11.4.2 General Inner Product Spaces
Any time you have a vector space which possesses an inner product, something satisfying the properties 1 -
3 above, it is called an inner product space. As usual, F will mean the field of scalars, either ℂ or
Here is a fundamental inequality called the Cauchy Schwarz inequality which holds in any inner
product space. First here is a simple lemma.
Lemma 11.4.3 If z ∈ F there exists θ ∈ F such that θz =
Proof: Let θ = 1 if z = 0 and otherwise, let θ =
. Recall that for
= x − iy
. In case z
is real, there is no change in the above. ■
Theorem 11.4.4 (Cauchy Schwarz)Let H be an inner product space. The following inequality holds for x
and y ∈ H.
Equality holds in this inequality if and only if one vector is a multiple of the other.
Proof: Let θ ∈ F such that
= 1 and
where t ∈ ℝ
. Then from the above list of properties of the inner
and this must hold for all t ∈ ℝ
. Therefore, if
= 0 it must be the case that
= 0 also since
otherwise the above inequality would be violated. Therefore, in this case,
On the other hand, if
0 for all t
means the graph of y
is a parabola which
opens up and it either has exactly one real zero in the case its vertex touches the
axis or it has no real
zeros. From the quadratic formula this happens exactly when
which is equivalent to 11.3.
It is clear from a computation that if one vector is a scalar multiple of the other that equality
holds in 11.3. Conversely, suppose equality does hold. Then this is equivalent to saying
= 0 and so from the quadratic formula, there exists one real zero to
Call it t0.
and so x = −θt0y. This proves the theorem. ■
Note that in establishing the inequality, I only used part of the above properties of the inner product. It
was not necessary to use the one which says that if
= 0 then
Now the length of a vector can be defined.
Definition 11.4.5 Let z ∈ H. Then
Theorem 11.4.6 For length defined in Definition 11.4.5, the following hold.
Proof: The first two claims are left as exercises. To establish the third,
One defines the distance between two vectors x,y in an inner product space as
This produces a
metric in the obvious way:
Not surprisingly we have the following theorem in which F will be either ℝ or ℂ.
Theorem 11.4.7 Fn is complete. Also, if K is a nonempty closed and bounded subset of Fn, then
K is compact. Also, if f : K → ℝ, it achieves its maximum and minimum on K.
Proof: Recall Example 11.1.23 which established completeness of Fn with the funny norm
and so the Cauchy sequences and limits are exactly the same for the two norms. Thus Fn is complete where
the norm is the one just discussed.
Now suppose K is closed and bounded. By the estimate on the norms just given, it is closed
and bounded with respect to
also because a point is a limit point with respect to one
norm if and only if it is a limit point with respect to the other. Now if B
also where this symbol denotes the ball taken with respect to
It suffices to verify sequential compactness thanks
to Theorem 11.1.38
it follows that Rexin
and so, taking 2
subsequences, there exists a subsequence still denoted with n
= ai ∈
for each i
. Hence xn → a
+ ib ≡ x
. Now, since K
closed, it follows that x ∈ K
and this shows sequential compactness which is equivalent to
The last claim is as follows. Let M ≡ sup
be a maximizing sequence so that
By compactness, there is a subsequence xnk → x ∈ K.
Then by continuity,
The existence of the minimum is similar. ■