Uniformliy Continuous

$f: S\rightarrow\mathbb{R}$

Defnition 1. The function $f$ is said to be continuous on $S$ $\iff$ $$\mathrm{\forall }{x}_0\in S,\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0suchthat\mathrm{\forall }x\in S,|x–x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon$$

Defnition 2. The function $f$ is said to be uniformly continuous on $S$ $\iff$ $$\mathrm{\forall }{x}_1,x_2\in S,\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0suchthat|x_1–x_2|<\delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$$

Example $f(x)=x^2$ is continuous but not uniformly continuous on $(0,\infty)$

Let $a=x_0 +1$ and $\delta=min(1, \varepsilon /2a)$

Choose $x\in S$

Assume $|x-x_0|<\delta$. Then $|x-x_0|<1$ so $x<x_0+1=a$ so $x,x_0<a$ so

$$|x^2-{x_0}^2|=(x+x_0)|x-x_0|\le 2a|x-x_0|<2a\delta \le 2a\frac{\epsilon }{2a}=\epsilon$$

Therefore $f$ is continuous on $(0,\infty)$

Let $\varepsilon=1$. Choose $\delta >0$. Let $x_1 = 1/\delta$ and $x_2=x_1 + \delta/2$.

Then $|x_2 -x_1|=\delta/2<\delta$ but

$$|{x_1}^2-{x_2}^2|=|(\frac{1}{\delta }{)}^2-(\frac{1}{\delta }+\frac{\delta }{2}{)}^2|<1+\frac{{\delta }^2}{4}>1=\epsilon$$

Therefore $f$ is not uniformly continuous on $(0,\infty)$

Theorem

If $x_1, x_2 \in S$ , $f$ satisfied an inequality of form $|f(x_1)-f(x_2)|\leq M|x_1 -x_2| \cdots(1)$, then $f$ is uniformly continuous on $S$

 (1) is called a Lipschitz ineqality and M is called the corresponding Lipschitz constant.

Theorem

If $f^{\prime}$ is bounded, then $f$ satisfied Lipschitz ineqality.

p324 exercises

86. Suppose that $f$ is continuous and nonnegative over $[a,b]$, as in the accompanying figure.

P is the partition of $[a,b]$

$$a=x_0,x_1,x_2,\cdots ,x_{k-1},x_k,\cdots ,x_{n-1},x_n=b$$ 

as shown, divide $[a,b]$ into $n$ subintervals of length $\Delta x_k=x_k-x_{k-1}$

a. If $m_k=$min$\{f(x)$ for $x$ in the $k$th subinterval$\}$, explain the connection between the lower sum.

$$L=m_1\mathrm{\Delta }{x}_1+m_2\mathrm{\Delta }{x}_2+\cdots +m_n\mathrm{\Delta }{x}_n$$

and the shaded regions in the first part of the figure.

b. If $M_k=$max$\{f(x)$ for $x$ in the $k$th subinterval$\}$, explain the connection between the upper sum.

$$U=M_1\mathrm{\Delta }{x}_1+M_2\mathrm{\Delta }{x}_2+\cdots +M_n\mathrm{\Delta }{x}_n$$

and the shaded regions in the second part of the figure.

c. Explane the connection between U-L and the shaded regions along the curve in the third part of the figure.

 

a. L and lower sum are same.

b. U and upper sum are same.

c. Given $\varepsilon >0$ 

Let $d=max(M_k-m_k)$

Because $f$ is continuous on $[a,b]$

We can choose points such that $\Delta x_k\leq d<\varepsilon$

$$U-L=(M_1-m_1)\mathrm{\Delta }{x}_1+(M_2-m_2)\mathrm{\Delta }{x}_2+\cdots +(M_n-m_n)\mathrm{\Delta }{x}_n$$

$$\le d\mathrm{\Delta }{x}_1+d\mathrm{\Delta }{x}_2+\cdots +d\mathrm{\Delta }{x}_n=d\sum _{k=1}^n\mathrm{\Delta }{x}_k=d(b-a)<\epsilon (b-a)$$

$$\therefore \underset{||p||\to 0}{lim}(U-L)=0$$

$f$ is integrable over $[a,b]$

87. Show that if $f$ is continuous on $[a,b]$ then $f$ is uniformly continuous on $[a,b]$

Given $\varepsilon >0$, let $x_1 =a$ 

We Choose $\delta>0$ such that $\forall x_2 , x_2-a <\delta \Rightarrow |f(x_1)-f(x_2)|<\varepsilon $  (Because $f$ is continuous at $x=a$)

Similary let $x_1 =b$

We Choose $\delta>0$ such that $\forall x_2 , b-x_2 <\delta \Rightarrow |f(x_1)-f(x_2)|<\varepsilon $  (Because $f$ is continuous at $x=b$)

and $x_1 \in (a,b)$

We Choose $\delta>0$ such that $\forall x_2 , |x_1-x_2| <\delta \Rightarrow |f(x_1)-f(x_2)|<\varepsilon $  (Because $f$ is continuous at $x=x_1$)