# On the equality case of the Hölder and Minkowski inequalites

I'm following the book Measure and Integral of Richard L. Wheeden and Antoni Zygmund. This is the problem 4 of chapter 8.

Consider $E\subseteq \mathbb{R}^n$ a measurable set. In the following all the integrals are taken over $E$, $1/p + 1/q=1$, with $1\lt p\lt \infty$.

I'm trying to prove that $$\int \vert fg\vert =\Vert f \Vert_p\Vert g \Vert_q$$ if and only if $\vert f \vert^p$ is multiple of $\vert g \vert^q$ almost everywhere. To do this, I want to consider the following cases: if $\Vert f \Vert_p=0$ or $\Vert g \Vert_q=0$, we are done. Then suppose that $\Vert f \Vert_p\ne 0$ and $\Vert g \Vert_q\ne 0$. If $\Vert f \Vert_p=\infty$ or $\Vert g \Vert_q=\infty$, we are done (I hope). If $0\lt\Vert f \Vert_p\lt\infty$ and $0\lt\Vert g \Vert_q\lt\infty$, proceed as follows.

When we are proving the Hölder's inequality, we use that for $a,b\geq 0$ $$ab\leq \frac{a^p}{p}+\frac{b^q}{q},$$ where the equality holds if and only if $b=a^{p/q}$. Explicitly $$\int\vert fg \vert\leq \Vert f \Vert_p \Vert g \Vert_q \int\left( \frac{\vert f \vert^p}{p\Vert f \Vert_p^p} + \frac{\vert g \vert^q}{q\Vert g \Vert_q^q}\right)=\Vert f \Vert_p \Vert g \Vert_q.$$ From here, we see that the equality in Hölder's inequalty holds iff $$\frac{\vert fg \vert}{\Vert f \Vert_p \Vert g \Vert_q}=\frac{\vert f \vert^p}{p\Vert f \Vert_p^p} + \frac{\vert g \vert^q}{q\Vert g \Vert_q^q}, \text{ a.e.}$$ iff $$\frac{\vert g \vert}{\Vert g \Vert_q}=\left( \frac{\vert f \vert}{\Vert f \Vert_p} \right)^{p/q},\text{ a.e.}$$ iff $$\vert g \vert^q\cdot \Vert f \Vert_p^p=\vert f \vert^p \cdot \Vert g \Vert_q^q,\text{ a.e.}$$ Q.E.D. But, assuming that $\Vert f \Vert_p\ne 0$ and $\Vert g \Vert_q\ne 0$, what about when $\Vert f \Vert_p=\infty$ or $\Vert g \Vert_q=\infty$? How can I deal with it?

In the case of Minkowski inequality, suppose that the equality holds and that $g\not \equiv 0$ (and then $\left( \int \vert f+g \vert^p\right)\ne 0$). I need to prove that $\Vert f \Vert_p$ is multiple of $\Vert g \Vert_q$ almost everywhere. I can reduce to the "Hölder's equality case". I can get $$\vert f \vert^p=\left( \int \vert f+g \vert^p\right)^{-1}\Vert f \Vert_p^p\vert f+g \vert^p$$ $$\vert g \vert^p=\left( \int \vert f+g \vert^p\right)^{-1}\Vert g \Vert_p^p\vert f+g \vert^p$$ almost everywhere, but again, using the finiteness and nonzeroness of $\Vert f \Vert_p$ and $\Vert g \Vert_p$.

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Your proof for Hölder looks fine. I don't think there's any meaningful way to deal with infinities here. Take a function $f$ which is nonzero only on $[0,2]$ and $g$ non-zero only on $[1,3]$, say. Arrange that $fg$ is non-integrable on $[1,2]$, $f$ is not $p$-integrable on $[0,1]$ and $g$ is not $q$-integrable on $[2,3]$. There's no way that $f$ is a multiple of $g$ or the other way around. –  t.b. Dec 2 '11 at 6:05
Got it. Thanks a lot @t.b. Can you please put this as an answer. Certainly this is not an unanswered question. –  leo Dec 2 '11 at 18:34

On leo's request I'm posting my comment as an answer.

Your treatment of the equality cases of Hölder's and Minkowski's inequalities are perfectly fine and clean. There's a small typo when you write that $\int|fg| = \|f\|_p\|g\|_q$ if and only if $|f|^p$ is a constant times of $|g|^q$ almost everywhere (you write the $p$-norm of $f$ and the $q$-norm of $g$ instead).

The case where either one $\|f\|_p$ or $\|g\|_q$ (or both) are infinite isn't part of this exercise and simply wrong. You can trisect $E = F \cup G \cup H$ into disjoint measurable sets of positive measure, take $f$ not $p$-integrable on $F$ and zero on $G$, take $g$ not $q$-integrable on $G$ and zero on $F$ and choose $fg$ non-integrable on $H$. Then certainly no power of $|f|$ is a constant multiple of a power of $|g|$ and vice versa, even though equality holds in the Hölder inequality.

A very nice “blackboard summary” of the equality case (for finite sequences) is given in Steele's excellent book The Cauchy–Schwarz Master Class. Let $a = (a_1,\ldots,a_n) \geq 0$ and $b = (b_1, \ldots, b_n) \geq 0$ and let $\hat{a}_i = \dfrac{a_i}{\|a\|_p}$ and $\hat{b}_i = \dfrac{b_i}{\|b\|_q}$. Then your argument is subsumed by the diagram (with an unfortunate typo in the upper right corner—no $p$th and $q$th roots there):

Mimicking this for functions, let us write $\hat{f} = \dfrac{|f|}{\|f\|_p}$ and $\hat{g} = \dfrac{|g|}{\|g\|_q}$ (assuming of course $\|f\|_p \neq 0 \neq \|g\|_q$), so $\int \hat{f}\vphantom{f}^p = 1$ and $\int \hat{g}^q =1$ and thus your argument becomes $$\begin{array}{ccc} \int |fg| = \left(\int|f|^p\right)^{1/p} \left(\int|g|^q\right)^{1/q} & & |f|^p = |g|^q \frac{\|f\|_{p}^p}{\|g\|_{q}^q} \text{ a.e.}\\ \Updownarrow\vphantom{\int_{a}^b} & & \Updownarrow \\ \int \hat{f}\,\hat{g} = 1 & & \hat{f}\vphantom{f}^p = \hat{g}^q \text{ a.e.} \\ \Updownarrow\vphantom{\int_{a}^b} & & \Updownarrow \\ \int \hat{f}\,\hat{g} = \frac{1}{p} \int \hat{f}\vphantom{f}^p + \frac{1}{q} \int \hat{g}^q & \qquad \iff \qquad & \hat{f}\,\hat{g} = \frac{1}{p} \hat{f}\vphantom{f}^p + \frac{1}{q} \hat{g}^q \text{ a.e.} \end{array}$$

I suggest that you draw a similar diagram for the equality case of Minkowski's inequality.

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Very nice answer. Thanks :) –  leo Dec 3 '11 at 18:01
nice diagramming, too! –  robjohn Mar 19 '12 at 2:50