MHB Pointwise and uniform convergence

[mathmari]

Hey!

I want to check the pointwise and uniform convergence for the following sequences or series of functions:

1. $f_n:[0, \infty)\rightarrow \mathbb{R}, f_n(x)=xe^{-nx}$ for all $n\in \mathbb{N}$
2. $f_n:[0, \infty)\rightarrow \mathbb{R}, f_n(x)=nxe^{-nx}$ for all $n\in \mathbb{N}$
3. $f_n:[0, 1]\rightarrow \mathbb{R}, f_n(x)=\sqrt[n]{n^2x}$ for all $n\in \mathbb{N}$
4. $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{nx-n^2}}$ for $x\in (0,1)$
5. $\displaystyle{\sum_{n=0}^{\infty}\frac{\sin (x)}{(1+x^4)^n}}$ for $x\in \mathbb{R}$

We have the following:

• The sequence of functions $(f_n)$ is on $D$ pointwise convergent iff for each $x\in D$ the sequence $(f_n(x))$ is convergent.
• The series of functions $\displaystyle{\sum_{n=1}^{\infty}f_n}$ is on $D$ pointwise konvergent iff for each $x\in D$ the sequence $(s_n(x))$ is convergent.
• $(f_n)$ converges on $D$ uniformly to $f: D\rightarrow \mathbb{R}$ iff $\forall \epsilon>0 \ \exists n_0=n_0(\epsilon)\in \mathbb{N}\ \ \forall n\geq n_0 \ \ \forall x\in D: |f_n(x)-f(x)|<\epsilon$.
• $\displaystyle{\sum_{n=1}^{\infty}f_n}$ converges on $D$ uniformly to $f: D\rightarrow \mathbb{R}$ iff $\forall \epsilon>0 \ \exists n_0=n_0(\epsilon)\in \mathbb{N}\ \ \forall n\geq n_0 \ \ \forall x\in D: |s_n(x)-f(x)|<\epsilon$.

I have done the following:

1. We have that $$f(x)=\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow\infty}\frac{x}{e^{nx}}=0$$
   For each $x\in [0, \infty)$ the sequence $(f_n(x))$ converges. So the sequence of function $(f_n)$ converges on $[0,\infty)$ pointwise to $f$.
   
   To check the uniform convergence we have that $|f_n(x)-f(x)|=\frac{x}{e^{nx}}$. What do we do next? (Wondering)
   
2. We have that $$f(x)=\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow\infty}\frac{nx}{e^{nx}}=\lim_{n\rightarrow\infty}\frac{nx}{\frac{\sum_{k=1}^{\infty}(nx)^k}{k!}}=\sum_{k=1}^{\infty}\frac{xk!}{\lim_{n\rightarrow \infty}\left (\frac{nx}{n}+\sum_{k=2}^{\infty}(nx)^{k-1}\right )}=0$$ Is this correct? (Wondering)
   For each $x\in [0, \infty)$ the sequence $(f_n(x))$ converges. So the sequence of function $(f_n)$ converges on $[0,\infty)$ pointwise to $f$.
   
   To check the uniform convergence we have that $|f_n(x)-f(x)|=\frac{nx}{e^{nx}}$. How could we continue? (Wondering)
   
3. We have that when $x=0$ then $$f(x)=\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}\sqrt[n]{n^2\cdot 0}=0$$
   When $x\in (0,1]$ we have that $$f(x)=\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}\sqrt[n]{n^2x}$$ Is this equal to $1$ ? (Wondering)

For the cases 4. and 5. do we use the Weierstrass criterium? Or do we use the definition? (Wondering)

 

[mathmari]

For the case 5. :

We have that for $x\neq 0$ :
$$\sum_{n=0}^{\infty}\frac{\sin (x)}{(1+x^4)^n}=\sin (x)\sum_{n=0}^{\infty}\left (\frac{1}{1+x^4}\right )^n=\sin (x)\frac{1}{1-\frac{1}{1+x^4}}=\sin (x)\frac{1+x^4}{1+x^4-1}=\sin (x)\frac{1+x^4}{x^4}$$
and for $x=0$ : $$\sum_{n=0}^{\infty}\frac{\sin (x)}{(1+x^4)^n}=0$$

What do we get from that about the pointwise and uniform convergence? (Wondering)

 

[Opalg]

For the case 5. :

We have that for $x\neq 0$ :
$$\sum_{n=0}^{\infty}\frac{\sin (x)}{(1+x^4)^n}=\sin (x)\sum_{n=0}^{\infty}\left (\frac{1}{1+x^4}\right )^n=\sin (x)\frac{1}{1-\frac{1}{1+x^4}}=\sin (x)\frac{1+x^4}{1+x^4-1}=\sin (x)\frac{1+x^4}{x^4}$$
and for $x=0$ : $$\sum_{n=0}^{\infty}\frac{\sin (x)}{(1+x^4)^n}=0$$

What do we get from that about the pointwise and uniform convergence? (Wondering)

That's good, and it tells you that the series converges pointwise.

To decide about uniform convergence, there are two useful techniques. The Weierstrass M-test is the most useful tool to prove that a series converges uniformly. To show that a series does not converge uniformly, it often helps to use the theorem that a uniformly convergent sequence or series of continuous functions has a continuous limit. So if each term in a sequence or series is continuous but the (pointwise) limit function is not continuous, then the convergence cannot be uniform.

 

[mathmari]

That's good, and it tells you that the series converges pointwise.

To decide about uniform convergence, there are two useful techniques. The Weierstrass M-test is the most useful tool to prove that a series converges uniformly. To show that a series does not converge uniformly, it often helps to use the theorem that a uniformly convergent sequence or series of continuous functions has a continuous limit. So if each term in a sequence or series is continuous but the (pointwise) limit function is not continuous, then the convergence cannot be uniform.

According to the definition of pointwise convergence of a series of functions we have to check if $s_k$ converges for each $x$.

So, we have that for $x\neq 0$ :
$$s_k=\sum_{n=0}^{k}\frac{\sin (x)}{(1+x^4)^n}=\sin (x)\sum_{n=0}^{k}\left (\frac{1}{1+x^4}\right )^n=\sin (x)\frac{1-\left (\frac{1}{1+x^4}\right )^{k+1}}{1-\frac{1}{1+x^4}}=\sin (x)\frac{1-\frac{1}{(1+x^4)^{k+1}}}{1-\frac{1}{1+x^4}}=\sin (x)\frac{(1+x^4)^{k+1}-1}{(1+x^4)^{k+1}-(1+x^4)^k}=\frac{\sin (x)}{(1+x^4)^k}\left (\frac{(1+x^4)^{k+1}-1}{1+x^4-1}\right )=\frac{\sin (x)}{x^4(1+x^4)^k}\left ((1+x^4)^{k+1}-1\right )$$
and for $x=0$ we have : $s_k=\sum_{n=0}^{k}\frac{\sin (x)}{(1+x^4)^n}=0$.

So, in each case $s_k$ is convergent. Therefore, $\sum_{n=0}^{\infty}\frac{\sin (x)}{(1+x^4)^n}$ convergence pointwise.

Is this correct? (Wondering)

For the case 5. :

We have that for $x\neq 0$ :
$$\sum_{n=0}^{\infty}\frac{\sin (x)}{(1+x^4)^n}=\sin (x)\sum_{n=0}^{\infty}\left (\frac{1}{1+x^4}\right )^n=\sin (x)\frac{1}{1-\frac{1}{1+x^4}}=\sin (x)\frac{1+x^4}{1+x^4-1}=\sin (x)\frac{1+x^4}{x^4}$$
and for $x=0$ : $$\sum_{n=0}^{\infty}\frac{\sin (x)}{(1+x^4)^n}=0$$

We have that $\frac{\sin (x)}{(1+x^4)^n}$ is continuous as a fraction of continuous functions, right? (Wondering)
To check if the limit function is continuous we have to check it at $x=0$ :
We have that $$\lim_{x\rightarrow 0}\sin (x)\frac{1+x^4}{x^4}=\lim_{x\rightarrow 0}\sin (x)\left (\frac{1}{x^4}+1\right )=\lim_{x\rightarrow 0}\frac{\sin (x)}{x^4}+\lim_{x\rightarrow 0}\sin (x)=\lim_{x\rightarrow 0}\frac{\sin (x)}{x}\frac{1}{x^3}=1\cdot \infty=\infty\neq 0$$

So, it is not continuous at $x=0$, and therefore the series does not converge uniformly. Is this correct? (Wondering)

 

[mathmari]

For 4. we have the following:

Since $0<x<1\Rightarrow 0<nx<n\Rightarrow -n<-nx<0\Rightarrow n^2-n<n^2-nx<n^2$ we have that $$\frac{1}{n^2-nx}<\frac{1}{n^2-n}=\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}$$

Since the series $\sum_{n=2}^{\infty}\left (\frac{1}{n-1}-\frac{1}{n}\right )$ converges, we have from the Weierstrass criterion that the sum $\sum_{n=2}^{\infty}\frac{1}{nx-n^2}$ converges uniformly.

So, the series $\sum_{n=1}^{\infty}\frac{1}{nx-n^2}=\frac{1}{x-1}+\sum_{n=2}^{\infty}\frac{1}{nx-n^2}$ converges also uniformly.
And from that it follows that the series converges also pointwise.
Is this correct? (Wondering)

Or dowe have to take into consideration that if $x\rightarrow 1$ then $\rightarrow \infty$ ? (Wondering)

 

[mathmari]

For the case 1. I have done the following:

From the definition ofthe exponential function we have that $e^{nx}=\sum_{k=1}^{\infty}\frac{(nx)^k}{k!}\geq \frac{(nx)^2}{2}$. So, $\frac{1}{e^{nx}}\leq \frac{2}{n^2x^2}$.

Therefore, we have that \begin{equation*}|f_n(x)-f(x)|=\left |\frac{x}{e^{nx}}\right |=\frac{x}{e^{nx}}\leq \frac{2x}{n^2x^2}=\frac{2}{n^2}\end{equation*}

It holds that $\lim_{n\rightarrow \infty}\frac{2}{n^2} =0$.

So, $(f_n)$ converges uniformly.
For the case 3. I have done the following:

The limit function is \begin{equation*}f(x)=\left\{\begin{matrix}
0 & x=0\\
1 & x \in (0,1]
\end{matrix}\right.\end{equation*}

We have that $\sqrt[n]{n^2x}$ is continuous, or not? (Wondering)

We have that \begin{equation*}\lim_{x\rightarrow 0}f(x)=1\neq 0\end{equation*}

So, $f$ is not continuous at $x=0$ and therefore $(f_n)$ does not converge uniformly. Is everything correct? (Wondering) Could you give me a hint for the case 2. ? (Wondering)