37.1. EMBEDDING THEOREMS FOR W m,p (Rn) 1277

Lemma 37.1.8 Suppose n≥ 2 and w j does not depend on the jth component of x, x j. Then

∫Rn

n

∏j=1

∣∣w j (x)∣∣dmn ≤

n

∏i=1

(∫Rn−1

∣∣w j (x)∣∣n−1 dmn−1

)1/(n−1)

.

In this inequality, assume all the functions are continuous so there can be no measurabilityquestions.

Proof: First note that for n = 2 the inequality reduces to the statement

∫ ∫|w1 (x2)| |w2 (x1)|dx1dx2 ≤

∫|w1 (x2)|dx2

∫|w2 (x1)|dx1

which is obviously true. Suppose then that the inequality is valid for some n. Using Fubini’stheorem, Holder’s inequality, and the induction hypothesis,

∫Rn+1

n+1

∏j=1

∣∣w j (x)∣∣dmn+1

=∫R

∫Rn|wn+1 (x)|

n

∏j=1

∣∣w j (x)∣∣dmndxn+1

=∫Rn|wn+1 (x)|

∫R

n

∏j=1

∣∣w j (x)∣∣dxn+1dmn

=∫Rn|wn+1 (x)|

(n

∏j=1

∫R

∣∣w j (x)∣∣n dxn+1

)1/n

dmn

=∫Rn|wn+1 (x)|

n

∏j=1

(∫R

∣∣w j (x)∣∣n dxn+1

)1/n

dmn

≤(∫

Rn|wn+1 (x)|n dmn

)1/n

·∫Rn

(n

∏j=1

(∫R

∣∣w j (x)∣∣n dxn+1

)1/n)n/(n−1)

dmn

(n−1)/n

=

(∫Rn|wn+1 (x)|n dmn

)1/n

·(∫Rn

n

∏j=1

(∫R

∣∣w j (x)∣∣n dxn+1

)1/(n−1)

dmn

)(n−1)/n