Young's Inequality

Form for Hölder exponents

If $a, b$ are non-negative reals, and $p,q$ are positive reals that satisfy $\frac{1}{p}+\frac{1}{q}=1$ , then the following inequality holds for all possible values of $a$ and $b$ . $\begin{align*} \frac{a^p}{p}+\frac{b^q}{q} \geq ab \end{align*}$ with equality iff $a^p=b^q$

Form for definite integrals

Suppose $f$ is a strictly increasing and continuous function on the interval $[0,t]$ where $t$ is a positive real number, and also $f(0)=0$ . Then the following inequality holds for all $a \in [0,c]$ and $b \in [0,f(c)]$ $\begin{align*} \int_0^a f(x)\text{d}x + \int_0^b f^{-1}(x) \text{d}x \geq ab \end{align*}$ with equality iff $f(a) = b$ .

Proof

The logarithm is concave and we know that $\frac{1}{p}+\frac{1}{q}=1$ , so by [Jensen's Inequality], we have $\log\left(\frac{a^p}{p}+\frac{b^q}{q}\right) \geq \frac{1}{p}\log(a^p) + \frac{1}{q}\log(b^q)$ $\log\left(\frac{a^p}{p}+\frac{b^q}{q}\right) \geq \log{a}+\log{b} = \log{ab}$ Young's Inequality then follows by exponentiation of both sides.

Page

Toolbox

Search

Young's Inequality

Form for Hölder exponents

Form for definite integrals

Proof