A doubt on Tao's proof of Minkowski's inequality

Question

Let $p \ge 1$ and $f,g \in L^p(X, \mu )$. Minkowski's inequality states that

$$\| f + g \|_{L^p} \le \| f \|_{L^p} + \| g \|_{L^p}$$

Following Tao's proof he says that because of the homogeneity of $L^p$ norms, i.e., for any complex number $c$ we have $ \|c f \|_{L^p} = |c| \| f \|_{L^p}$, we can normalize $ \|f \|_{L^p} + \| g \|_{L^p} $ to be equal to one. Then, again from homogeneity, we can write $f= (1 - \theta) F$ and $g = \theta G$ for some $0<\theta<1$, where $F,G \in L^p$ and $\| F\|_{L^p} = \| G\|_{L^p} = 1$.

The proof then follows from the convexity of the function $x \rightarrow |x|^p$ on $\mathbb{C}$ since

$$ |(1-\theta) F(x) + \theta G(x)|^p \leq (1-\theta) |F(x)|^p + \theta |G(x)|^p .$$

Questions:

1. How do we not lose any generality by normalizing $ \|f \|_{L^p} + \| g \|_{L^p} $ to be equal to one?

2. How can we then also write $f= (1 - \theta) F$ and $g = \theta G$ for some $0<\theta<1$ , where $F,G \in L^p$ and $\| F\|_{L^p} = \| G\|_{L^p} = 1$ without any loss of generality?

Basically, I'm a bit lost in how the homogeneity is being used to not lose generality.

Answer 1

It's a classic trick. Let's break it up into steps.

Step 1: If $\|f\|_p$ or $\|g\|_p = 0$, then the result follows immediately. This is due to the fact that $\|\cdot\|_p$ is a norm, so $\|f\|_p = 0$ means $f = 0$ and hence $\|f+ g\|_p = \|g\|_p$.

Step 2: By Step 1, we may assume that both $\|f\|_p$ and $\|g\|_p$ are greater than $0$. Suppose we know the inequality assuming $\|f\|_p + \|g\|_p = 1$. Let $$F := \frac{f}{\|f\|_p + \|g\|_p}, \ \ G := \frac{g}{\|f\|_p + \|g\|_p }$$ By homogeneity, we have $$\|F\|_p + \|G\|_p = \frac{\|f\|_p}{\|f\|_p + \|g\|_p} + \frac{\|g\|_p}{\|f\|_p + \|g\|_p} = \frac{\|f\|_p + \|g\|_p}{\|f\|_p + \|g\|_p} = 1.$$ So by homogeneity again, we know that $$ \|F + G\|_p \leq \|F\|_p + \|G\|_p = 1\implies \frac{\|f + g\|_p}{\|f\|_p + \|g\|_p} \leq 1 \implies \|f+g\|_p \leq \|f\|_p + \|g\|_p.$$ Our result follows if we can show the result in the case where $\|f\|_p + \|g\|_p = 1.$

Step 3: Now we need to show that the inequality holds assuming that $\|f\|_p + \|g\|_p = 1.$ Let $$F := \frac{f}{\|f\|_p}, G := \frac{g}{\|g\|_p}.$$ Notice in Terry's proof he's setting $\theta = \|g\|_p$. In this case, we have $\theta G = g$. Notice also that $$\|f\|_p + \|g\|_p = 1 \implies \|f\|_p = 1 - \|g\|_p = 1 - \theta,$$ so $(1-\theta)F = \|f\|_p F = f.$ Now by convexity, we get that $$ |(1-\theta)F + \theta G|^p \leq (1-\theta) |F|^p + \theta |G|^p,$$ hence $$ \int_X |(1-\theta)F + \theta G|^p d\mu \leq (1-\theta) \int_X |F|^p d\mu + \theta \int_X |G|^p d\mu,$$ or $$ \|f + g\|_p^p = \|(1-\theta)F + \theta G\|_p^p \leq (1-\theta) \|F\|_p^p + \theta \|G\|_p^p.$$ Now notice that $$ \|F\|_p^p = \left(\frac{\|f\|_p}{\|f\|_p}\right)^p = 1,$$ and similarly $\|G\|_p^p = 1.$ We get that $$ \|f + g\|_p^p = \|(1-\theta)F + \theta G\|_p^p \leq (1-\theta) + \theta = 1 = \|f\|_p + \|g\|_p.$$ By Step 2, we see the inequality holds.

Answer 2

0. By the non-degeneracy property we may take $\|f\|_{p}$ and $\|g\|_{p}$ to be non-zero.

1. If $C:=\|f\|_p+\|g\|_p\ne 1$, the normalized case allows you to conclude $$ \|\frac{f}{C}+\frac{g}{C}\|_p\le 1 $$ which, by the homogeneity of the norms, gives you the desired inequality: $$ \|f+g\|_p\le C=\|f\|_p+\|g\|_p $$

2. Now assume that $\|f\|_p+\|g\|_p=1$. You can then let $\theta=\|g\|_p$ and write $$ g=\theta G,\quad G=\frac{g}{\|g\|_p},\quad f = (1-\theta)F,\quad F = \frac{f}{\|f\|_p} $$

Answer 3

Lemma: If Minkowski's inequality holds for all pairs $((1-\theta)F,\theta G)$ such that

• $F$ and $G$ have norm $1$;
• $\theta \in (0,1)$;

then Minkowski's inequality holds for all pairs of functions.

Proof: Let $f,g \in L^p$. If one of $\Vert f \Vert_p$ or $\Vert g \Vert_p$ equals zero, then the corresponding function itself (almost everywhere) vanishes, and the result is trivial. Therefore, we assume that both norms are nonzero.

Let $F := \frac{\frac{f}{\Vert f\Vert_p + \Vert g \Vert_p}}{\Vert \frac{f}{\Vert f\Vert_p + \Vert g \Vert_p} \Vert_p}$, $G := \frac{\frac{g}{\Vert f\Vert_p + \Vert g \Vert_p}}{\Vert \frac{g}{\Vert f\Vert_p + \Vert g \Vert_p} \Vert_p}$.

Then both $F$ and $G$ have norm $1$. Let $\theta := \Vert \frac{g}{\Vert f\Vert_p + \Vert g \Vert_p} \Vert_p$. Then $1-\theta = \Vert \frac{f}{\Vert f\Vert_p + \Vert g \Vert_p} \Vert_p$. Therefore, by assumption, Minkowski's inequality holds for $(\frac{f}{\Vert f \Vert_p + \Vert g \Vert_p},\frac{g}{\Vert f \Vert_p + \Vert g \Vert_p})$ so we are done.