Homework 1

SHI YUE 999027873

Let $g_n(x) = \frac{nx + \sin(nx)}{2n}$. Find the pointwise limit of $g_n$ on $\mathbb{R}$. Is the convergence uniform on $[-10,10]$? Is the convergence uniform on $\mathbb{R}$?

Answer
1. Since $\lim_{n\to\infty} g_n(x) = \lim_{n\to\infty} \frac{nx + \sin(nx)}{2n} = \lim_{n\to\infty} \frac{x + \sin(nx)/n}{2} = \frac{x}{2},$ then the pointwise limit of $g_n(x)$ on $\mathbb{R}$ is $\frac{x}{2}$.
2. To check the uniform convergence, let's use the definition: $\forall\varepsilon>0,\exists N\in\mathbb{N},\forall n\geq N:\left|g_{n}(x)-\frac{x}{2} \right|<\varepsilon$
  
  Consider $\left| g_n(x) - \frac{x}{2} \right| = \left| \frac{nx + \sin(nx)}{2n} - \frac{x}{2} \right| = \left| \frac{nx + \sin(nx) - nx}{2n} \right| = \left| \frac{\sin(nx)}{2n} \right| \leq \frac{1}{2n}\leq \frac{1}{2N}$, then we should take $N = \frac{1}{2\varepsilon}$.
  
  Thus, we get $\left|g_{n}(x)-\frac{x}{2}\right|\leq\frac{1}{2\frac{1}{2\varepsilon}}=\varepsilon $ as desired
  
  Therefore, $g_n$ is uniformly convergent on $[-10,10]$ and $\mathbb{R}$.
For each $n \in \mathbb{N}$, define $f_n$ on $\mathbb{R}$ by $f_n(x) = \begin{cases} 1 & \text{if } |x| \geq \frac{1}{n} \\ n|x| & \text{if } |x| < \frac{1}{n}. \end{cases}$

Find the pointwise limit of $(f_n)$ on $\mathbb{R}$ and decide whether or not the convergence is uniform.

Compute $\lim_{n\to\infty}f_{n}\left(x\right)=\lim_{n\to\infty} \begin{cases} 1 & \text{if }|x|\geq\frac{1}{n} \\ n|x| & \text{if }|x|<\frac{1}{n} \end{cases}= \begin{cases} 1 & \text{if }\left|x\right|>0 \\ 0 & \text{if }\left|x\right|\leq0 \end{cases}= \begin{cases} 1 & \text{if }x\neq0 \\ 0 & \text{if }x=0 \end{cases}$

Thus the pointwise limit is $f(x)= \begin{cases} 1 & \text{if }x\neq0 \\ 0 & \text{if }x=0 \end{cases}$

Then by continuous limit theorem, $f_n$ is not uniform convergent since if $f_n$ is uniform convergent and we know $f_n(x)$ is continuous, then $f$ must be continuous which will be a contradiction
Assume $(f_n)$ and $(g_n)$ are uniformly convergent sequences of functions.
1. Show that $(f_n + g_n)$ is a uniformly convergent sequence of functions.
  
  We know $f_n$ and $g_n$ are uniformly convergent, then we have
  1. $\forall\varepsilon_{1}=\frac{\varepsilon}{2}>0,\exists N_{1}\in\mathbb{N},\forall n\geq N_{1}:|f_{n}(x)-f(x)|<\varepsilon_{1}=\frac{\varepsilon}{2},\forall x\in A$
  2. $\forall\varepsilon_{2}=\frac{\varepsilon}{2}>0,\exists N_{2}\in\mathbb{N},\forall n\geq N_{2}:|g_{n}(x)-g(x)|<\varepsilon_{2}=\frac{\varepsilon}{2},\forall x\in A$
  Then consider $|f_n(x)+g_n(x)-(f(x)+g(x))|=|(f_n(x)-f(x))+(g_n(x)-g(x))|$
  
  $\leq|f_{n}(x)-f(x)|+|g_{n}(x)-g(x)|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}= \varepsilon$
  
  Thus if we take $\forall\varepsilon>0,\exists N=\max\left(N_{1},N_{2}\right),\forall n\geq N:|f_{n} (x)+g_{n}(x)-(f(x)+g(x))|<\varepsilon,\forall x\in A$
  
  Thus $(f_n + g_n)$ is a uniformly convergent sequence of functions. 2. Give an example to show that the product $(f_n g_n)$ may not converge uniformly.
  
  Assume $f_{n}(x)=\frac{1}{n}$ and $g(x)=x$ is uniform convergent\, but$f_{n}(x)g_{n}(x)=\frac{x}{n}$ is not uniform convergent 3. Prove that if there exists an $M > 0$ such that $|f_n| \leq M$ and $|g_n| \leq M$ for all $n \in \mathbb{N}$, then $(f_n g_n)$ does converge uniformly.
  
  We know $f_n$ and $g_n$ are uniformly convergent, then we have
  1. $\forall\varepsilon_{1}=\frac{\varepsilon}{2}>0,\exists N_{1}\in\mathbb{N},\forall n\geq N_{1}:|f_{n}(x)-f(x)|<\varepsilon_{1}=\frac{\varepsilon}{2\left(\varepsilon_{2}+M\right)} ,\forall x\in A$
  2. $\forall\varepsilon_{2}=\frac{\varepsilon}{2M}>0,\exists N_{2}\in\mathbb{N},\forall n\geq N_{2}:|g_{n}(x)-g(x)|<\varepsilon_{2}=\frac{\varepsilon}{2M},\forall x\in A$
  Then consider $\left|f_{n}\left(x\right)g_{n}\left(x\right)-f\left(x\right)g\left(x\right)\right |=|f_{n}(x)g_{n}(x)-f_{n}\left(x)g\left(x\right)+f_{n}\left(x\right)g\left(x\right )-f\left(x\right)g\left(x\right)|\leq|f_{n}(x\right)(g_{n}\left(x\right)-g\left(x ))\left|+\right|g\left(x\right)(f_{n}(x\right)-f(x))|$
  
  $\leq\left|f_{n}\left(x\right)\right|\left|g_{n}\left(x\right)-g\left(x\right)\right |+\left|g\left(x\right)\right|\left|f_{n}\left(x\right)-f\left(x\right)\right|$ and since $|f_n| \leq M$ and $|g_n| \leq M$
  
  Since we know (ii), then $|g(x)|\leq\varepsilon_{2}+|g_{n}(x)|\leq\varepsilon_{2}+M$, thus it is bounded
  
  Thus if we take $\forall\varepsilon>0,\exists N=\max\left(N_{1},N_{2}\right),\forall n\geq N:|f_{n} (x)g_{n}(x)-f\left(x\right)g\left(x\right)|\leq\varepsilon,\forall x\in A$
  
  Thus $(f_n g_n)$ is a uniformly convergent sequence of functions.
Consider the sequence of functions defined by $g_n(x) = \frac{x^n}{n}$
1. Show that $(g_n)$ converges uniformly on $[0, 1]$ and find $g = \lim g_n$. Show that $g$ is differentiable and compute $g'(x)$ for all $x \in [0, 1]$.
  
  Since $\lim_{n\to\infty}\frac{x^{n}}{n}=0$ on $[0,1]$, then let's check uniform convergence
  
  Consider $|\frac{x^{n}}{n}|\leq\left|\frac{1}{n}\right|$, thus $\forall\varepsilon>0,\exists\frac{1}{\varepsilon}=N\in\mathbb{N},\forall n\geq N :\left|\frac{x^{n}}{n}-0\right|\leq|\frac{1}{n}|\leq|\frac{1}{N}|<\varepsilon$
  
  Thus this is uniform convergence and $g(x)=0$
  
  NTP: $\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}$ exists.
  
  Since $\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}=\lim_{h\to0}\frac{0-0}{h} =0$ exists, thus $g$ is differentiable
  
  $g'(x)=0$ as above 2. Show that $(g'_n)$ converges on $[0, 1]$. Is the convergence uniform? Set $h = \lim g'_n$ and compare $h$ and $g'$. Are they the same?
  
  $g_{n}^{\prime}(x)=\left(\frac{x^{n}}{n}\right)^{\prime}=\frac{nx^{n-1}}{n}=x^{n-1}$
  
  Since $\lim_{n\to\infty}x^{n-1}=0$ on $[0,1]$, then let's check uniform convergence
  
  Consider $|x^{n-1}|\leq\left|1^{n-1}\right|=1$, thus suppose it's uniform convergent, then
  
  $\forall\varepsilon>0,\exists N\in\mathbb{N},\forall n\geq N:\left|x^{n-1}\right| \leq1<\varepsilon$ which is a contradiction since $\varepsilon$ cannot be any
  
  Thus this is not uniform convergence
  
  $h=\lim g_{n}^{\prime}= \begin{cases} 1\quad\text{if }x=1 \\ 0\quad\text{if }x\in[0,1) \end{cases}$, thus they are not same

‍