Young's Inequality

# Young's Inequality

Recall from the Conjugate Indices page that if $p > 1$ then the conjugate index of $p$ is the number $q > 1$ such that:

(1)\begin{align} \quad \frac{1}{p} + \frac{1}{q} = 1 \end{align}

And if $p = 1$ we define its conjugate index to be $q = \infty$.

We now prove an important result known as Young's inequality regarding conjugate indices.

Theorem 1 (Young's Inequality): If $a, b \geq 0$ and $p, q >1$ are such that $\frac{1}{p} + \frac{1}{q} = 1$ then $\displaystyle{ab \leq \frac{a^p}{p} + \frac{b^q}{q}}$. |

**Proof:**There are a few cases to consider.

**Case 1:**Suppose that $a = 0$ or $b = 0$. The trivially:

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

**Case 2:**Suppose that $a \neq 0$ and $b \neq 0$. Let:

\begin{align} \quad s = p \ln a \quad \mathrm{and} \quad t = p \ln b \end{align}

- Note that $s$ and $t$ are well-defined since $a, b > 0$. Therefore:

\begin{align} \quad \frac{s}{p} = \ln a \quad \mathrm{and} \quad \frac{t}{p} = \ln b \end{align}

- So $e^{s/p} = a$ and $e^{t/p} = a$ and also:

\begin{align} \quad e^s = a^p \quad \mathrm{and} \quad e^t = a^p \quad (*) \end{align}

We note that the function $f(x) = e^x$ is convex on $\mathbb{R}$. So for all $x, y \in \mathbb{R}$ and for all $s \in [0, 1]$ we have that:

(6)\begin{align} \quad f(mx + (1 - m)y) \leq mf(x) + (1-m)f(y) \quad \Leftrightarrow \quad e^{mx + (1-m)y} \leq me^x + (1-m)e^y \end{align}

- Take $m = \frac{1}{p}$. Then $1 - m = \frac{1}{q}$. Take $x = s$ and $y = t$. Then:

\begin{align} \quad e^{\frac{s}{p} + \frac{t}{q}} \leq \frac{e^s}{p} + \frac{e^t}{q} \quad (**) \end{align}

- Now we have that:

\begin{align} \quad ab = e^{s/p}e^{t/p} = e^{\frac{s}{p} + \frac{t}{p}} \overset{(**)} \leq \frac{e^s}{p} + \frac{e^t}{q} \overset{(*)} = \frac{a^p}{p} + \frac{b^q}{q} \quad \blacksquare \end{align}