Math 2.0

Real functions Classical inequalities Theorem

Hermite–Hadamard inequality

Builds on Convex functionRiemann integral

Statement

Let a<ba < b be real numbers and let f ⁣:[a,b]→ℝf \colon [a, b] \to \R be a convex function. Then ff is Riemann integrable on [a,b][a, b] and f(a+b2)≤1b−a∫abf(x) dx≤f(a)+f(b)2.f\Big( \frac{a + b}{2} \Big) \le \frac{1}{b - a} \int_a^b f(x) \, dx \le \frac{f(a) + f(b)}{2}.

Remarks

Proof idea: a convex function on [a,b][a, b] is bounded and is continuous on (a,b)(a, b), so it is integrable. For the left inequality, convexity gives f(a+b2)≤12(f(x)+f(a+b−x))f\big(\frac{a+b}{2}\big) \le \frac{1}{2}\big(f(x) + f(a + b - x)\big); integrate over x∈[a,b]x \in [a, b]. For the right inequality, the graph lies below the chord: f(x)≤b−xb−af(a)+x−ab−af(b)f(x) \le \frac{b - x}{b - a} f(a) + \frac{x - a}{b - a} f(b); integrate this bound, whose mean value is f(a)+f(b)2\frac{f(a) + f(b)}{2}.

Example: f(x)=x2f(x) = x^2 on [0,1][0, 1] gives 14≤13≤12\frac{1}{4} \le \frac{1}{3} \le \frac{1}{2}.

For a concave function both inequalities are reversed.

Going deeper

For a student

The inequality gives error bounds for the midpoint rule and the trapezoidal rule: for a convex integrand the midpoint rule underestimates the integral and the trapezoidal rule overestimates it. It is a two-sided refinement of Jensen's inequality for the uniform distribution on [a,b][a, b].

Reading path

To understand this page from the ground up, read these 7 pages in order. The foundations come first.

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  1. Convex function definition
  2. Partition of an interval definition
  3. Mesh of a partition definition
  4. Riemann sum definition
  5. Riemann integrable function definition
  6. Riemann integral definition
  7. Hermite–Hadamard inequality statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Hermite–Hadamard_inequality
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