Foundations of Data Science

In vector form, |x||y| ≥ x T y, the inequality states that the dot product of two vectors
is at most the product of their lengths. The Cauchy-Schwartz inequality is a special case
of Hölder’s inequality with p = q = 2.
Young’s inequality
For positive real numbers p and q where 1 p + 1 q
= 1 and positive reals x and y,
1
p xp + 1 q yq ≥ xy.
The left hand side of Young’s inequality, 1 p xp + 1 q yq , is a convex combination of x p and y q
since 1 and 1 sum to 1. ln(x) is a concave function for x > 0 and so the ln of the convex
p q
combination of the two elements is greater than or equal to the convex combination of
the ln of the two elements
ln( 1 p xp + 1 q yp ) ≥ 1 p ln(xp ) + 1 q ln(yq ) = ln(xy).
Since for x ≥ 0, ln x is a monotone increasing function, 1 p xp + 1 q yq ≥ xy..
Hölder’s inequalityHölder’s inequality
For positive real numbers p and q with 1 p + 1 q = 1,
n∑
|x i y i | ≤
i=1
( n∑
i=1
|x i | p ) 1/p ( n∑
i=1
) 1/q
|y i | q .
Let x ′ i = x i / ( ∑ n
i=1 |x i| p ) 1/p and y i ′ = y i / ( ∑ n
i=1 |y i| q ) 1/q . Replacing x i by x ′ i and y i by
y i ′ does not change the inequality. Now ∑ n
i=1 |x′ i| p = ∑ n
i=1 |y′ i| q = 1, so it suffices to prove
∑ n
i=1 |x′ iy ′ i| ≤ 1. Apply Young’s inequality to get |x ′ iy ′ i| ≤ |x′ i |p
right hand side sums to 1 p + 1 q
= 1 finishing the proof.
For a 1 , a 2 , . . . , a n real and k a positive integer,
+ |y′ i |q
p q
(a 1 + a 2 + · · · + a n ) k ≤ n k−1 (|a 1 | k + |a 2 | k + · · · + |a n | k ).
Using Hölder’s inequality with p = k and q = k/(k − 1),
|a 1 + a 2 + · · · + a n | ≤ |a 1 · 1| + |a 2 · 1| + · · · + |a n · 1|
( n∑
) 1/k
≤ |a i | k (1 + 1 + · · · + 1) (k−1)/k ,
i=1
. Summing over i, the
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