4Hilbert spaces
II Linear Analysis
4.3 Orthonormal systems and basis
Definition
(Orthonormal system)
.
Let
E
be a Euclidean space. A set of unit
vectors {e
α
}
α∈A
is called an orthonormal system if he
α
, e
β
i = 0 if α 6= β.
We want to define a “basis” in an infinite dimensional vector space. The idea
is that these should be orthonormal systems “big enough” to span everything.
In finite-dimensions, this was easy, since there is the notion of dimension — if
we have
n
dimensions, then we just take an orthonormal system of
n
vectors,
and we are done.
If we have infinite dimensions, this is trickier. If we have many many
dimensions and vectors, we can keep adding things to our orthonormal system,
but we might never get to such a “basis”, if our “basis” has to be uncountable.
Hence we have the idea of “maximality”.
Definition
(Maximal orthonormal system)
.
Let
E
be a Euclidean space. An
orthonormal space is called maximal if it cannot be extended to a strictly larger
orthonormal system.
By Zorn’s lemma, a maximal orthonormal system always exists. We will
later see that in certain nice cases, we can construct a maximal orthonormal
system directly, without appealing to Zorn’s lemma. The advantage of an explicit
construction is that we will understand our system much more.
One important thing we would like to do is given an orthonormal system,
decide whether it is maximal. In general, this is difficult, and Zorn’s lemma is
completely useless.
Now suppose we are nicer and have a Hilbert space. What we would like to
say is that if we have a maximal orthonormal system, then its span is the whole
space
H
. However, this doesn’t really work. The span of a set
S
only allows us
to take finite linear combinations, but by completeness of
H
, we want to have
the infinite sums, i.e. the limits as well. So what we really have is the following.
Proposition.
Let
H
be a Hilbert space. Let
S
be a maximal orthonormal
system. Then span S = H.
While this might seem difficult to prove at first, it turns out the proof is
pretty short and simple.
Proof. Recall that S
⊥
= (span S)
⊥
. Since H is a Hilbert space, we have
H = span S ⊕ (span S)
⊥
= span S ⊕ S
⊥
.
Since S is maximal, S
⊥
= {0}. So done.
How about the converse? It is also true. In fact, it is true even for Euclidean
spaces, and the proof is easy.
Proposition.
Let
E
be Euclidean, and let
S
be an orthonormal system. If
span S = E, then S is maximal.
Proof.
S
⊥
= (span S)
⊥
= E
⊥
= {0}.
So in a Hilbert space, we have an if and only if condition — a system is
maximal if and only if the closure of the span is everything. In other words,
given any vector
v ∈ H
, we can find a sequence
v
i
in the span of the maximal
system that converges to
v
. This sequence is clearly not unique, since we can
just add a random term to the first item.
However, we can do something better. Consider our space
`
2
, and the element
(1
,
1
2
,
1
4
, ···
). There is a very natural way to write this as the limit of the sequence:
(1, 0, 0, ···),
1,
1
2
, 0, ···
,
1,
1
2
,
1
4
, 0, ···
, ··· .
What we are doing is that we are truncating the element at the
n
th component
for each
n
. Alternatively, the
n
th term is what we get when we project our
v
onto the space spanned by the first
n
“basis” vectors. This is a nice and natural
way to produce the sequence.
Definition
(Hilbert space basis)
.
Let
H
be a Hilbert space. A maximal or-
thonormal system is called a Hilbert space basis.
Recall that at the beginning, we said we needed Zorn’s lemma to get an
orthonormal system. In many cases, we can find a basis without using Zorn’s
lemma. This relies on the Gram-Schmidt procedure.
Proposition.
Let
{x
i
}
n
i=1
,
n ∈ N
be linearly independent. Then there exists
{e
i
}
n
i=1
such that {e
i
}
n
i=1
is an orthonormal system and
span{x
1
, ··· , x
j
} = span{e
1
, ··· , e
j
}
for all j ≤ n.
Proof. Define e
1
by
e
1
=
x
1
kx
1
k
.
Assume we have defined {e
i
}
j
i=1
orthonormal such that
span{x
1
, ··· , x
j
} = span{e
1
, ··· , e
j
}.
Then by linear independence, we know that
x
j+1
6∈ span{x
1
, ··· , x
j
} = span{e
1
, ··· , e
j
} = F
j
.
We now define
˜
x
j+1
= x
j+1
− P
F
j
(x
j+1
),
where P
F
j
is the projection onto F
j
given by
P
F
j
=
j
X
i=1
hx, e
i
ie
i
.
Since F
j
is a closed, finite subspace, we know that
x
j+1
− P
F
j
x
j+1
⊥ F
j
.
Thus
e
j+1
=
˜
x
j+1
k
˜
x
j+1
k
is the right choice. We can also write this in full as
e
j+1
=
x
j+1
−
P
j
i=1
hx
j
e
j
ie
i
kx
j+1
−
P
j
i=1
hx
j
e
j
ie
i
k
.
So done.
Note that projection into the first
n
basis is exactly what we did when we
wrote an element in `
2
as the limit of the sequence.
This is a helpful result, since it is a constructive way of producing orthonormal
systems. So if we are handed a set of vectors, we can just apply this result, plug
our vectors into this ugly formula, and get a result. Of course, we want to apply
this to infinite spaces.
Proposition.
Let
H
be separable, i.e. there is an infinite set
{y
i
}
i∈N
such that
span{y
i
} = H.
Then there exists a countable basis for span{y
i
}.
Proof.
We find a subset
{y
i
j
}
such that
span{y
i
}
=
span{y
i
j
}
and
{y
i
j
}
are
independent. This is easy to do since we can just throw away the useless
dependent stuff. At this point, we do Gram-Schmidt, and done.
Example. Consider H = `
2
and the sequence {e
i
}
i∈N
be defined by
e
i
= (0, 0, ··· , 0, 1, 0, ···),
with the zero in the ith column.
Note that
x ⊥ {e
i
}
i∈N
if and only if each component is zero, i.e.
x
=
0
. So
{e
i
} is maximal, and hence a basis.
Example. Consider H = L
2
, the completion of C(S
1
) under the L
2
norm, i.e.
hf, gi =
Z
π
−π
f ¯g dx.
Trigonometric polynomials are dense in
C
(
S
1
) with respect to the supremum
norm due to Stone-Weierstrass. So in fact
span
n
1
√
2π
e
inx
o
for
n ∈ N
is dense
in
C
(
S
1
). Hence it is dense in
C
(
S
1
) under the
L
2
norm since convergence
under the supremum norm implies convergence under
L
2
. In particular, it is
dense in the
L
2
space since
L
2
is the completion of
C
(
S
1
). Moreover, this set is
orthonormal in C(S
1
) under the L
2
norm. So
n
1
√
2π
e
inx
o
is a basis for L
2
.
Note that in these two examples, we have exhibited two different ways of
constructing a basis. In the first case, we showed that it is maximal directly. In
the second case, we show that its span is a dense subset of the space. By our
proposition, these are equivalent and valid ways of proving that it is a basis.