NOTES ON HILBERT SPACES

GESTUR ÓLAFSSON

1. Hilbert spaces

A Hilbert space is a Banach space where the norm comes from an inner product (·, ·).

Definition 1. Let V be a real or complex vector space. An inner product on V is a map

V × V ↦ 𝕂

with the following properties:

1. For fixed y ∈ V the map V ∋ v ↦ (v, y) ∈ 𝕂 is linear;

2. (x, y) = \overline{(y, x)} for all x, y ∈ V;

3. For all x ∈ V we have (x, x) ≥ 0 and (x, x) = 0 if and only if x = 0.

A space V with inner product (·, ·) is called an inner product space.

Let us make few remarks on this:

Remark 1. If 𝕂 = ℝ, then (2) reads (x, y) = (y, x) and the map x ↦ (y, x) is linear. If 𝕂 = ℂ, then it follows from (2) and (1) that the map x ↦ (y, x) is conjugate linear, i.e.,

(y, λx_{1} + x_{2}) = \overline{λ} (y, x_{1}) + (y, x_{2})

Remark 2. Condition (3) says that we can define a norm on V by ||x|| := \sqrt{(x, x)} .

Lemma 1 (Cauchy-Schwarz). Let V be an inner product space. Then for all x, y ∈ V :

|(x, y)| ≤ ||x|| ||y||

and |(x, y)| = ||x|| ||y|| if and only if x and y are linearly dependent.

Proof. The claim is obvious if y = 0. So assume that y ≠ 0. Let t ∈ ℂ. Then

||x − ty||^{2} = (x − ty, x − ty)
= ||x||^{2} − t (y, x) − \overline{t (y, x)} + |t|^{2} ||y||^{2} .

Take t = (x, y) / ||y||^{2}. Then

0 ≤ ||x − ty||^{2} = ||x||^{2} − \begin{matrix}|(x, y)| & 2 \\ ||y|| & 2 \end{matrix} . (1.1)

Hence ||x||^{2} ||y||^{2} ≥ |(x, y)|^{2}. Assume that we have equality. Then (1.1) implies that ||x − ty||^{2} = 0 or ty = x.

Example 1. Let I = [a, b] a closed interval in ℝ or I = ℝ. Let C_{c}(I) be the set of continuous functions on I such that Supp(f) is compact. Let φ : I → ℝ such that φ(x) > 0 for all x ∈ I. Then

(f, g)_{φ} := ∫_{I} f(x)\overline{g(x)} φ(x) dx

defines an inner product on C_{c}(I). Here ∫_{I} f(x) dx stands for the usual Riemann integral.

Example 2. Let V = ℂ^{n}. Define an inner product on V by

((z_{1}, …, z_{n}), (y_{1}, …, y_{n})) = ∑_{j=1}^{n} z_{j}\overline{y_{j}} .

Then

||z|| = \sqrt{∑_{j=1}^{n} |z_{j}|^{2}} for z = (z_{1}, …, z_{n}) ∈ ℂ^{n} .

Definition 2. Let V be a vector space with inner product (·, ·). Then V is a Hilbert space if (V, ||·||) is complete, i.e., a Banach space.

Example 3. The space ℂ^{n} is an Hilbert space and any finite dimensional Hilbert space is isomorphic to ℂ^{n}.

Example 4. Let X ⊂ ℝ be a measurable subset. Let H = L^{2}(X) with the inner product

(f, g) = ∫_{X} f(t)\overline{g(t)} dt

More generally let φ : X → ℝ^{+} be integrable. For measurable functions f, g : X → ℝ let

(f, g)_{φ} := ∫_{X} f(t)\overline{g(t)}φ(t) dt

Let H_{φ} be the space of all measurable functions with (f, f)_{φ} < ∞. Then H_{φ} is a Hilbert space.

Example 5. Define an inner product on ℓ^{2} by

((a_{n})_{n} , (b_{n})_{n}) = ∑_{n=0}^{∞} a_{n}\overline{b_{n}}

Then ℓ^{2} becomes a Hilbert space (prove this).

We will later show that each inner product space can be isometrically embedded into a Hilbert space such that the image is dense. But first we will recall some simple facts.

Definition 3. Let (H, (·, ·)) be an inner product space:

1. The vectors u, v ∈ H are orthogonal if (u, v) = 0.

2. Let S, T ⊂ H. Then S and T are orthogonal if (s, t) = 0 for all s ∈ S and all t ∈ T.

3. A set S ⊂ H is orthogonal if for all s, t ∈S, s ≠ t we have (s, t0 = 0.

4. A set S ⊂ H is orthogonormal if S is orthogonal and ||s|| = 1 for all s ∈ S.

5. Let S ⊂ H then

S^{⊥} = {x ∈ H | ∀s ∈ S : (x, s) = 0}

6. Let S, T ⊂ H. Then

S + T = {s + t | s ∈ S, t ∈ T }

7. Assume that H is an Hilbert space. Let S and T be closed subspaces of H. Then S and T are Hilbert spaces. If S ∩ T = {0} then the sum S + T is denoted by S ⊕ T and called the direct sum of S and T. In this case each element s + t ∈ S ⊕ T has a unique expression of this form.

Lemma 2. Let H be a Hilbert space and let S ⊂ H be a closed subspace. Let u ∈ H then there exits an v ∈ S such that

||u − v|| = min_{s∈S} ||u − s|| .

Lemma 3. Let S ⊂ H. Then S^{⊥} is a closed subspace of H and (S^{⊥})^{⊥} is the closed subspace of H generated by S. If S is a closed subspace of H, then

H = S ⊕ S^{⊥} .

Proof. Let {u_{n}} be a sequence in S^{⊥} converging to u ∈ H. Let s ∈ S, then

(u, s) = lim (u_{n}, s) = 0 .

Hence u ∈ S^{⊥} and S^{⊥} is closed. Let u, v ∈ S^{⊥}, λ ∈ 𝕂 and s ∈ S. Then

(λu + v, s) = λ(u, s) + (v, s) = 0 .

Hence S^{⊥} is a subspace. We leave it as an exercise to show that cl (S) = S^{⊥⊥}. Assume now that S is closed. Then S ⊕ S^{⊥} is a subspace of H. Let u ∈ H. Let v ∈ S be as in Lemma 2. We claim that u − v ∈ S^{⊥}. Assume not, then there is an s ∈ S, ||s|| = 1, and t = (u − v, s) ≠ 0. Then

||u − v − ts||^{2} = (u − v − ts, u − v − ts)
= ||u − v||^{2} − |(u − v, s)|^{2}
< ||u − v||^{2}

contradicting our choice of v.

It follows now that u−v ∈ S^{⊥} and u = v+(u−v) ∈ S⊕S^{⊥}. Hence S⊕S^{⊥} = H.

From now on H will always stand for a Hilbert space with inner product (·, ·).

Lemma 4. Let x and y be orthogonal. Then ||x + y||^{2} = ||x||^{2} + ||y||^{2}. In particular if x and y have norm one, then ||x + y|| = \sqrt{2} .

Proof. We have ||x + y||^{2} = (x + y, x + y) = (x, x) + (y, y) because x and y are orthogonal.

Lemma 5. Assume that H is separable. Let S be a orthonormal subset of H. Then S is finite or countable infinite.

Proof. Assume that S is not finite. Let 0 < ϵ < 1. Let s, t ∈S, s ≠ t. Suppose that y ∈ B_{ϵ}(s) ∩ B_{ϵ}(t). Then

2 = ||s − t||^{2}
= ||s − y + y − t||^{2}
≤ ||s − y||^{2} + ||y − t||^{2}
< 2ϵ < 2

A contradiction. Hence the balls B_{ϵ}(s), s ∈ S, are disjoint. As H is assumed separable, we can at most have countable many disjoint open balls in H which implies that S is at most countable.

Lemma 6 (Bessel’s inequality). Let {e_{j}}_{j∈I}, I a finite or countable index set. Let x ∈ H. Then

∑_{j∈I}(x, e_{j})^{2} ≤ ||x||^{2}

Proof. Let J ⊂ I be finite. Then

0 ≤ ||x − ∑_{j∈J}(x, e_{j})e_{j}||^{2}

= ||x||^{2} − 2 ∑_{j∈J}(x, e_{j})^{2} + ∑_{i,j∈J}(x.e_{i})(x, e_{j}) (e_{i}, e_{j})

= ||x||^{2} − ∑_{j∈J}(x, e_{j})^{2}

because (x.e_{i})(x, e_{j}) (e_{i}, e_{j}) = (x.e_{i})^{2}δ_{ij}. As this holds for any finite subset of I the lemma follows.

Our aim is to find a orthonormal subset S = {e_{j}} such that each x ∈ H can be written in a unique way as

x = ∑ a_{j}e_{j}

To avoid problems we will for simplicity assume that H is separable. Before we construct S let us notice the following:

Lemma 7. Let S = {e_{j}}_{j∈J} be a orthonormal set in H. Then ||x − ∑_{j∈J} a_{j}e_{j}|| is minimal for a_{j} = (x, e_{j}).

Proof. We have

||x − ∑_{j} a_{j}e_{j}||^{2} − ||x − ∑_{j}(x, e_{j})e_{j}||^{2} = ∑_{j} (x, e_{j})^{2} − 2a_{j} (x, e_{j}) + a_{j}^{2}

= ∑_{j} ((x, e_{j}) − a_{j})
≥ 0

Hence ||x − ∑_{j} a_{j}e_{j}|| ≥ ||x − ∑_{j}(x, e_{j})e_{j}||and ||x − ∑_{j} a_{j}e_{j}|| = ||x − ∑_{j}(x, e_{j})e_{j}|| if and only if a_{j} = (x.e_{j}) for all j.

Lemma 8. Let S = {e_{j}}_{j∈J} be a orthonormal set in H. Then x − ∑_{j∈J} (x, e_{j}) e_{j} is perpendicular to S.

Proof. Let e_{k} ∈ S. Then

(x − ∑_{j∈J} (x, e_{j}) e_{j}, e_{k}) = (x, e_{k}) − ∑_{j}(x.e_{j})(e_{j}, e_{k})
= (x, e_{k}) − (x, e_{k})
= 0

It follows by the above lemma that x = ∑_{e∈S}(x, e)e for each x ∈ H if and only if S is maximal orthonormal subset of H. In that case we call S for an orthonormal basis for H. The existence of an orthonormal basis follows from the next theorem.

Theorem 1. Let T be a orthonormal subset of H. Then there exists a orthonormal basis S for H such that T ⊂ S.

Proof. Let X be the set of orthonormal subsets of H that contains T. Define an ordering on X by inclusion. By the Hausdorff Maximum Principle there exists a maximal linearly ordered subset T ⊂ X. Let

S = ⋃_{L∈T} L

Then S is a maximal orthonormal subset containing T.

Proposition 1. Let H be a separable Hilbert space. Let T be an orthonormal subset of H. Then there exists an orthonormal basis S = {e_{j}} containing T. We have for each x ∈ H:

1. x = ∑_{j}(x, e_{j})e_{j} and;

2. ||x||^{2} = ∑_{j} (x, e_{j})^{2}

The next theorem answer the question when a sum ∑_{j} b_{j}e_{j} converges in H.

Theorem 2. Let S = {e_{j}}_{j∈J} be an orthonormal set. Then ∑_{j∈J} b_{j}e_{j} converges in H if and only if ∑_{j∈J} b_{j}^{2} converges.

Proof. We can assume that J = ℕ because the statement is obviously true for J finite. Assume that ∑_{j=1}^{∞} b_{j}e_{j} converges. Then s_{n} := ∑_{j=1}^{n} b_{j}e_{j} is a Cauchy sequence in H. Let ϵ > 0 be given. Then there exists a N ∈ ℕ such that ||s_{n} − s_{m}||^{2} < ϵ for all n, m ≥ N and assume that m > n. Then

||s_{n} − s_{m}||^{2} = ∑_{j=n+1}^{m} b_{j}^{2}

= (∑_{j=1}^{n} b_{j}^{2}) − (∑_{j=1}^{m} b_{j}^{2})

It follows that {(∑_{j=1}^{n} b_{j}^{2})}_{n} is a Cauchy sequence, and hence convergent.

Assume now that {(∑_{j=1}^{n} b_{j}^{2})}_{n} is a Cauchy sequence. Then the above calculation shows that {s_{n}}_{n} is a Cauchy sequence. As H is complete it follows that lim_{n→∞} s_{n} = ∑_{j=1}^{∞} b_{j}e_{j} exists.

We define a metric on V by d(x, y) = ||x − y||. This metric defines a topology on V such that the maps

V × V ∋ (x, y) ↦ x + y ∈ V
𝕂 × V ∋ (λ, x) ↦ λx ∈ V

are continuous (prove this!). Hence V is a topological vector space. Recall that the (continuous) dual of V – denoted by V^{∗} – is the space of all linear maps f : V → 𝕂. The space V^{∗} becomes a Banach space if we define

||f|| := sup_{||x||=1} |f(x)| .

Lemma 1 implies that for a given u ∈ V the map x ↦ (x, u) is linear and satisfies |(x, u)| ≤ ||x|| ||u||. Hence it is in V^{∗}. Define ι^{∗} : V → V^{∗} and ι^{∗∗} : V → V^{∗∗} by

i^{∗}(u)(v) := (v, u)

and

ι^{∗∗}(u)(f) := f(u) .

We notice the following simple facts:

Lemma 9. Let the notation be as above. Then the following holds:

1. ι^{∗} is conjugate linear and ι^{∗∗} is linear.

2. Let u ∈ V then ||ι^{∗}(u)|| = ||u||. In particular it follows that ι^{∗} is injective.

3. Let u ∈ V then ||ι^{∗∗} (u)|| = ||u||. In particular it follows that ι^{∗∗} is injective.

4. ||ι^{∗}|| = 1. In particular it follows thatι^{∗} is continuous.

5. ||i^{∗∗}|| = 1. In particular it follows that ι^{∗} is continuous.

Proof. (1) This is clear.

(2) We have

||ι^{∗}(u)|| = sup_{||x||=1} |(x, u)| ≤ sup_{||x||=1} ||x|| ||u|| = ||u|| .

So ||ι^{∗}(u)|| ≤ ||u||. If u = 0 then there is nothing to prove as ι^{∗}(0) = 0. So assume that u ≠ 0. Let x = ||u||^{−1} u. Then ||x|| = 1 and

ι^{∗} (u) (x) = ||u||

which implies that ||ι^{∗}(u)|| ≥ ||u||. Thus ||ι^{∗} (u)|| = ||u||.

(3) We have

||ι^{∗∗}(u)|| = sup_{||f||=1} |f(u)| ≤ ||f|| ||u|| .

We may assume that u ≠ 0. As we have already proven that ι^{∗} (u) ∈ V^{∗} and ||ι^{∗}(u)|| = ||u|| we can take f = ||u||^{−1} ι^{∗}(u). Then ||f|| = 1 and

|ι^{∗∗}(u)(f)| = ||u|| .

Hence ||ι^{∗∗} (u)|| = ||u||.

(4) and (5) follows from (2) respectively (3).

Proposition 2 (Riesz representation Theorem). Let H be a separable Hilbert space. Then ι^{∗} : H → H^{∗} is bijective, i.e. let f ∈ H^{∗}, then there exists an unique u ∈ H such that for all x ∈ H we have:

f(x) = (x, u) .

Proof. Let S = {e_{j}}_{j∈J} be an orthonormal basis for H. Let f ∈ H^{∗}. To find u let us notice that if such an u exists then u = ∑ (u, e_{j}) e_{j} and f(e_{j}) = \overline{(e_{j}, u)}. Hence u = ∑ \overline{f (e_{j})} e_{j}. So we have to show that

u := ∑ \overline{f(e_{j})} e_{j}

exists in H. If H is finite dimensional there is nothing to proof, so let us assume that J = ℕ. Let n ∈ ℕ. Let u_{n} := ∑_{j=1}^{n} \overline{f(e_{j})}e_{j}. Then

||u_{n}||^{2} = ∑_{j=1}^{n} |f(e_{j})|^{2}

= f (∑_{j=1}^{n} \overline{f(e_{j})} e_{j})
= f (u_{n})
≤ ||f|| ||u_{n}||

Hence

||u_{n}|| ≤ ||f||

It follows that ∑ |f(e_{j})|^{2} exists and hence also lim u_{n} = ∑_{j=1}^{∞} \overline{f(e_{j})}e_{j}. Let x = ∑(x, e_{j})e_{j} ∈ H. Then

f(x) = f(∑(x, e_{j})e_{j})
= ∑ (x, e_{j}) f(e_{j})  f continuous and linear
= (x, ∑ \overline{f(e_{j})}e_{j})
= (x, u)

Let now V be an inner product space and ι^{∗∗} : V → V^{∗∗}. Recall that V^{∗∗} is a Banach space. Let H be the closure of ι^{∗∗} (V) in V^{∗∗}. Then H is a Banach space. Let u, v ∈ H. Let {u_{n}}_{n}, {v_{n}}_{n} be Cauchy sequence in V such that lim ι^{∗∗}(u_{n}) = u and lim ι^{∗∗}(v_{n}) = v. Then there is an M > 0 such that ||u_{m}|| , ||u_{n}|| ≤ M for all m, n and hence

|(u_{n}, v_{n}) − (u_{m}, v_{m})| = |(u_{n} − u_{m}, v_{n})| + |(u_{m}, v_{n} − v_{m})|
≤ ||u_{n} − u_{m}|| ||v_{n}|| + ||v_{n} − v_{m}|| ||u_{m}||
≤ M (||u_{n} − u_{m}|| + ||v_{n} − v_{m}||) .

Hence {(u_{n}, v_{n})}_{n} is a Cauchy sequence in ℝ. Define

(u, v) := lim_{n→∞} (u_{n}, v_{n}) .

We leave it to the reader to show that (u, v) is independent of the Cauchy sequences used in the definition and that (·, ·) is an inner product on H such that ||u||^{2} = (u, u) for all u ∈ H, where ||·|| is the norm on H coming from V^{∗∗}. As ι^{∗∗} is an isometry and Im(ι^{∗∗}) is dense in H we have actually proved the following.

Theorem 3. Let V be an inner product space with inner product (·, ·)_{V}. Then there exists a Hilbert space (H, (·, ·)_{H}) and a linear map φ : V → H such that the following holds:

1. For all u, v ∈ V we have (u, v)_{V} = (φ(u), φ(v))_{H}. In particular φ is an isometry.

2. φ (V) is dense in H.

Department of Mathematics, LSU, Baton Rouge, LA 7083

   E-mail address: olafsson@math.lsu.edu