Proof of Conjugate Exponents and Young's Inequality

Each result cited is universally quantified over the data appearing in its own statement, and is applied here to the data named in the statement above.

Claim 1. Let $1<p$. Then $0<1<p$, so $p$ is positive and its inverse $1/p$ is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field. Multiplying the strict inequality $1<p$ by the positive number $1/p$, using claim 10 of that lemma, gives $1/p<1$. Adding $-1/p$ to both sides of $1/p<1$, by claim 1 of the same lemma, gives $0<1-1/p$. Hence $1-1/p$ is nonzero, and we may set

$$q=\Bigl(1-\frac{1}{p}\Bigr)^{-1},$$

which is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field. Then $1/q=1-1/p$, so $1/p+1/q=1$, which proves existence.

For uniqueness, suppose $q'$ is a real number with $1/p+1/q'=1$; implicitly $q'\ne0$. Then $1/q'=1-1/p=1/q$, and multiplying by $qq'$ gives $q=q'$. So $q$ is the only such number.

Since $1/q=1-1/p$ and $0<1/p$, we get $1/q<1$; multiplying this by the positive number $q$ gives $1<q$. From $1/q=(p-1)/p$, multiplying by the nonzero numbers $p$ and $q$ gives $p=(p-1)q$, that is $q=p/(p-1)$, the number $p-1$ being positive because $1<p$. Multiplying $1/p+1/q=1$ by $pq$ gives $q+p=pq$. Finally the identity $1/q+1/p=1$ is the defining relation with the roles of $p$ and $q$ exchanged, and $1<q$, so by the uniqueness just proved $p$ is the unique real number bearing this relation to $q$.

Claim 2. Let $p,q$ be conjugate exponents and let $a,b\in\mathbb{R}_{+}$. By claim 1 of Properties of Real Powers the numbers $a^{p}$ and $b^{q}$ are nonnegative, and $1/p$ and $1/q$ are positive by claim 1 above, so $a^{p}/p$ and $b^{q}/q$ are nonnegative by claim 5 of Elementary Arithmetic in an Ordered Field and their sum is nonnegative by claim 2 of that lemma.

Suppose first that $a=0$ or $b=0$. Then $ab=0$, and the right-hand side is nonnegative, so the asserted inequality holds.

Suppose now that $0<a$ and $0<b$. Then $a^{p}$ and $b^{q}$ are positive by claim 1 of Properties of Real Powers, so the real numbers $s=\log(a^{p})$ and $u=\log(b^{q})$ are defined, where $\log$ is the natural logarithm. Put $\theta=1/q$, so that $1-\theta=1/p$ by claim 1, and $0\le\theta$ and $\theta\le1$ because $0<1/q<1$.

Apply A Real Function with Nonnegative Second Derivative is Convex on an Interval with $J=\mathbb{R}$, $g=\exp$ and $g_{1}=\exp$. The set $\mathbb{R}$ is order-convex, and each $t\in\mathbb{R}$ satisfies $t-1<t<t+1$ with $t-1,t+1\in\mathbb{R}$, so every point of $J$ is interior to it. By claim 3 of Basic Properties of the Exponential Function the function $\exp$ is differentiable at every point with derivative $\exp$, so $g'=g_{1}$ and $g_{1}'=\exp$ everywhere; and $0\le\exp(t)$ for every $t$ by claim 2 of the same theorem. The conclusion of that result, applied to the points $s,u$ and the number $\theta$, reads

$$\exp\bigl((1-\theta)s+\theta u\bigr)\le(1-\theta)\exp(s)+\theta\exp(u).$$

We identify both sides. By The Natural Logarithm we have $\exp(\log t)=t$ for every positive $t$, so $\exp(s)=a^{p}$ and $\exp(u)=b^{q}$, and the right-hand side equals $\frac{1}{p}a^{p}+\frac{1}{q}b^{q}$, which is $\frac{a^{p}}{p}+\frac{b^{q}}{q}$. For the left-hand side, claim 1 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities gives $\log(a^{p})=p\log a$ and $\log(b^{q})=q\log b$, whence

$$(1-\theta)s+\theta u=\frac{1}{p}\,p\log a+\frac{1}{q}\,q\log b=\log a+\log b=\log(ab),$$

the last equality by the identity $\log(st)=\log s+\log t$ recorded in The Natural Logarithm, applicable since $a$ and $b$ are positive. As $ab$ is positive, $\exp(\log(ab))=ab$. Substituting these identifications into the displayed inequality gives

$$ab\le\frac{a^{p}}{p}+\frac{b^{q}}{q},$$

as claimed.