MA3422 (Functional Analysis 2) Tutorial sheet 2
[January 30, 2015]
Name: Solutions
1. In L 2 [0 , 1] take f ( t ) = t and compute the orthogonal projection of f onto span { 1 , e
2 πit
, e
4 πit
}
Solution: If we write φ n
( t ) = e
2 πint for n = 0 , 1 , 2 , then there are orthonormal in and so the orthogonal projection of f onto their span is
L
2
[0 , 1] h f, φ
0 i φ
0
+ h f, φ
1 i φ
1
+ h f, φ
2 i φ
2
We need to compute these inner products h f, φ
1 i =
Z
1
0 h f, φ
0 i =
Z
1 t 1 dt = [ t
2
/ 2]
1
0
0
=
1
2 te
− 2 πit dt =
Z
1
0 t d e
− 2 πit
− 2 πi
= t e
− 2 πit
− 2 πi
1
0
−
Z
1 e
− 2 πit
0
− 2 πi
1 dt = −
2 πi h f, φ
2 i =
Z
1 te
− 4 πit dt =
Z
1
0 0 t d e
− 2 πit
− 4 πi
= t e
− 4 πit
− 4 πi
1
0
−
Z
1 e
− 4 πit
0
− 4 πi
1 dt = −
4 πi
So the projection is
1
2
−
1
2 πi e
2 πit −
1
4 πi e
4 πit
=
1
2
+ i
1
2 π e
2 πit
+
1
4 π e
4 πit
2. In L 2 [0 , 1] take r n
( t ) = sgn(sin 2 n +1 πt ) (where sgn( x ) = x/ | x | if x = 0 and sgn(0) = 0 ).
[These are usually called Rademacher functions . The value when sin 2 n +1
πt = 0 is not really important.]
(a) Plot r
0
, r
1 and r
2
.
Solution: (In these graph there should be a point at (0 , 0) and another at (1 , 0) and there should be no vertical lines — there should be points on the x -axis instead. The endpoints of the horizontal liens at y = ± 1 should be excluded.)

1.0
0.5
0.2
0.4
0.6
0.8
1.0
0.5
1.0
1.0
0.5
0.2
0.4
0.6
0.8
1.0
0.5
1.0
1.0
0.5
0.2
0.4
0.6
0.8
1.0
0.5
1.0
(b) Show that { t ∈ [0 , 1] : r n
( t ) = 1 } has measure 1 / 2 .
Solution: r n
( t ) has frequency 2 n , or period 1 / 2 n . Between 0 and 1 / 2 n , sin(2 n +1 is positive for 0 < t < 1 / 2 n +1 and negative for 1 / 2 n +1
πt )
< t < 1 / 2 n
. So r n
( t ) = 1 on the left half, r n
( t ) = − 1 on the right half (zero at the midpoint and the endpoints).
This pattern repeats 2 n times. So
{ t ∈ j − 1
,
2 n j
2 n
: r n
( t ) = 1 } is an interval of length 1 / 2 n +1 for each j = 1 , 2 , . . . , 2 n . Taking the union gives total length 1 / 2 .
(c) Show that h r
0
, r
1 i = 0 .
Solution: On each interval where r
0
( t ) is constant, r
1
( t ) = 1 for the first half and r
1
( t ) = − 1 for the second half, giving integral of the product = 0 .
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Aside: It is common think of [0 , 1] with length measure as a probability space and then sometimes to use probabilistic language to describe things. The Rademacher functions are then random variables taking the values ± 1 (almost surely — ignoring the set of measure zero, or of probability zero, where they are 0). The fact that they take the values 1 and − 1 with probability 1 / 2 for each makes them identically distributed random variables. The argument that h r
0
, r
1 i = 0 can actually be souped up a little to show that h r n
, r m i = 0 if n = m (and h r n
, r n i = 1 always also). So the
Rademacher functions are othonormal, or in probabilistic terminology independent identically distributed random variables of mean 0 and variance 1. As they take only the values ± 1 they are often viewed as random ± 1 signs and used then for places where averages are required over all such signs.
If x
1
, x
2
, . . . , x n are vectors in a normed space or a Banach space, then
Z
1
0 n
X r j
( t ) x j dt j =1 is the same as the average of n
X
σ j x j j =1 over all 2 n choices of signs σ
1
, σ
2
, . . . , σ n
∈ { 1 , − from the parallelogram identity that this average is
1 } . In a Hilbert space, it follows v n u u
X t k x j k 2 j =1 but in other normed spaces the answer would often not be that. The biggest or smallest possible ratio between the average and that is used as a way to detect differences between normed spaces.
Richard M. Timoney
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