From Euclidean to Hilbert Spaces. Edoardo Provenzi
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[1.2]
In
where v = (v1, v2, . . . , vn), w = (w1, w2, . . . , wn) ∈
The symbol
THEOREM 1.1.– Let (V, 〈 , 〉) be an inner product space. We have:
1) 〈v, 0V 〉 = 0 ∀v ∈ V ;
2) if 〈u, w〉 = 〈v, w〉 ∀w ∈ V , then u and v must coincide;
3) 〈v, w〉 = 0 ∀v ∈ V
PROOF.–
1) 〈v, 0V 〉 = 〈v, 0V + 0V 〉 = 〈v, 0V 〉 + 〈v, 0V 〉 by linearity, i.e. 〈v, 0V 〉 − 〈v, 0V 〉 = 0 = 〈v, 0V 〉.
2) 〈u, w〉 = 〈v, w〉 ∀w ∈ V implies, by linearity, that 〈u − v, w〉 = 0 ∀w ∈ V and thus, notably, considering w = u − v, we obtain 〈u − v, u − v〉 = 0, implying, due to the definite positiveness of the inner product, that u − v = 0V , i.e. u = v.
3) If w = 0V , then 〈v, w〉 = 0 ∀v ∈ V using property (1). Inversely, by hypothesis, it holds that 〈v, w〉 = 0 = 〈v, 0V 〉 ∀v ∈ V , but then property (2) implies that w = 0V .
Finally, let us consider a typical property of the complex inner product, which results directly from a property of complex numbers.
THEOREM 1.2.– Let (V, 〈 , 〉) be a complex inner product space. Thus:
PROOF.– Consider any complex number z = a + ib, so −iz = b − ia, hence b = ℑ (z) = ℜ (−iz). Taking z = 〈v, w〉, we obtain ℑ (〈v, w〉) = ℜ (−i〈v, w〉) = ℜ (〈v, iw〉) by sesquilinearity.
1.2. The norm associated with an inner product and normed vector spaces
If (V, 〈, 〉) is an inner product space over
Note that ‖v‖ is well defined since 〈v, v〉 ≽ 0 ∀v ∈ V . Once a norm has been established, it is always possible to define a distance between two vectors v, w in V : d(v, w) = ‖v − w‖.
The vector v ∈ V such that ‖v‖= 1 is known as a unit vector. Every vector v ∈ V can be normalized to produce a unit vector, simply by dividing it by its norm.
NOTABLE EXAMPLES.–
Three properties of the norm, which should already be known, are listed below. Taking any v, w ∈ V , and any α ∈
1) ‖v‖≽ 0, ‖v‖= 0
2) ‖αv‖= |α|‖v‖(homogeneity);
3) ‖v + w‖≼ ‖v‖+ ‖w‖(triangle inequality).
DEFINITION 1.4 (normed vector space).– A normed vector space is a pair (V, ‖ ‖) given by a vector space V and a function, called a norm,
A norm ‖ ‖ is Hilbertian if there exists an inner product 〈 , 〉 on V such that
Canonically, an inner product space is therefore a normed vector space. Counterexamples can be used to show that the reverse is not generally true.
Note that, by definition, 〈v, v〉 = ‖v‖ ‖v‖, but, in general, the magnitude of the inner product between two different vectors is dominated by the product of their norms. This is the result of the well-known inequality shown below.
THEOREM 1.3 (Cauchy-Schwarz inequality).– For all v, w ∈ (V, 〈 , 〉) we have:
PROOF.– Dozens of proofs of the Cauchy-Schwarz inequality have been produced. One of the most elegant proofs is shown below, followed by the simplest one:
– first proof: if w = 0V , then the inequality is verified trivially with 0 = 0. If w ≠ 0V , then we can define