Proof of Conjugate Exponents and Young's Inequality

Conjugate exponents. Since $p>1$, we have $p-1>0$, so $q=p/(p-1)$ is defined and

$$\frac{1}{q}=\frac{p-1}{p}=1-\frac{1}{p} .$$

Conversely, if a real number $q'$ satisfies $1/p+1/q'=1$, then $1/q'=(p-1)/p$, and so $q'=p/(p-1)=q$. Finally $p>p-1>0$ gives $q>1$.

A convexity inequality for the exponential function. Let $x,y,\theta\in\mathbb{R}$ with $0<\theta<1$, and put $m=\theta x+(1-\theta)y$. Since $\exp(m)>0$ and $\exp(u)\ge1+u$ for every real $u$ (Basic Properties of the Exponential Function (iii) and Basic Properties of the Exponential Function (v)), Basic Properties of the Exponential Function (ii) gives

$$\exp(x)=\exp(m)\exp(x-m)\ge\exp(m)(1+x-m),\qquad \exp(y)=\exp(m)\exp(y-m)\ge\exp(m)(1+y-m).$$

Multiplying the first inequality by $\theta>0$ and the second by $1-\theta>0$ and adding, and using $\theta(x-m)+(1-\theta)(y-m)=m-m=0$, we obtain

$$\theta\exp(x)+(1-\theta)\exp(y)\ge\exp\bigl(\theta x+(1-\theta)y\bigr).$$

Young's inequality. If $a=0$ or $b=0$, then $ab=0$, while $a^{p}\ge0$ and $b^{q}\ge0$ by Properties of Real Powers (ii); so the inequality holds. Let now $a,b>0$, and put

$$x=\log a^{p}=p\log a,\qquad y=\log b^{q}=q\log b,\qquad \theta=\frac{1}{p},$$

by Properties of Real Powers (iii) ($a^{p},b^{q}>0$ by Properties of Real Powers (ii)). Then $0<\theta<1$ and $1-\theta=1/q$, so $\theta x+(1-\theta)y=\log a+\log b=\log(ab)$ by Properties of the Natural Logarithm: Values at 1 and e, Products, Quotients, Monotonicity, Continuity, Derivative and the Bound log t at most t - 1 (ii). By The Natural Logarithm (ii), $\exp(x)=a^{p}$, $\exp(y)=b^{q}$ and $\exp(\log(ab))=ab$, and the convexity inequality becomes

$$\frac{a^{p}}{p}+\frac{b^{q}}{q}\ge ab .$$