Young's inequality.

Question

I am referring to the inequality: Young's inequality The standard version for increasing functions.

I read the article of Young and also a generalization of this claim in Hardy, Littlewood and Polya's Inequalities.

But I don't see that Young proves rigorously his claim, and in Zygmund's Trigonometric Series they say that the proof is easily seen geometrically, i.e by looking at a picture, but surely this is not rigorous proof (in high school geometry pictures aren't a viable approach for a rigorous proof).

I don't have as of yet access to the Henstock's book in the references of the wikipedia page.

Does someone have a rigorous proof for this claim?

Answer 1

Zygmund is right, and the proof suggested by a picture is easily made rigorous. The function $f(x)=x^{p-1}$ has inverse function $g(y)=y^{q-1}$. For $a>0$, the region $A=\{(x,y): 0\le x\le a, 0\le y\le f(x)\}$ has area $a^p/p$; the region $B=\{(x,y): 0\le y\le b, 0\le x\le g(y)\}$ has area $b^q/q$. The rectangle $\{(x,y): 0\le x\le a,0\le y\le b\}$ has area $ab$ and is a subset of $A\cup B$, whence the inequality.

Answer 2

I will use Zygmund's notation: let $\phi(u)$ and $\psi(v)$ be continuous strictly increasing functions on the non-negative real numbers such that $\phi(0)=\psi(0)=0$ and $\phi$ and $\psi$ are inverses. What will be shown is that $$ab\le\Phi(a)+\Psi(b)$$ where $$\Phi(x)=\int_0^x\phi(u)~du\qquad\Psi(y)=\int_0^y\psi(v)~dv$$ and $a$ and $b$ are non-negative, and that equality holds if and only if $b=\phi(a)$.

The first step is to show $$\Phi(x)+\Psi(\phi(x))-x\phi(x)\leq\Phi(x)+\Psi(y)-xy,$$ where equality holds if and only if $y=\phi(x)$. Suppose $y < \phi(x)$. Since $\psi$ is the inverse of $\phi$ and a strictly increasing function, $\psi(v) < x$ when $y < v < \phi(x)$. So $$\Psi(\phi(x))-\Psi(y) = \int_y^{\phi(x)}\psi(v)~dv < \int_y^{\phi(x)}x~dv=x(\phi(x)-y).$$ The case $y > \phi(x)$ has an analogous proof.

The second step is to show $$\Phi(x)+\Psi(\phi(x))-x\phi(x)=0.$$ This trivially holds when $x=0$. What follows is the demonstration that the left-hand side is differentiable and that its derivative identically vanishes. Let $f(x)$ be the function on the left, let $x_0$ and $x_1$ be non-negative numbers with $x_0 < x_1$, and let $y_0$ and $y_1$ be $\phi(x_0)$ and $\phi(x_1)$, respectively. By adding and subtracting $x_0y_1$ and refactoring, $$f(x_1)-f(x_0)=\int_{x_0}^{x_1}\phi(u)-y_1~du+\int_{y_0}^{y_1}\psi(v)-x_0~dv.$$ Let $I_0$ and $I_1$ be the integrals on the right, respectively. Considerations similar to those made in the first step yield $$-(x_1-x_0)(y_1-y_0)\le I_0\le0\le I_1\le(x_1-x_0)(y_1-y_0).$$ Hence, $$\Bigl|\frac{f(x_1)-f(x_0)}{x_1-x_0}\Bigr|\le|\phi(x_1)-\phi(x_0)|.$$ Evidently, this inequality holds even when the assumption $x_0 < x_1$ is removed.

Since $\phi(u)$ is continuous, $f(x)$ is differentiable and $f'(x)=0$. This completes the proof.