13.1 The p Norms
In addition to
mentioned above, it is common to consider the so
for x ∈ ℂn.
Definition 13.1.1 Let x ∈ ℂn. Then define for p ≥ 1,
The following inequality is called Holder’s inequality.
Proposition 13.1.2 For x,y ∈ ℂn,
The proof will depend on the following lemma.
Lemma 13.1.3 If a,b ≥ 0 and p′ is defined by
Proof of the Proposition: If x or y equals the zero vector there is nothing to prove.
Therefore, assume they are both nonzero. Let A =
Then using Lemma 13.1.3
Theorem 13.1.4 The p norms do indeed satisfy the axioms of a norm.
Proof: It is obvious that
does indeed satisfy most of the norm axioms. The only
one that is not clear is the triangle inequality. To save notation write
in place of
in what follows. Note also that
Then using the Holder inequality,
so dividing by
It only remains to prove Lemma 13.1.3.
Proof of the lemma: Let p′ = q to save on notation and consider the following
Note equality occurs when ap = bq.
Alternate proof of the lemma: For a,b ≥ 0, let b be fixed and
If b = 0, it is clear that f
0 for all a.
Then assume b >
It is clear since p >
This is negative for small a and then eventually is positive. Consider the minimum value of f
which must occur at a > 0 thanks to the observation that the function is initially strictly
decreasing. At this point,
and so ap = bq at the point where this function has a minimum. Thus at this value of
0 for all a ≥
0 and this proves the inequality. Equality occurs when ap
may be considered as the operator norm of A
taken with respect to
case when p
this is just the spectral norm. There is an easy estimate for
in terms of the
entries of A.
Theorem 13.1.5 The following holds.
1 and let A
are the columns of A.
and so by Holder’s inequality,