MA3422 (Functional Analysis 2) Tutorial sheet 1
[January 23, 2015]
Name: Solutions
1. If V is an inner product space with inner product h·, ·i, show that the Cauchy-Schwarz
inequality
|hx, yi| ≤ kxkkyk
(x, y ∈ V )
holds (for the norm arising from the inner product). [Hint: kx + λyk2 ≥ 0. If y 6= 0,
rewrite using h·, ·i and choose λ = −hx, yi/kyk2 .]
Solution: First, if y = 0 then the inequality holds because hx, yi = hx, 0i = 0hx, 0i = 0 =
kxkkyk.
If y 6= 0, consider (as per the hint)
0 ≤
=
=
=
=
kx + λyk2
hx + λy, x + λyi
hx, xi + hx, λyi + hλy, xi + hλy, λyi
hx, xi + λ̄hx, yi + λhy, xi + λλ̄hy, yi
kxk2 + λ̄hx, yi + λhx, yi + |λ|2 kyk2
Taking λ − hx, yi/kyk2 (as in the hint) we get
0 ≤ kxk2 −
|hx, yi|2 |hx, yi|2 |hx, yi|2
−
+
kyk2
kyk2
kyk2
kyk4
or
0 ≤ kxk2 −
|hx, yi|2
kyk2
which simplifies to
|hx, yi|2 ≤ kxk2 kyk2
Taking square roots we get the result.
2. If V is an inner product space, show that the triangle inequality holds.
Solution: For x, y ∈ V we have
kx + yk2 =
=
=
≤
≤
hx + y, x + yi
hx, xi + hx, yi + hy, xi + hy, yi
kxk2 + 2 Rehx, yi + kyk2
kxk2 + 2|hx, yi| + kyk2
kxk2 + 2kxkkyk + kyk2
(using Cauchy-Schwarz)
= (kxk + kyk)2
Taking square roots we get kx + yk ≤ kxk + kyk.
3. In H = L2 [−1, 1] use the Gram-Schmidt process to find an orthonormal basis for
span{1, x, x2 }.
Solution: Let ψ1 (x) = 1, ψ2 (x) = x and ψ3 (x) = x2 .
The first step in Gram-Schmidt is to take φ1 = ψ1 /kψ1 k and so we need to compute
sZ
1
p
√
kψ1 k = hψ1 , ψ1 i =
12 dx = 2.
−1
√
So φ1 (x) = 1/ 2.
For φ2 we compute ψ2 − hψ2 , φ1 i and normalise. But
Z 1
Z
ψ2 (x)φ1 (x) dx =
hψ2 , φ1 i =
1
1
x √ dx = 0
2
−1
−1
(because the integrand is odd). so φ2 is got by normalising ψ2 − hψ2 , φ1 i = ψ2 and
s sZ
r
1
3 1
p
2
x
x2 dx =
kψ2 k = hψ2 , ψ2 i =
=
.
3 −1
3
−1
Thus
r
3
x
2
For φ3 we compute ψ3 − hψ3 , φ2 i − hψ3 , φ1 i and normalise. Now
Z 1 r
3
x dx = 0
x2
hψ3 , φ2 i =
2
−1
φ2 (x) =
and
√
3 1
1
x
1
2
hψ3 , φ1 i =
x √ dx = √
=2 √ =
3
2
3 2 −1
3 2
−1
Z
1
2
Thus
ψ3 − hψ3 , φ2 i − hψ3 , φ1 i = x2 −
To compute the norm (squared) we need
Z 1
Z
2
2
(x − 1/3) dx =
−1
1
−1
5
x4 − (2/3)x2 + 1/9 dx
x
2
=
− x3 +
5
9
2 4 2
=
− + =
5 9 9
2
1
3
1
x
9 −1
8
45
√
√ √
2 2
So that the norm is 8/ 45 = √ . Thus
3 5
√
√ 5 3 2 1
3 5 2
φ3 (x) = √ (x − 1/3) = √
x −
2 2
2
2 2
So the answer is {φ1 , φ2 , φ3 }.
(The first 3 Legendre polynomials are 1, x and (3/2)x2 − (1/2). Not the same normalisation, but equal to 1 at x = 1 instead of L2 [−1, 1] norm 1.)
Richard M. Timoney
3