Young's Inequality

Recall from the Conjugate Indices 1/p + 1/q = 1 page that if $1 \leq p \leq \infty$ then the conjugate index of $p$, denoted $q$ is defined as:

• If $1 < p < \infty$ then $q$ is the number such that $\displaystyle{\frac{1}{p} + \frac{1}{q} = 1}$.

• If $p = 1$ then $q = \infty$.

• If $p = \infty$ then $q = 1$.

We will now prove the very important Young's inequality

Theorem 1 (Young's Inequality): Let $1 < p < \infty$ and let $q$ be the conjugate index of $p$. Then for all $a, b \geq 0$, $\displaystyle{ab \leq \frac{a^p}{p} + \frac{b^q}{q}}$.

• Proof: If $a$ or $b$ is equal to $0$ then the inequality trivially holds. So assume that $a, b > 0$.

• Consider the function $f(x) = x^{p-1}$. Since $p > 1$, $p - 1 > 0$ and so $f$ is a strictly increasing function on $[0, a]$ and the inverse of $f$ is $f^{-1}(y) = y^{\frac{1}{p-1}}$.

• The area between $f$, the $x$-axis and the lines $x = 0$ and $x = a$ is given by the following integral:

(1)

\begin{align} \quad \int_0^a f(x) \: dx = \int_0^a x^{p-1} \: dx = \frac{x^p}{p} \biggr |_0^a = \frac{a^p}{p} \end{align}

• The area between $f$, the $y$-axis, and the lines $y = 0$ and $y = b$ is given by the following integral:

(2)

\begin{align} \quad \int_0^b f^{-1}(y) \; dx = \int_0^b y^{\frac{1}{p - 1}} \: dx = \frac{y^{\frac{1}{p - 1} + 1}}{\frac{1}{p - 1} + 1} \biggr |_{y=0}^{y=b} = \frac{b^{\frac{1}{p - 1} + 1}}{\frac{1}{p - 1} + 1} = \frac{b^{\frac{p}{p-1}}}{\frac{p}{p-1}} = \frac{b^q}{q} \end{align}

• The sum of these areas is greater than or equal to the area of the rectangle $ab$, that is:

(3)

\begin{align} \quad ab \leq \frac{a^p}{p} + \frac{b^q}{q} \quad \blacksquare \end{align}