1
Vade Mecum for Dierentiability
To study the dierentiability of a function f : E ! R, where E
RN , the
following are good guidelines. You should begin to distinguish good points and
bad points.
1.1
G
Wednesday, March 30, 2011
Exercise 139 Let E ) RN , let x ! E , let y ! RN \ E . Prove that the segment
S joining x and y intersects # E .
Theorem 140 A bounded set E ) RN is PeanoJordan measurable if
Friday, April 01, 2011
The next theorem shows that the integral of a nonnegative function is given
by the volume of the subgraph.
Theorem 141 Let R ) RN be a rectangle and let f : R * [0, /) be a boun
(v) There are uncountable sets that have Lebesgue measure zero. One such
example is given by the Cantor set.
Exercise 116 Prove that the boundary of a rectangle R ) RN has Lebesgue
measure zero.
Exerc
Note that P ! is a renement of both P and Q. Thus, by what we just proved
in Step 1,
L (f, P ) % L (f, P ! ) % U (f, P ! ) % U (f, Q).
(24)
Hence, L (f, P ) % U (f, Q) for all partitions P and Q of R.
Wednesday, March 16, 2011
Proof. We prove that (i) implies (ii). Assume that g is a conservative vector
eld with scalar potential f : U * R, let x, y ! U and let -1 : [a, b] * RN and
-2 : [c, d] * RN
Monday, March 21, 2011
11
Integration
Given N bounded intervals I1 , . . . , IN ) R, a rectangle in RN is a set of the
form
R := I1 1 1 IN .
The elementary measure of a rectangle is given by
meas R :=
Friday, March 18, 2011
Lemma 106 (Di!erentiation under the Integral Sign) Let U ) RN be
an open set, let f : U 1 [a, b] * R be a continuous function, let i = 1, . . . , N
"f
and assume that there exis
Friday, February 25, 2011
The opposite inequality does not hold in general, but we have the following
lemma.
Lemma 85 Assume that f : [c, d] * RN is Riemann integrable. Then for
t0 ! [c, d],
;A
;A
Ad
Monday, March 28, 2011
Denition 134 A set E ) RN is called a pluri-rectangle if it can be written
as a nite union of rectangles.
Exercise 135 Prove that a pluri-rectangle can be written as a nite unio
The previous theorem unfortunately does not work for polar coordinates.
The problem is that polar coordinates are not one-to-one.
Corollary 156 Let E ) RN be a PeanoJordan measurable set, let g : E *
18
18.1
Supplements to Section 1
Normed Spaces
Denition 128 A vector space over R is a nonempty set V , whose elements
are called vectors, together with two operations, addition and multiplication by
Wednesday, April 27, 2011
Another important application is given by the areas formulas in R2 . Let
U ) R2 be an open, bounded set and assume that its boundary # U is the range
of a closed, simple, reg
Wednesday, April 13, 2011
Exercise 163 Prove that the set
$
%
M := (x, y ) ! R2 : x2 = y 2
is not a manifold.
Example 164 (Torus) A torus is a 2-dimensional surface M obtained from a
rectangle of R2 b
Friday, April 22, 2011
Proof. Step 3: We prove that in a neighborhood of every point x ! ! U on the
boundary there exists an open cube Q (x, rx ) centered at x and of side-length
rx such that U " Q (x
=
A
"U
.k f 1 dHN #1 ,
where we have used the fact that .k f is zero outside Q (xk , rxk ). Now
N
+ # (.k fi )
div (.k f ) =
i=1
# xi
N
+
=
N
fi
i=1
N
+
+ # fi
#.k
+ .k
# xi
# xi
i=1
#.k
=
fi
+ .k div
Monday, April 04, 2011
Example 147 Lets calculate the integral
!
x (1 ! y ) dxdy,
E
where
"
#
E := (x, y ) " R2 : y # x, x2 + y 2 # 1, x $ 0, y $ 0 .
We can rewrite E as follows,
$
E=
%
%
2
(x, y ) "
Friday, April 15, 2011
Carnival, no classes.
Monday, April 18, 2011
16
Surface Integrals
To dene the integral of a function over a surface, we use local charts. Given
1 % k < N we dene
$
%
#N,k := ' !
Wednesday, April 20, 2011
We are ready to prove the divergence theorem.
Theorem 175 (Divergence Theorem) Let U ) RN be an open, bounded,
regular set and let f : U * RN be such that f is bounded and co
Wednesday, March 02, 2011
Next we show that the curve integral does not depend on the particular
parametric representation.
Proposition 92 Let # be a piecewise C 1 curve and let ! : [a, b] * RN and
"
Denition 79 Given an interval I R, a function : I RN is said to be
of class C 1 in I if is dierentiable in I and is continuous.
Denition 80 Given a function : [a, b] RN , we say that f is piecewise
C
Monday, March 14, 2011
Next we introduce the notion of an oriented curve.
Denition 95 Given a curve # with parametric representations ! : I * RN
and " : J * RN , we say that ! and " have the same orie
Friday, January 21, 2011
4
Higher Order Derivatives
Let E RN , let f : E R and let x0 E . Let i cfw_1, . . . , N and assume
f
that there exists the partial derivatives xi (x) for all x E . If j cfw_1
N
A multi-index ! is a vector ! = (!1 , . . . , !N ) " (N0 ) . The length of a multiindex is dened as
|!| := !1 + + !N .
Given a multi-index !, the partial derivative
!
! x!
is dened as
"
" |"|
:=
,
"
Wednesday, January 12, 2011
2
Elementary Geometry of RN
A vector v ! RN of norm one is called a direction. Given a point x0 ! RN and
a direction v, the line through x0 in the direction v is given by
!
Wednesday, January 19, 2011
The next exercise shows that the previous conditions are su!cient but not
necessary for di"erentiability.
Exercise 17 Let
f (x, y ) :=
!"
#
1
x2 + y 2 sin x+y
0
if (x, y )
and assume that x0 is an accumulation point of the set E ! L. The directional
derivative of f at x0 in the direction v is dened as
!f
f (x0 + tv) " f (x0 )
(x0 ) := lim
,
t!0
!v
t
provided the limit e
0 as (x, y ) # (0, 0). We claim that f is not di!erentiable at (0, 0). We need to
consider to prove that the quotient
f (x, y ) ! f (0, 0) ! $f (0, 0) (x, y )
!
2
2
(x ! 0) + (y ! 0)
does not tend to
Monday, January 17, 2011
If all the partial derivatives of f at x0 exist, the vector
!
"
!f
!f
(x0 ) , . . . ,
(x0 ) ! RN
! x1
! xN
is called the gradient of f at x0 and is denoted by "f (x0 ) or grad