Show the sharp constant $\frac{p^{1/p} q^{1/q} r^{1/r}}{p'^{1/p'} q'^{1/q'} r'^{1/r'}} \leq 1$

Question

I am reading Chapter 4 of Lieb-Loss. I want to know how to show the sharp constant in Young's convolution inequality is less than 1 (inequality with constant 1 is much easier to prove.)

That is, for $p,q,r \geq 1$ and $1/p+1/q+1/r = 2$, I want to prove that

$\frac{p^{1/p} q^{1/q} r^{1/r}}{p'^{1/p'} q'^{1/q'} r'^{1/r'}} \leq 1$, where $1/p'+1/p=1$. (Similar for $q'$ and $r'$.)

I first try to show $p^{1/p} p'^{-1/p'} \leq 1$ for all $p \geq 1$. Soon, I find it's not true for $p = 3$, so I think the condition $1/p+1/q+1/r = 2$ must be used somewhere.

Thanks for any suggestions.

Answer 1

I think I solve this problem by Lagrange multiplier:

Let $1/p:= x, 1/q = y$ and $1/r:= z$, then the constraints are $x+y+z=2 , 0 \leq x,y,z \leq 1$ and we want to maximize $(1-x)^{1-x} x^{-x}(1-y)^{1-y} y^{-y}(1-z)^{1-z} z^{-z}$ which is equivalent to maximize

$(1-x)\log(1-x)-x\log x + (1-y)\log(1-y)-y\log y+ (1-z)\log(1-z)-z\log z$

By lagrange multiplier, we have the relation that

$-\log (1-x) - 2-\log x = -\log (1-y) - 2-\log y = -\log (1-z) - 2-\log z $, that is, $x(1-x) = y(1-y) = z(1-z).$

Since the parabola t = s(1-s) on $s \in [0,1]$ meets each horizontal line at most twice, we know either $x=y$ or $x=1-y$. After some discussions, we know the maxima of the sharp constant $(1-x)^{1-x} x^{-x}(1-y)^{1-y} y^{-y}(1-z)^{1-z} z^{-z}$ is 1.

Answer 2

This is equality for $r=1$, $p=q=1/2$.