21.3. UNIFORM CONVEXITY OF Lp 547

Proof: If∫| f g|dµ = ∞, there is nothing to prove. Hence assume this is finite. Then∫

| f |p dµ =∫|g|−p | f g|p dµ

This makes sense because, due to the hypothesis on g it must be the case that g equals 0only on a set of measure zero, since p/(p−1)< 0.

Then by the usual Holder inequality, one of the exponents being 1/p > 1, the otherbeing 1/(1− p) also larger than 1 with p+(1− p) = 1,

∫| f |p dµ ≤

(∫| f g|dµ

)p(∫ ( 1

|g|p)1/(1−p)

dµ

)1−p

=

(∫| f g|dµ

)p(∫|g|p/p−1 dµ

)1−p

Now divide by(∫|g|p/p−1 dµ

)1−pand then take the pth root. ■

Here is the backwards Minkowski inequality. It looks just like the ordinary Minkowskiinequality except the inequality is turned around.

Corollary 21.3.3 Let 0 < p < 1 and suppose∫|h|p dµ < ∞ for h = f ,g. Then(∫

(| f |+ |g|)p dµ

)1/p

≥(∫| f |p dµ

)1/p

+

(∫|g|p dµ

)1/p

Proof: If∫(| f |+ |g|)p dµ = 0 then there is nothing to prove since this implies | f | =

|g|= 0 a.e. so assume this is not zero.∫(| f |+ |g|)p dµ =

∫(| f |+ |g|)p−1 (| f |+ |g|)dµ

Since p < 1, (| f |+ |g|)p ≤ | f |p + |g|p and so∫ ((| f |+ |g|)p−1

)p/p−1dµ < ∞.

Hence the backward Holder inequality applies and it follows that∫(| f |+ |g|)p dµ =

∫(| f |+ |g|)p−1 | f |dµ +

∫(| f |+ |g|)p−1 |g|dµ

≥(∫ (

(| f |+ |g|)p−1)p/p−1

dµ

)(p−1)/p[(∫

| f |p dµ

)1/p

+

(∫|g|p dµ

)1/p]

=

(∫(| f |+ |g|)p dµ

)(p−1)/p[(∫

| f |p dµ

)1/p

+

(∫|g|p dµ

)1/p]

and so, dividing gives the desired inequality. ■Consider the “easy” Clarkson inequalities.