Page 73 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes
f = sup |fg|d , 1/p + 1/q = 1
p
g 1
q
where it is easy to see that the right-hand side satisfies the triangular inequality.
Like Holder's inequality, the Minkowski inequality can be specialized to sequences and vectors
by using the counting measure:

1/p 1/p 1/p
n n n
|x y | p |x | p |y | p
k k k k
k 1 k 1 k 1
for all real (or complex) numbers x , ..., x , y , ..., y and where n is the cardinality of S (the
1 n 1 n
number of elements in S).
Thus, we may conclude that
If p > 1, then Minkowski's integral inequality states that

1/p 1/p 1/p
b b b
p
p
p
|f(x) g(x)| dx |f(x)| dx |g(x)| dx
a a a
Similarly, if p >1 and a , b > 0, then Minkowski's sum inequality states that
k k
n 1/p n 1/p n 1/p
|a b | p |a | p |b | p
k k k k
k 1 k 1 k 1
Equality holds iff the sequences a , a , ... and b , b , ... are proportional.
1 2 1 2
6.1.1 Proof of Minkowski Inequality Theorems

p
p
Theorem 1: State and prove Minkowski inequality. If f and g L (1 p < ), then f + g L and
f + g f + g .
p p p
or
p
p
Let 1 p . Prove that for every pair f, g L {0, 1}, the function f + g L {0, 1} and that
f + g f + g . When does equality occur?
p p p
Or
Suppose 1 p . Prove that for any two functions f and g in L [a, b]
p
1 1 1
b p b p b p
p
p
|f g| p |f| dx |g| dx
a a a
Proof: When p = 1, the desired result is obvious.

If p = , then
|f| f a.e.
|g| g a.e.

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