0 ≤ kx − αtyk2 = kxk2− 2 Re(hx, αtyi) + t2kyk2
= kxk2− 2t Re(α hx, yi) + t2kyk2
= kxk2− 2t |hx, yi| + t2kyk2.
This is a real-valued quadratic polynomial in the variable t. Since this poly-nomial is nonnegative, it can have at most one real root. This requires that the discriminant of the polynomial be at most zero, so we must have
¡−2 |hx, yi|¢2− 4 kxk2kyk2 ≤ 0.
The desired result follows by rearranging this inequality. ⊓⊔
By combining the Polar Identity with the Cauchy–Bunyakovski–Schwarz Inequality, we can now prove that the induced seminorm k · k satisfies the Triangle Inequality.
Lemma 3.4. Let H be a vector space. If h·, ·i is a semi-inner product on H, then k · k is a seminorm on H, and if h·, ·i is an inner product on H, then k · k is a norm on H.
Proof. The only property of a seminorm that is not obvious is the Trian-gle Inequality. To prove this, we use the Polar Identity and the Cauchy–
Bunyakovski–Schwarz Inequality to compute that kx + yk2 = kxk2+ 2 Rehx, yi + kyk2
≤ kxk2+ 2 |hx, yi| + kyk2
≤ kxk2+ 2 kxk kyk + kyk2
= ¡kxk + kyk¢2.
The Triangle Inequality follows by taking square roots. ⊓⊔
3.3 Hilbert Spaces
The question of whether every Cauchy sequence in a given inner product space must converge is very important, just as it is in a metric or normed space. We give the following name to those inner product spaces that have this property.
Definition 3.5 (Hilbert Space). An inner product space H is called a Hilbert space if it is complete with respect to the induced norm. ♦
Thus, an inner product space is a Hilbert space if and only if every Cauchy sequence in H converges to an element of H. Equivalently, a Hilbert space is
an inner product space that is a Banach space with respect to the induced norm.
Using real scalars, Rd is a Hilbert space with respect to the usual dot product. Likewise, Cd is a Hilbert space with respect to the dot product if we use complex scalars.
Here is an example of an infinite-dimensional Hilbert space (compare this to Example 1.2, which introduced the space ℓ1).
Example 3.6. Given a sequence of real or complex scalars x = (xk)k∈N = (x1, x2, . . . ), we define the ℓ2-norm of x to be
kxk2 = k(xk)k∈Nk2 = µ∞
X
k=1
|xk|2
¶1/2
. (3.2)
Note that we have not yet proved that k · k2 is a norm in any sense; we will address this issue below.
We say that a sequence x = (xk)k∈Nis square summable if kxk2< ∞, and we let ℓ2 denote the space of all square summable sequences, i.e.,
ℓ2 =
½
x = (xk)k∈N : kxk22 =
∞
X
k=1
|xk|2 < ∞
¾ .
We will define an inner product on ℓ2. First we recall that the arithmetic-geometric mean inequality implies that if a, b ≥ 0, then
ab ≤ a2 2 +b2
2.
Consequently, if we choose any two vectors x = (xk)k∈N and y = (yk)k∈N in ℓ2, then
∞
X
k=1
|xkyk| ≤
∞
X
k=1
µ |xk|2
2 +|yk|2 2
¶
= kxk22
2 + kyk22
2 < ∞. (3.3) Therefore we can define
hx, yi =
∞
X
k=1
xkyk, x, y ∈ ℓ2, (3.4)
because equation (3.3) tells us that this series of scalars converges absolutely.
In particular, for x = y we have
3.3 Hilbert Spaces 39
hx, xi =
∞
X
k=1
|xk|2 = kxk22. (3.5)
We can check that the function h·, ·i satisfies all of the requirements of an inner product. For example, simply by definition we have 0 ≤ hx, xi < ∞ for every x ∈ ℓ2. Further, if hx, xi = 0 then it follows from equation (3.5) that xk = 0 for every k, so x = 0. This establishes the nonnegativity and uniqueness requirements of an inner product, and the conjugate symmetry and linearity in the first variable requirements are easily checked as well.
Thus h·, ·i is an inner product on ℓ2, and therefore ℓ2 is an inner product space with respect to this inner product. Further, equation (3.5) shows that the norm induced from this inner product is precisely the ℓ2-norm k · k2, so Lemma 3.4 implies that k · k2 really is a norm on ℓ2.
Is ℓ2 a Hilbert space? This is slightly more difficult to check, but an argu-ment very similar to the one used in Theorem 1.9 shows that every Cauchy sequence in ℓ2converges to an element of ℓ2(we assign the details as Problem 3.29). Consequently ℓ2is complete, so it is a Hilbert space. ♦
The space ℓ1 is a proper subspace of ℓ2. For, if x = (xk)k∈N belongs to ℓ1 then we must have |xk| < 1 for all large enough k, say k > N, and therefore
∞
X
k=N +1
|xk|2 ≤
∞
X
k=N +1
|xk| ≤ kxk1 < ∞.
Hence x ∈ ℓ2. On the other hand, the sequence x = ¡1
k
¢
k∈N = ¡1,12,13, . . .¢
belongs to ℓ2, but it does not belong to ℓ1. Now, since ℓ1 is a subspace of ℓ2, we can define an inner product on ℓ1 simply by restricting equation (3.4) to vectors in ℓ1. However, the norm induced from this inner product is k · k2
rather than k · k1. Although ℓ1is complete with respect to the norm k · k1, it is not complete with respect to the norm k · k2 (see Problem 3.30). Therefore ℓ1 is an inner product space with respect to the inner product defined in equation (3.4), but it is not a Hilbert space.
Here is a different example of an inner product space.
Example 3.7. Let C[a, b] be the space of all continuous functions of the form f : [a, b] → C. All functions in C[a, b] are bounded and Riemann integrable.
If we choose f, g ∈ C[a, b], then the product f (x) g(x) is continuous and Riemann integrable, so we can define
hf, gi = Z b
a
f (x) g(x) dx, f, g ∈ C[a, b]. (3.6)
We can easily see that h·, ·i satisfies the nonnegativity, conjugate symmetry, and linearity in the first variable requirements stated in Definition 3.1, and hence is at least a semi-inner product on C[a, b]. According to Problem 3.28, the uniqueness requirement also holds, and therefore h·, ·i is an inner product on C[a, b]. The norm induced from this inner product is
kf k2 = hf, f i1/2 = µZ b
a
|f (x)|2dx
¶1/2
, f ∈ C[a, b].
We call kf k2 the L2-norm of the function f. ♦
Although we have defined an inner product on C[a, b], we will prove that C[a, b] is not complete with respect to the norm k · k2 that is induced from that inner product (in contrast, note that Theorem 2.16 implies that C[a, b]
is complete with respect to a different norm—the uniform norm k · ku).
Lemma 3.8. C[a, b] is not a Hilbert space with respect to h·, ·i.
Proof. For simplicity of presentation we will take a = −1 and b = 1. As illustrated in Figure 3.1 for the case n = 5, define
fn(x) =
−1, −1 ≤ x ≤ −n1, linear, −1n < x < n1, 1, n1 ≤ x ≤ 1;
f5
-1.0 -0.5 0.5 1.0
-1 -0.5 0.5 1
Fig. 3.1 Graph of the function f5.
Given positive integers m < n we compute that
kfm− fnk22 = Z 1
−1
|fm(x) − fn(x)|2dx
= Z 1/m
−1/m
|g(x) − fn(x)|2dx
≤ Z 1/m
−1/m
1 dx = 2 m.
3.3 Hilbert Spaces 41
Therefore, if we fix ε > 0 then for all m, n large enough we will have kfm− fnk2< ε. This shows that {fn}n∈Nis a Cauchy sequence in C[−1, 1].
However, we will prove that there is no function g ∈ C[−1, 1] such that kg − fnk2→ 0 as n → ∞.
Suppose that there was some g ∈ C[−1, 1] such that kg − fnk2 → 0. Fix 0 < c < 1, and suppose that h is not identically 1 on [c, 1]. Then
C = Z 1
c
|g(x) − 1| dx > 0.
On the other hand, fn is identically 1 on [c, 1] for all n > 1/c, so for all large enough n we have
kg − fnk22 = Z 1
−1
|g(x) − fn(x)|2dx
≥ Z 1
c
|g(x) − fn(x)|2dx
= Z 1
c
|g(x) − 1| dx ≥ C.
Since C > 0, it follows that kg−fnk2→/ 0, which is a contradiction. Therefore g must be identically 1 on [c, 1]. This is true for every c > 1, so g(x) = 1 for all 0 < x ≤ 1. A similar argument shows that g(x) = −1 for all −1 ≤ x < 0.
There is no continuous function that takes these values, so we have obtained a contradiction.
Thus, although {fn}n∈N is Cauchy in C[−1, 1], there is no function in C[−1, 1] that this sequence can converge to with respect to the induced norm k · k2. Therefore C[−1, 1] is not complete with respect to this norm. ⊓⊔
Each incomplete normed space that we encountered prior to C[a, b] was contained in some larger complete space. For example, c00is incomplete with respect to the norm k · k1, but it is a subset of ℓ1, which is complete with respect to k · k1. Likewise, Cc(R) is incomplete with respect to the uniform norm, yet it is contained in C0(R), which is a Banach space with respect to that norm. It is likewise true that the inner product space C[a, b] is con-tained in a larger Hilbert space, the Lebesgue space L2[a, b]. However, defining L2[a, b] is beyond the scope of this manuscript, as it requires the theory of the Lebesgue integral. We refer to texts such as [Fol99], [SS05], or [WZ77] for discussion of the Lebesgue integral and the Hilbert space L2[a, b].