Problem 1. By CauchyShwartz inequality
1 1

0
f (t) sin tdt
0
f (t) sin tdt
1 1
f (t)2 dt
0 0
sint2 dt
1 = 2
1
f (t)2 dt .
0
Also, equality holds if and only if f (t) = C sin t. Indeed, consider functions f and g. Then we have (f + tg, f + tg
Lecture 2: Jordans theorem and the structure of functions of bounded
variation
Todays lecture will mostly follow Carothers. I will cover Lemma 13.2,Lemma 13.3,
Theorem 13.4, Theorem 13.5, Corollary 13.6 and Proposition 13.12. I include in this le
some ext
Solutions to Problem set 2
14.5. Let = x and let f (x) = 1 for x Q and 1 for x Q. Then clearly f is not
/
integrable since the dierence between the inf and sup of f in any interval is 2, but f  is
a constant function.
14.8 I assume [x] means the oor of
19.1 First let
fj,k = j[ k1 , k ] ,
j2
j2
with 0 k < j 2 . Now let fn consist of the fj,k s ordered so that function with smaller j
precede functions with larger j and the sequence increases k with j xed. This sequence
diverges pointwise for points in [0,
Problem set 5 solutions
17.11 The function G is Borel measurable since the sets cfw_G > are Borel sets.
When 0 these are the emptyset. When 0 < 1, these are Gc which is closed
and when > 1 these are the whole of [0, 1]. The function G is not Riemann inte
16.8 Let J1 , . . . , Jn , . . . be a covering of E by intervals of any length. We will replace
this covering by a covering by intervals of length less that . If Jl has length smaller than
, we keep it. If Jl has length greater than , we let k be the sma
Solutions to Problem set 1
b
13.5. Part (i) of Lemma: If Va f = 0 then so is V (f, P ) for the partition P = a, x, b.
Then f (x) = f (a) = f (b). But since f is arbitrary, this means f is constant. On the
other hand, if f is constant, then V (f, P ) = 0 f
Solutions to Problem set 3
15.3. By integration by parts we have
1
and
1
f (x) sin nxdx =
f (x) cos nxdx =
1
n
cos nxdf (x),
1
n
sin nxdf (x),
where the boundary terms cancel because f ( ) = f ( ). The right hand sides give a
1
bound of n V f .
15.5 a) S
Problem 1. It is easy to see that the tangent vector to the curve (r(t), (t) in polar coordi nates has components r and r. So, the length of a curve is
b
l=
a
(r(t)2 + (r(t)(t)2 dt.
Using this formula we compute length of the curve (2 sin t, t):
1
l=
0
MA108 Homework 2
Seung Woo Shin January 25th 2009
1
x First, note that by Theorem 13.9, v (x) = Va f is right continuous on [a, b]. Fix > 0 and let P = a = x0 < x1 < x2 < . . . < xn = b. Then, obviously there is y > 0 such that Vx < /2n whenever x y  <
Problem 2. ikx ikx ik Let g(x) = f (x + ). If f (x) = fk then g(x) = e fk = ke ke ikx ik e fk , where fk = e fk . If f (x) = k (ak sin kx+bk cos kx) then by the above k argument we get g(x) = k (k sin kx + k cos kx), where ak = ak cos k bk sin k a b and k
Lecture 1: Introduction to integration theory and bounded variation
What is this course about? Integration theory. The rst question you might have is
why there is anything you need to learn about integration.
You took a class called Math 1a and learned so