Page 11 - Vector Analysis

§1.3 Normed Vector Spaces 7

we find that for 1 ă p ă 8,

ˇˇ(xr, yr)ˇˇ ˇ n xk yk ˇ n |xk| |yk| n ( 1 |xk|p 1 |yk|p1 )
ˇ }x}p }y}p1 ˇ }x}p }y}p1 p 1 }y}pp11
ˇ ÿ ˇ ÿ ÿ

= k=1 ď k=1 ď k=1 p }x}pp +

= 1 n 1 n = }x}pp + }y}p1 =1
p}x}pp p 1}y}pp11 p}x}pp p 1}y}pp11
ÿ |xk|p + ÿ |yk|p1

k=1 k=1

which conclude the case for 1 ă p ă 8. The proof for the case that p = 1 or p = 8 is

trivial, and is left to the reader. ˝

Corollary 1.24 (Minkowski inequality). Let 1 ď p ď 8. Then

}x + y}p ď }x}p + }y}p @ x, y P Fn .

Proof. We only prove the case that 1 ă p ă 8. First we note that

}x y}pp n yk |p n yk|p´1(|xk| )
|yk|
ÿ ÿ

+ = |xk + ď |xk + +

k=1 k=1

nn

= ÿ |xk + yk|p´1|xk| + ÿ |xk + yk|p´1|yk| .

k=1 k=1

( )( )
Let u = |x1|, |x2|, ¨ ¨ ¨ , |xn| and v = |x1 +y1|p´1, |x2 +y2|p´1, ¨ ¨ ¨ , |xn +yn|p´1 . By Hölder’s

inequality,

n (n )1
ÿ |xk + yk|p´1|xk| = (u, v) ď }u}p}v}p1 = }x}p ÿ |xk + yk|(p´1)p1 p1

k=1 k=1

(n ) p´1
= }x}p ÿ |xk + yk|p p = }x}p}x + y}pp´1 .

k=1

n

Similarly, we have ř |xk + yk|p´1|yk| ď }y}p}x + y}pp´1; thus

k=1

}x + y}pp ď ( + ) + y}pp´1
}x}p }y}p }x

which concludes the Minkowski inequality. ˝

Theorem 1.25. Let 1 ď p ď 8, and p1 be the conjugate of p; that is, 1 + 1 = 1. Then
p p1

}x}p = sup ˇˇ(x, y)ˇˇ @ x P Fn .

}y}p1 =1