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Inner product spaces
Definition 1. A vector space (over R) is a set V , whose elements are referred to as vectors, together with operations of vector addition and scalar
multiplication– that is, given u, v ∈ V and cR there exist vectors u + v ∈ V
and cv ∈ V which we refer to as the sum of u and v and a scalar multiple of
v– which satisfy the following properties for all u, v, w ∈ V and c, d ∈ R:
(i) Associativity of addition:
(u + v) + w = u + (v + w),
(ii) Commutativity of addition:
u + v = v + u,
(iii) Identity element of addition: there exists a vector 0 ∈ V which we call
the zero vector with the property that v + 0 = v for all v ∈ V ,
(iv) Inverse elements of addition: for each v ∈ V there exists a vector which
we write as −v such that v + (−v) = 0,
(v) Compatibility of scalar multiplication with field multiplication:
(cd)v = c(dv),
(vi) Identity element of scalar multiplication: The scalar 1 ∈ R satisfies 1v = v
for all v ∈ V ,
(vii) Distributivity of scalar multiplication with respect to vector addition:
c(u + v) = cu + cv,
(viii) Distributivity of scalar multiplication with respect to field addition:
(c + d)u = cu + du.
Definition 2. An inner product space is a vector space V together with an
operation called an inner product which assigns to a pair of vectors v, w ∈ V
a scalar denoted hv, wi ∈ R such that this pairing satisfies the following three
properties: for all u, v, w ∈ V and c ∈ R we have
(i) symmetry: hu, vi = hv, ui,
(ii) linearity: hu + cv, wi = hu, wi + chv, wi,
(iii) non-degeneracy: hv, vi ≥ 0 with equality if and only if v = 0.
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Example 1. The Euclidean space Rm equipped with the usual dot product is
an inner product space.
Example 2. Let I be an interval. Let V be set of functions f : I → R. For
f, g ∈ V define a new function f +g called their sum by (f +g)(x) = f (x)+g(x),
and given c ∈ R define another new function cf called the scalar multiple by
(cf )(x) = cf (x). The set V together with these operations of addition and
scalar multiplication is a vector space. To prove this, one must check that the
8 properties in the definition above are satisfied.
Example 3. Let I be a closed interval. The set of continuous functions f :
I → R, which we often denote C(I), is a vector space as well. Consider the
following pairing of functions f, g ∈ C(I) which yields a real number:
Z
hf, gi = f (x)g(x) dx
I
The space C(I) together with this pairing is an inner product space. To prove
this, one must check that C(I) is a vector space and that the pairing defined
above satisfies the properties of symmetry, linearity, and non-degeneracy. We
show that this is the case here. Let f, g, h ∈ C(I) and c ∈ R. Observe that
Z
Z
hf, gi = f (x)g(x) dx = g(x)f (x) dx = hf, gi.
I
I
This proves symmetry. Next, observe that
Z
hf + cg, hi = (f + cg)(x)h(x) dx
ZI
= (f (x) + cg(x))h(x) dx
ZI
= (f (x)h(x) + cg(x)h(x)) dx
ZI
Z
= f (x)h(x) dx + c g(x)h(x) dx
I
I
= hf, hi + chg, hi.
This proves linearity. Finally, observe that
Z
hf, f i = f (x)2 dx ≥ 0
I
where the inequality follows because the integrand is non-negative. It is also
clear that the integral of a non-negative function is only 0 when the function
is the constant 0 function. This shows equality holds above only when f is the
0 function. Thus non-degeneracy is proved.
Remark 1. A closed interval I was used in the example above (as opposed to
say an open interval) so as to guarantee that all f ∈ C(I) are bounded, which
in turn guaranteed that the integral in the pairing defined was always finite.
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One great reason to give C(I) the structure of an inner product space is
that it allows us to prove inequalities like the one below, which otherwise would
be quite difficult to prove.
Proposition 1 (Cauchy-Schwarz inequality in C(I)). Let I be a closed interval
and let f, g ∈ C(I).
sZ
sZ
Z
f (x)2 dx
f (x)g(x) dx ≤
I
I
g(x)2 dx.
(1)
I
Proof. Our proof of the Cauchy-Schwarz inequality in Rm depended only on
the threeR generic properties of an inner product. As such, since the pairing
hf, gi = I f (x)g(x) dx satisfies these properties, the same proof holds.