Understanding the condition $\frac{1}{p} + \frac{1}{q} = 1$ in the estimate $xy \le \frac{1}{p}x^p + \frac{1}{q}y^q$

Question

I just read a proof of Holder's inequality in measure theory, which boils down to the following inequality: $$xy \le \frac{1}{p}x^p + \frac{1}{q}y^q$$ where $x,y\ge 0$ and $\frac{1}{p} + \frac{1}{q} = 1$.

This inequality can be proved by considering the difference, viewed as a function in $x$, and showing, using calculus, that its minimum is 0.

My question is: Why the heck should this be true? What is the meaning behind the condition that $\frac{1}{p} + \frac{1}{q} = 1$? To me, this reminds me of the theory of triangle groups, where the properties of the triangle group $\Delta(p,q,r) := \langle a,b,c|a^p = b^q = c^r = abc = 1\rangle$ is very sensitive to whether $\frac{1}{p} + \frac{1}{q} + \frac{1}{r}$ is less than, greater than, or equal to 1. I don't know if there's any resemblance?

Answer 1

This is related to the concept of Fenchel-Legendre convex conjugate. More precisely, the functions $$ \Phi(x)=\frac 1p x^p,\qquad \Psi(y)=\frac 1q y^q $$ are convex conjugate from each other if and only if $\frac 1p+\frac 1q=1$, which by definition means $$ \Psi(y)=\Phi^*(y)=\sup\limits_{x}\Big( xy - \Phi(x)\Big). $$ The standard real-analysis computation simply checks this in case $\frac 1p +\frac 1q=1$. In fact there is a generalized Young's inequality, which reads $$ xy\leq \phi(x)+\phi^*(y) $$ for any convex function $\phi$. (The proof is completely similar to the standard one)

Answer 2

One reasonable explanation, to me, is that the above is nothing but the convexity inequality $f\big(t u+(1-t)v\big)\le tf(u)+(1-t)f(v)$ for the function $-\log$, changing the names of the variables and exponentiating. So what we see in $\frac1p+\frac1q=1$ is: coefficients of a convex combination. Thus $\log\big(\frac1px^p + \frac1q y^q\big)\ge\frac1p\log(x^p)+ \frac1q\log(x^q)=\log(xy).$

It is also worth mentioned the nice elementary proof based on definite integrals: we write the duality relation as $\displaystyle\frac1{p-1}=q-1$ and what it tells now is: $t\mapsto t^{q-1}$ is the inverse function of $t\mapsto t^{p-1}$. So their hypo-graphs cover the rectangle $[0,x]\times[0,y]$. Thus $xy\le \int_0^x t^{p-1}dt+ \int_0^y t^{q-1}dt$. (Note that the derivatives of the functions $\frac{t^p}p$ and $\frac{t^q}q$ being inverses of each other, in fact means they are a pair of convex conjugate functions, so a deeper explanation for the latter proof is given by Leo Monsaingeon's answer).

Answer 3

You may think the following way. Ignore the sharp constants and study when $xy\le C(x^p+y^q) $ for some $C>0$ and arbitrary positive $x, y$. This is equivalent to asking when $xy\le C\max(x^p, y^q)$ (with a different optimal constant). In other words, when at least one of the bounds $x\le C y^{q-1}$ or $y\le Cx^{p-1}$ always holds. Taking, say, $x=2Cy^{q-1}$ so that the first bound does not hold for sure, we need $y\le C(2C y^{q-1})^{p-1}$ for all positive $y$. This reads as $1\le C_1 y^{pq-p-q}$ for some $C_1$. If $pq-p-q\ne 0$, this does not hold either for large or for small $y$. So, the only hope is the case $pq=p+q$.

Answer 4

This is actually the weighted AM-GM inequality for $n = 2$ in disguise. Recall that this inequality says that if $w_1, w_2$ are two non-negative weights such that $w_1 + w_2 = 1$ and $x_1, x_2 \ge 0$, then

$$w_1 x_1 + w_2 x_2 \ge x_1^{w_1} x_2^{w_2}.$$

Now set $w_1 = \frac{1}{p}, w_2 = \frac{1}{q}, x_1 = x^p, x_2 = y^q$. The significance of the condition $w_1 + w_2 = 1$ is to ensure that the LHS is a weighted version of the arithmetic mean $\frac{x_1 + x_2}{2}$ (which we get when $w_1 = w_2 = \frac{1}{2}$) while the RHS is a weighted version of the geometric mean $\sqrt{x_1 x_2}$ (same). In particular it is necessary for the inequality to become an equality when $x_1 = x_2$, as in that case any mean should just return the common value. It is also necessary in order for the inequality to have a scaling symmetry $(x_1, x_2) \mapsto (\lambda x_1, \lambda x_2)$ (equivalent to the one Terence Tao discusses in the comments but a little easier to see in this different set of coordinates).

The weighted AM-GM inequality actually follows from the ordinary one, as follows. Recall that the ordinary AM-GM inequality says that if $y_1, \dots y_n \ge 0$ then

$$\frac{y_1 + \dots + y_n}{n} \ge \sqrt[n]{y_1 \dots y_n}.$$

This is classic and has many proofs, many of which don't involve calculus; I particularly like Cauchy's proof which reduces to the case that $n$ is a power of $2$, and then reduces further to the case $n = 2$, where it is just the observation that $(\sqrt{y_1} - \sqrt{y_2})^2 \ge 0$.

Now apply the ordinary AM-GM inequality to a collection of $k_i$ copies of another set of variables $x_i$. This gives

$$\frac{k_1 x_1 + \dots + k_n x_n}{k_1 + \dots + k_n} \ge \sqrt[k_1 + \dots + k_n]{x_1^{k_1} \dots x_n^{k_n}}.$$

Setting $w_i = \frac{k_i}{k_1 + \dots + k_n}$ we get weighted AM-GM for rational weights. Now weighted AM-GM for arbitrary weights follows by continuity. This argument does not require computing a single derivative!