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
Good Points
1. A good point should be an interior point.
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 and only
if its boundary is PeanoJordan measurable and
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 bounded
function. Then f is Riemann integrable over R if an
(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.
Exercise 117 Prove that in Denition 114, the rectangles Rn c
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. Taking the supremum
over all partitions P of R, we get
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 be as in (ii). Then by the previous theorem
A
A
A
A
g=
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 := length I1 length IN ,
where if In has endpoints an % b
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 exists " xi (x, y ) for every x ! U and y ! [a, b] and that
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
;d
;
d
;
;
f (t) dt; $
"f (t)" dt ' 2
"f (t) ' f (t0 )"
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 union of
disjoint rectangles.
Remark 136 In view of the pre
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 * RN
"g
be such that there exist " xi in E for all i, j =
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
scalars,
V V !V
R V !V
and
(v; w) 7! v + w
(t; v) 7! tv
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, regular curve # with parametric representation ! : [a, b]
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 by identifying opposite sides. Consider the chart - : (0
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, rx ) is either of the form (36) or (37).
Fix x0 ! ! U
=
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 f
# xi
i=1
and so
n
+
div (.k f ) =
nN
+
n
fi
k=1 i=1
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 ) " R : 0 # y #
, y # x # 1 ! y2
2
2
&
and since the functi
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 := ' ! Nk : 1 % '1 < '2 < < 'k % N .
0
Given a k -dimensional
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 continuous in
U and there exist the partial derivatives o
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
" : [c, d] * RN be two parametric representations. Given
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 1 if there exists a partition P = cfw_t0 , . . . , tn
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 orientation if the parameter change h : I * J is increasing
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
and x0 is an accumulation point of E L, wh
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
"
" |"|
:=
,
"1
"
"x
" x1 " x"N
N
!0f
! x0
where x = (x1 , . . . , xN
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
!
"
L := x ! RN : x = x0 + tv, t ! R .
Given two points
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 ) != (0, 0) or x + y != 0,
otherwise.
Prove that f is di!
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 exists in R. In the special case in which v = ei , the d
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 zero as (x, y ) # (0, 0). Take y = x2 . Then
"
#
"
#
"
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 f (x0 ) or
Df (x0 ). Note that part (ii) of the previo