15.1. BASIC INEQUALITIES AND PROPERTIES 401

Lemma 15.1.5 Suppose x,y ∈ C. Then

|x+ y|p ≤ 2p−1 (|x|p + |y|p) .

Proof: The function f (t) = t p is concave up for t ≥ 0 because p > 1. Therefore, thesecant line joining two points on the graph of this function must lie above the graph of thefunction. This is illustrated in the following picture.

|x| |y|m

(|x|+ |y|)/2 = m

Now as shown above, (|x|+ |y|

2

)p

≤ |x|p + |y|p

2

which implies|x+ y|p ≤ (|x|+ |y|)p ≤ 2p−1 (|x|p + |y|p)

and this proves the lemma.Note that if y = φ (x) is any function for which the graph of φ is concave up, you could

get a similar inequality by the same argument.

Corollary 15.1.6 (Minkowski inequality) Let 1≤ p < ∞. Then(∫| f +g|p dµ

)1/p

≤(∫| f |p dµ

)1/p

+

(∫|g|p dµ

)1/p

. (15.1.2)

Proof: If p = 1, this is obvious because it is just the triangle inequality. Let p > 1.Without loss of generality, assume(∫

| f |p dµ

)1/p

+

(∫|g|p dµ

)1/p

< ∞

and (∫| f +g|p dµ)1/p ̸= 0 or there is nothing to prove. Therefore, using the above lemma,∫

| f +g|pdµ ≤ 2p−1(∫| f |p + |g|pdµ

)< ∞.

Now | f (ω)+g(ω)|p ≤ | f (ω)+g(ω)|p−1 (| f (ω)|+ |g(ω)|). Also, it follows from thedefinition of p and q that p−1 = p

q . Therefore, using this and Holder’s inequality,∫| f +g|pdµ ≤