1.3. Simple estimation techniques 9 Proposition 1.4 (H¨ older inequality). If I is an interval and f and g are continuous functions on I then I |f(x)g(x)| dx I |f(x)|p dx 1/p I |g(x)|q dx 1/q , where 1 p, q with 1/p + 1/q = 1. Proof. The case where I |f(x)|p dx = 0 or I |g(x)|q dx = 0 is trivial. Multiplying f and g by suitable positive real numbers, we may therefore assume that I |f(x)|p dx = I |g(x)|q dx = 1 by homogeneity. Now |f(x)g(x)| = (|f(x)|p)1/p (|g(x)|q)1/q = (|f(x)|p)1/p (|g(x)|q)1−1/p, so log(|f(x)g(x)|) = 1 p log (|f(x)|p) + 1 1 p log (|g(x)|q). But the logarithm function is strictly concave, that is to say, all the chords lie strictly below the graph except for their endpoints. Hence 1 p log (|f(x)|p) + 1 1 p log (|g(x)|q) log 1 p log (|f(x)|p) + 1 1 p log (|g(x)|q) , and so |f(x)g(x)| 1 p log (|f(x)|p) + 1 1 p log (|g(x)|q). Integrating over I yields I |f(x)g(x)| dx 1 p ·1 + 1 1 p ·1 = 1, and this proves the inequality. The case p = 2 is especially important this is the Cauchy-Schwarz inequality. The older inequality extends by induction to estimate integrals of products of more than two functions. We have I |f1(x)···fn(x)| dx I |f1(x)|p1 dx 1/p1 · · · I |fn(x)|pn dx 1/pn if 1 p1,...,pn with 1/p1 + · · · + 1/pn = 1.
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