620-295 Real Analysis with Applications, Lecture 21

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 16 February 2010

Derivatives, the Mean Value Theorem and the Intermediate Value Theorem

The function $f$ is differentiable at   $x=c$ if $$\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\text{ exists.}$$

The derivative of $f$ at $x=c$ is $$f\text{'}\left(c\right)=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c},$$ if $f$ is differentiable at $x=c.$

If $f:\left[a,b\right]\to ℝ$ and $g:\left[a,b\right]\to ℝ$ are functions and $l\in ℝ$ define

$\begin{array}{rcl}\left(f+g\right):\left[a,b\right]\to ℝ& \text{by}& \left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right),\\ \left(lf\right):\left[a,b\right]\to ℝ& \text{by}& \left(lf\right)\left(x\right)=lf\left(x\right)\\ fg:\left[a,b\right]\to ℝ& \text{by}& \left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\end{array}$

If $c\in \left[a,b\right]$ and $f\text{'}\left(c\right)$ exists and $g\text{'}\left(c\right)$ exists then

$\begin{array}{rcll}\text{(a)}& \left(f+g\right)\text{'}\left(c\right)& =& f\text{'}\left(c\right)+g\text{'}\left(c\right)\\ \text{(b)}& \left(lf\right)\text{'}\left(c\right)& =& lf\text{'}\left(c\right)\\ \text{(c)}& \left(fg\right)\text{'}\left(c\right)& =& f\text{'}\left(c\right)g\left(c\right)+f\left(c\right)g\text{'}\left(c\right).\end{array}$

		Proof (a)
	To show: $\left(f+g\right)\text{'}\left(c\right)=f\text{'}\left(c\right)+g\text{'}\left(c\right).$ $\begin{array}{rcl}\left(f+g\right)\text{'}\left(c\right)& =& \underset{x\to c}{\lim }\frac{\left(f+g\right)\left(x\right)-\left(f+g\right)\left(c\right)}{x-c}\\ & =& \underset{x\to c}{\lim }\frac{f\left(x\right)+g\left(x\right)-\left(f\left(c\right)+g\left(c\right)\right)}{x-c}\\ & =& \underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)+g\left(x\right)-g\left(c\right)}{x-c}\\ & =& \underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}+\frac{g\left(x\right)-g\left(c\right)}{x-c}\\ & =& \underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}+\underset{x\to c}{\lim }\frac{g\left(x\right)-g\left(c\right)}{x-c}\\ & =& f\text{'}\left(c\right)+g\text{'}\left(c\right),\end{array}$ if you believe that $\underset{x\to c}{\lim }A\left(x\right)+B\left(x\right)=\underset{x\to c}{\lim }A\left(x\right)+\underset{x\to c}{\lim }B\left(x\right).$$\square$

Let $A:\left[a,b\right]\to ℝ$ and $B:\left[a,b\right]\to ℝ$ and assume that $\underset{x\to c}{\lim }A\left(x\right)$ exists and $\underset{x\to c}{\lim }B\left(x\right)$ exists. Then $$\underset{x\to c}{\lim }A\left(x\right)+B\left(x\right)$$ exists and $$\underset{x\to c}{\lim }A\left(x\right)+B\left(x\right)=\underset{x\to c}{\lim }A\left(x\right)+\underset{x\to c}{\lim }B\left(x\right).$$

		Proof.
	Let $l_1=\underset{x\to c}{\lim }A\left(x\right)$ and $l_2=\underset{x\to c}{\lim }B\left(x\right).$ To show: $\underset{x\to c}{\lim }A\left(x\right)+B\left(x\right)=l_1+l_2.$ To show: If $\epsilon \in {ℝ}_{\ge 0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $x\in B_{\delta }\left(c\right)$ then $\left(A\left(x\right)+B\left(x\right)-\left(l_1+l_2\right)\right)<\epsilon .$ Assume $\epsilon \in {ℝ}_{>0}.$ We know: there exists ${\delta }_1\in {ℝ}_{>0}$ such that if $x\in B_{{\delta }_1}\left(c\right)$ then $\left(A\left(x\right)-l_1\right)<\frac{\epsilon }{2}.$ We know: there exists ${\delta }_2\in {ℝ}_{>0}$ such that if $x\in B_{{\delta }_2}\left(c\right)$ then $\left(A\left(x\right)-l_2\right)<\frac{\epsilon }{2}.$ Let $\delta =\min \left({\delta }_1,{\delta }_2\right).$ To show: $\left(A\left(x\right)+B\left(x\right)-\left(l_1+l_2\right)\right)<\epsilon .$ $\begin{array}{rcl}\left(A\left(x\right)+B\left(x\right)-\left(l_1+l_2\right)\right)& =& \left(\left(A\left(x\right)-l_1\right)+\left(B\left(x\right)-l_2\right)\right)\\ & \le & \left(A\left(x\right)-l_1\right)+\left(B\left(x\right)-l_2\right)\\ & \le & \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .\end{array}$ So $\underset{x\to c}{\lim }A\left(x\right)+B\left(x\right)=\underset{x\to c}{\lim }A\left(x\right)+\underset{x\to c}{\lim }B\left(x\right).$$\square$

(Intermediate Value Theorem) If $f:\left[a,b\right]\to ℝ$ is continuous and $w$ is between $f\left(a\right)$ and $f\left(b\right)$ then there exists $c\in \left(a,b\right)$ such that $f\left(c\right)=w.$

(Mean Value Theorem) If $f:\left[a,b\right]\to ℝ$ is continuous for $x\in \left[a,b\right]$ and differentiable for $x\in \left(a,b\right)$then there exists $c\in \left(a,b\right)$ such that $$f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$$

Functions $f:\left[a,b\right]\to ℝ$

Let $a,b\in ℝ.$ Let $c\in \left[a,b\right].$

The function $f$ is continuous at $x=c$ if $f$ satisfies:

if $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $x\in B_{\delta }\left(c\right)$ then $f\left(x\right)\in B_{\epsilon }\left(f\left(c\right)\right),$

where $B_{\delta }\left(c\right)=\left\{x\in \left[a,b\right]|\left|x-c\right|<\delta \right\}=\left(c-\delta ,c+\delta \right).$

The function $f$ is continuous at $x=c$ if $f$ satisfies:

if $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that $B_{\delta }\left(c\right)\subseteq f^{-1}\left(B_{\epsilon }\left(f\left(c\right)\right)\right).$

The function $f$ is continuous at $x=c$ if $f$ satisfies:

if $\epsilon \in {ℝ}_{>0}$ and $V=B_{\epsilon }\left(f\left(c\right)\right)$ then $c$ is an interior point of $f^{-1}\left(V\right).$

A set $E\subseteq \left[a,b\right]$ is connected if there do not exist open sets $A,B\subseteq \left[a,b\right]$ with

1. $A\ne \varnothing$ and $B\ne \varnothing$
2. $A\cup B=\left[a,b\right]$
3. A ∩B=∅.

If $f:X\to Y$ is continuous and $X$ is connected then $f\left(X\right)$ is connected.

		Proof
	Proof by contradiction. Assume $f\left(X\right)$ is not connected. Let $A,B$ be open in $f\left(X\right)$ such that $A\cup B=f\left(X\right)$ and $A\cap B=\varnothing$ Then let $$C=f^{-1}\left(A\right)$$ and $D=f^{-1}\left(B\right).$ Then $\begin{array}{rcl}C\cup D& =& f^{-1}\left(A\right)\cup f^{-1}\left(B\right)=f^{-1}\left(A\cup B\right)=f^{-1}\left(f\left(X\right)\right)=X.\\ C\cap D& =& f^{-1}\left(A\right)\cap f^{-1}\left(B\right)=f^{-1}\left(A\cap B\right)=f^{-1}\left(\varnothing \right)=\varnothing .\end{array}$ $C\ne \varnothing$ since $A\ne \varnothing$ and $A\subseteq f\left(X\right).$ $D\ne \varnothing$ since $B\ne \varnothing$ and $B\subseteq f\left(X\right).$ So $X$ is not connected. Contradiction. So $f\left(X\right)$ is connected. $\square$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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