Download MATH 645 - HOMEWORK 1: Exercise 1.1.3. Show that

MATH 645 - HOMEWORK 1:
JEREMY WEST
Exercise 1.1.3. Show that the space H in Example 1.8 is a Hilbert space.
Proof. Since sums and scalar multiples of absolutely continuous functions are absolutely continuous, H is clearly a vector space. Furthermore, we have previously
verfied that the given inner product is a valid inner product on L2 (0, 1). We need
only show that H is complete with respect to the norm derived from the inner
product
Z 1
hf, gi =
f 0 (t)g 0 (t)dt.
0
Let fn ∈ H for n ∈ N be a Cauchy sequence. Since it is Cauchy with respect
to the inner product on L2 (0, 1), the sequence fn0 is Cauchy in L2 (0, 1). Since
L2 is complete, there exists a limit fn → f 0 ∈ L2 (0, 1) (where we have used f 0 for
notational convenience, not because we are claiming it is the derivative of anything,
yet). Define
Z x
f (x) =
f 0 (t)dt.
0
Note that since (0,1) is a finite measure space, f 0 ∈ L2 implies f 0 ∈ L1 so that such
a function is well-defined. Furthermore, from a result of measure theory we know
that f (x) defined in such a manner is absolutely continuous and the fundamental
d
theorem of Calculus guarantees that dx
f (x) = f 0 . Since fn0 → f 0 and H shares a
2
metric with L we see that fn → f in H as desired. Thus H is a Hilbert space. Exercise 1.1.6. Let u be a semi-inner product on X and put N = {x ∈ X :
u(x, x) = 0}.
(a) Show that N is a linear subspace of X .
(b) Show that if
hx + N , y + N i ≡ u(x, y)
for all x + N and y + N in the quotient space X /N , then h·, ·i is a welldefined inner product on X /N .
Proof.
(a) Let x, y ∈ N and a1 , a2 ∈ F be given. Then by Cauchy-Schwarz,
|u(a1 x + a2 y, a1 x + a2 y)|
≤ u(a1 x, a1 x)u(a2 y, a2 y)
= |a1 |2 u(x, x)|a2 |2 u(y, y)
=
0.
Hence N is a linear subspace of X . Note further that the Cauchy-Schwarz
inequality and this same argument implies that u(x, y) = u(y, x) = 0 for
x ∈ N and any y ∈ X .
1
2
JEREMY WEST
(b) To show h·, ·i is well-defined we take x1 , x2 ∈ x + N and y1 , y2 ∈ y + N and
show that
u(x1 , y1 )
=
u(x2 + n1 , y2 + n2 ) where n1 , n2 ∈ N
=
u(x2 , y2 ) + u(n1 , y2 ) + u(x2 , n2 ) + u(n1 , n2 )
= u(x2 , y2 )
which shows that hx + N , y + N i is well-defined. To show it is an inner
product we need,
hx + N , x + N i = 0 → x + N = 0 + N .
To see this note that if
hx + N , x + N i = u(x, x) = 0
we have x ∈ N so that x + N = N .
Exercise 1.1.8. If G = {z ∈ C : 0 < |z| < 1} show that every f ∈ L2a (G) has a
removable singularity at z = 0.
Proof. We must show that limz→0 zf (z) = 0 for all f ∈ L2a (G). By corollary 1.12,
for |z| < 1/2 we have that
|f (z)| ≤
so that
1
2
√ kf k2 = √ kf k2
|z|/2 π
|z| π
2
|zf (z)| = |z||f (z)| ≤ √ kf k2 .
π
To be continued...
Exercise 1.2.1. Let H be a Hilbert space and suppose f and g are linearly independent vectors in H with kf k = kgk = 1. Show that ktf + (1 − t)gk < 1 for
0 < t < 1. What does this say about {h ∈ H : khk ≤ 1}.
Proof. By the proof of the triangle inequality and the result on equality from the
Cauchy-Schwarz inequality we see that
ktf + (1 − t)gk ≤ |t|kf k + |(1 − t)|kgk = 1
where equality occurs if and only if some nontrivial linear combination of f and g
is zero. Since these vectors are linearly independent this is impossible, hence
ktf + (1 − t)gk < 1
for 0 < t < 1. This shows that {h ∈ H : khk ≤ 1} is a linear subspace of H.
Exercise 1.2.2. If M ≤ H and P + PM , show that I − P is the orthogonal
projection of H onto M⊥ .
Proof. Let h ∈ H be given. We need only show that h − (I − P )h = P h ⊥ M⊥ .
However, since P h ∈ sM by definition, hP h, mi = 0 for all m ∈ M⊥ , thus I − P
satisfies the definition.
Exercise 1.2.4. If A ⊂ H, let ∨A ≡ the intersection of all closed linear subspaces
of H that contain A. ∨A is called the closed linear span of A. Prove the following:
MATH 645 - HOMEWORK 1:
3
(a) ∨A ≤ H and ∨A is the smallest closed linear subspace of H that contains
A.
Pn
(b) ∨A = the closure of { k=1 αk fk : n ≥ 1, αk ∈ F, fk ∈ A}.
(a) Since the intersection of closed spaces is closed we have that ∨A is
closed. Furthermore, for a1 , a2 ∈ F and x1 , x2 ∈ ∨A we see that a1 x1 +a2 x2
is an element of every closed linear subspace containing A and hence is in
∨A. There, ∨A ≤ H. That it is the smallest one containing A follows
immediatelyP
from the definition.
n
(b) Call B = { k=1 αk fk : n ≥ 1, αk ∈ F, fk ∈ A}. Clearly B ⊂ ∨A since
∨A is a linear space containing A and B consists of linear combinations
of A. Since ∨A is closed it therefore also contains the closure of B, hence
B ⊂ ∨A.
For the other direction, it is clear to see that B is a linear manifold of
H. It’s closure is therefore a closed linear subspace of H. Since A ⊂ B we
see that ∨A ⊂ B since ∨A is the smallest closed linear subspace containing
A.
Proof.
Exercise 1.2.5. Prove Corollary 2.10: If A ⊂ H, then (A⊥ )⊥ is the closed linear
span of A in H.
Proof. Note that if A ⊂ B then B ⊥ ≤ A⊥ since if f ⊥ B, clearly f ⊥ A. Therefore,
since A ⊂ ∨A we have that (∨A)⊥ ≤ A⊥ and hence (A⊥ )⊥ ≤ ((∨A)⊥ )⊥ = ∨A by
Corollary 2.9. Since it is clear that A ⊂ (A⊥ )⊥ and we know that (A⊥ )⊥ is a linear
subspace, we conclude that (A⊥ )⊥ = ∨A, the smallest linear subspace containing
A.
Exercise 1.2.6. Prove Corollary 2.11: If Y is a linear manifold in H, then Y is
dense in H iff Y ⊥ = (0).
Proof. Suppose Y ⊥ = (0). Then (Y ⊥ )⊥ = H = ∨Y by Corollary 2.10. Since Y is a
linear space, we see that ∨Y = Y = H so that Y is dense in H.
Now suppose Y is dense in H. Note that by the definition of the inner product,
the function fy (x) = hy, xi is a continuous function on H. Let y ∈ Y ⊥ be given.
Since Y is dense, there exists a sequence xn ∈ Y with xn → y. Then, hy, xn i = 0
since y ∈ Y ⊥ . But, by continuity we have,
0 = lim hy, xn i = hy, yi
n→∞
which implies that y = 0 so that Y ⊥ = (0).