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
ℝ.
Here is a fundamental inequality called the Cauchy Schwarz inequality which holds in any inner
product space. First here is a simple lemma.
Lemma 11.4.3If z ∈ F there exists θ ∈ F such that θz =
|z|
and
|θ|
= 1.
Proof: Let θ = 1 if z = 0 and otherwise, let θ =
z
|z|
. Recall that for z = x + iy,z = x − iy and
zz =
|z|
^{2}. 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 xand y ∈ H.
1∕2 1∕2
|(x,y)| ≤ (x,x) (y,y) (11.3)
(11.3)
Equality holds in this inequality if and only if one vector is a multiple of the other.
Proof: Let θ ∈ F such that
|θ|
= 1 and
θ(x,y) = |(x,y)|
Consider p
(t)
≡
(x + θty, x+ tθy)
where t ∈ ℝ. Then from the above list of properties of the inner
product,
= 0 also since
otherwise the above inequality would be violated. Therefore, in this case,
|(x,y)| ≤ (x,x)1∕2(y,y)1∕2.
On the other hand, if
(y,y)
≠0, then p
(t)
≥ 0 for all t means the graph of y = p
(t)
is a parabola which
opens up and it either has exactly one real zero in the case its vertex touches the t axis or it has no real
zeros. From the quadratic formula this happens exactly when
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
4
|(x,y)|
^{2}− 4
(x,x)
(y,y)
= 0 and so from the quadratic formula, there exists one real zero to p
(t)
= 0.
Call it t_{0}. Then
- - -
p(t0) ≡ (x+ θt0y,x+ t0θy) = ||x+ θty||2 = 0
and so x = −θt_{0}y. 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
(x,x)
= 0 then x = 0.
Now the length of a vector can be defined.
Definition 11.4.5Let z ∈ H. Then
|z|
≡
(z,z)
^{1∕2}.
Theorem 11.4.6For length defined in Definition 11.4.5, the following hold.
|z| ≥ 0 and |z| = 0 if and only if z = 0 (11.5)
(11.5)
If α is a scalar, |αz| = |α||z| (11.6)
(11.6)
|z + w| ≤ |z|+ |w |. (11.7)
(11.7)
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
|x − y |
. This produces a
metric in the obvious way:
d(x,y) ≡ |x− y |
Not surprisingly we have the following theorem in which F will be either ℝ or ℂ.
Theorem 11.4.7F^{n}is complete. Also, if K is a nonempty closed and bounded subset of F^{n}, thenK 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 F^{n} with the funny norm
||x||∞ ≡ max {|xi|,i = 1,2,⋅⋅⋅,n}
However,
√1-|x| ≤ ||x|| ≤ |x|
n ∞
and so the Cauchy sequences and limits are exactly the same for the two norms. Thus F^{n} 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
(0,r)
⊇ K, then
B_{∞}
(0,r)
⊇ K also where this symbol denotes the ball taken with respect to
||⋅||
_{∞} rather than
|⋅|
. Hence K ⊆∏_{j=1}^{n}
([− r,r]+ [− ir,ir])
. It suffices to verify sequential compactness thanks
to Theorem 11.1.38. Letting
{xn}
⊆ K, it follows that Rex_{i}^{n} is in
[− r,r]
and Imx_{i}^{n} is in
[− r,r]
and so, taking 2n subsequences, there exists a subsequence still denoted with n such that
lim_{n→∞}Rex_{i}^{n} = a_{i}∈
[− r,r]
,lim_{n→∞}Imx_{i}^{n} = b_{i} for each i. Hence x^{n}→ a + ib ≡ x. Now, since K is
closed, it follows that x ∈ K and this shows sequential compactness which is equivalent to
compactness.
The last claim is as follows. Let M ≡ sup
{f (x) : x ∈ K }
and let x_{n} be a maximizing sequence so that
M = lim_{n→∞}f
(xn)
. By compactness, there is a subsequence x_{nk}→ x ∈ K. Then by continuity,
M = lim_{k→∞}f