Page 66 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 5: Spaces, Hölder

Notes
b b
p
q
|f| dx |g| dx
1 1
= a b a b
p q
p
q
|f| dx |g| dx
a a
1 1
= 1 .
p q
b
Hence |f(x) g(x)|dx 1.
a
Putting the values of f (x) and g (x), we get

b
|f(x) g(x)|dx
1 or fg f g … (4)
f g p q
a p q
q
Now f L [a, b], g L [a, b]
p
b b
q
p
|f| dx and |g| dx
a a
f and g
p q
Therefore, from (4), we have

fg < f L [a, b]
1 g
Also the equality will hold when A = B
p
q
i.e. |f (x)| = |g (x)| , a.e.
|f| p |g| q
i.e. if p = q , a.e.
f g
p q
q p
p
q
or if g |f| = f |g| , a.e.
q p
or if we have got some non-zero constants ,
p
q
|f| = |g| , a.e.
Hence the theorem.
5.1.7 Riesz-Hölder's Inequality for 0 < p < 1

If 0 < p < 1 and p and q are conjugate exponents, and f L and g L , then
q
p
|fg| f p g q , provided |g| q 0 .

(In this case, the inequality is reversed than that of the case for 1 p < .)
1 1
Proof: Conjugacy of p, q 1
p q

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