8/13/2019 Real Analysis - Berkeley - Class style

http://slidepdf.com/reader/full/real-analysis-berkeley-class-style 1/2

MATH H104 LECTURE 23, NOVEMBER 15, 2005

LECTURER: YUVAL PERES.

SCRIBE: BRIAN SHOTWELL

We move on to your 2nd (or 3rd) exposure to Calculus.

Definition 0.1.   The  derivative of a function  f (x)  is denoted by  f (x), which is defined by 

f (x) = limh→0

f (x + h)− f (x)

h  = L.

(provided the limit exists). Recall that the limit  L  exists if for all   > 0, then there exists a δ > 0  such that if  0 <| h |< δ   then  |   f (x+h)−f (x)

h  − L |< .

We state the following two properties of derivatives of functions, whose proofs can befound in Pugh:

(f  + g)(x) = f (x) + g(x).

(f g)(x) = f (x)g(x) + f (x)g(x).

Theorem 0.2.   The   Chain Rule  states that if we have a function  f   : (a, b)  →  R  which is differentiable at   x  ∈   (a, b), and another function   g   : (c, d)  →   R   which is differentiable at 

y = f (x) ∈ c, d, then  ψ =  g ◦ f  is differentiable at  x  and  ψ

(x) = g

(y)f 

(x).Proof.   ψ(z ) = g(f (z )) for all  z  ∈ (a, b). We are given  > 0. Before we start finding  δs, let’sfirst look at  ψ(x + h) − ψ(x) = g(f (x + h)) − g(f (x)):

ψ(x + h)− ψ(x)

h  =

 g(f (x + h)) − g(f (x))

h  =

 g(y + hR) − g(y)

h  =

 f (x + h)− f (x)

h  .

(aside): For  1  >  0 chosen later, there is a  δ > 0 such that  | h | δ  =⇒| R − f (x) |< 1.

Assuming that the limit of products is the product of limits, if f (x) = 0, then [ g(y+hR)−g(y)hR

  ]R →g(y)f (x) (as h → 0).

If  f (x) = 0, then given  1   choose  α > 0 so that

0 <| β   |< α =⇒|  g(y + β ) − g(y)β    − g(y) |< 1.

| hR |< α =⇒| g(y + hR)− g(y) − hRg(y) |≤ e1hR

=⇒| ψ(x + h) − ψ(x)

h  −Rg(y) |≤ 1R ≤ /2.

Choose 1 = /2 and ensure δ < α small enough so that | h |< δ . Thus | h |< δ  =⇒| R |< 1,and therefore  | Rg(y) |< /2.  

Date : December 14, 2005.

1

8/13/2019 Real Analysis - Berkeley - Class style

http://slidepdf.com/reader/full/real-analysis-berkeley-class-style 2/2

2 Yuval Peres

Theorem 0.3.   The   Mean Value Theorem  says that if  f   : [a, b] → R  is continuous, and 

if  f  is differentiable on  (a, b), then there exists a  θ ∈ (a, b)  with  f (θ) =   f (b)−f (a)b−a

  .

Proof.   Let  g(x) = f (x) − cx, where  c =   f (b)−f (a)

b−

a

  .   g(b) − g(a) = f (b) − f (a) − c(b − a) = 0.We have the folloing three cases:

Case 1:max g ∈ [a, b] > g(a)

Then find  θ ∈ (a, b) with  g(θ) =  max g ∈ [a, b].   g(x + h) ≤ g(θ) for all small  h.

limh→0

g(θ + h) − g(θ)

h  =

  g(θ) ≤ 0, using  h > 0g(θ) ≥ 0, using  h < 0

Case 2: Case 1 fails but  min g ∈ [a, b] < g(a).Case 3: Both 1 and 2 fail.   g(x)  ≡   g(a) such that for all  x  ∈   [a, b),   g = 0  ∈   (a, b).   f 

differentiable at  x =⇒ f   is continuous at x.  

We will now note a few properties concerning the relationship between continuity anddifferentiability. First, although differentiability implie continuity, continuity does not implydifferentiability (look at the function  f (x) =| x |  at  x = 0.

Now, suppose f   : R → R is differentiable in  R. Must f  be continuous? NO. For example,look at the function  f  defined as follows:

f (x) =

  0, if   x = 0x2sin(1/x), if   x = 0

For  x = 0, f (x) = 2xsin(1/x)− cos(1/x).   limf (x) as  x → 0 does not exist.

University of California, Berkeley

E-mail address :  peres@stat.berkeley.edu