to you? Because you are infinite?
Exactly. Now you are understanding me better. Of course, you shall never be able to understand me because I am too great.