I'm following the book Measure and Integral of Richard L. Wheeden and Antoni Zygmund. This is 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 $\Vert g \Vert_p$.

  • $\begingroup$ 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. $\endgroup$
    – t.b.
    Commented Dec 2, 2011 at 6:05
  • $\begingroup$ Got it. Thanks a lot @t.b. Can you please put this as an answer. Certainly this is not an unanswered question. $\endgroup$
    – leo
    Commented Dec 2, 2011 at 18:34
  • $\begingroup$ You have used a argument in your proof, which is that if two integral coincide, then the corresponding two integrands are equal a.e. . Unfortunately, this argument is wrong. Consider $f(x)=x, x \in (0,1),$ vanishing outside $(0,1)$ and $g(x)=-x, x \in (-1,0),$ vanishing outside $(-1,0)$. Their integrals on $\mathbb R$ are same, but they are different on a measurable set with positive measure, i.e. (-1,1). $\endgroup$
    – Sam Wong
    Commented Mar 29, 2018 at 14:04
  • 1
    $\begingroup$ @SamWong I think this can be replaced by the result that if integration of a positive function is zero then the function is zero a.e. And the positive function can be obtained by just taking the left identity to right. Now the positivity is ensured by the Hölder's inequality on $\mathbb{R}$. $\endgroup$ Commented Sep 5, 2019 at 14:13

3 Answers 3


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):

Steele's diagram for Hölder equality

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.

  • $\begingroup$ Very nice answer. Thanks :) $\endgroup$
    – leo
    Commented Dec 3, 2011 at 18:01
  • 1
    $\begingroup$ nice diagramming, too! $\endgroup$
    – robjohn
    Commented Mar 19, 2012 at 2:50
  • 4
    $\begingroup$ @t.b. Can you explain to me why this step in the diagram is true? $ \int \hat{f}\,\hat{g} = \frac{1}{p} \int \hat{f}\vphantom{f}^p + \frac{1}{q} \int \hat{g}^q \Leftrightarrow \; \hat{f}\,\hat{g} = \frac{1}{p}\hat{f}\vphantom{f}^p + \frac{1}{q} \hat{g}^q $. I see why the "$ \Leftarrow $" direction is true but not the "$ \Rightarrow $" direction. $\endgroup$
    – alpastor
    Commented Feb 22, 2016 at 20:01
  • $\begingroup$ @Filburt All integrands are nonnegative can demonstrate nothing. If two integrals are equal, to prove their integrands are equal, what we need is the difference of those two integrands is greater than or equal to zero, not the non-negativity of these two integrands. $\endgroup$
    – Sam Wong
    Commented Mar 29, 2018 at 14:16
  • 1
    $\begingroup$ @alpastor: Everything in sight is nonnegative. Now, there is also Young's inequality for products of nonnegative numbers: $a\,b \le \frac{a^p}{p} + \frac{b^q}{q}$. Applying Young's inequality to $a = \hat f$ and $b = \hat g$, we see that $\hat f\,\hat g \le \frac{\hat f^p}{p} + \frac{\hat g^q}{q}$ a.e.. Now, a fact from measure theory: if $\int F = \int G$, and $0 \le F \le G$ a.e., then $F = G$ a.e., demonstrating the impilcation "$\Rightarrow$". $\endgroup$
    – Alex Ortiz
    Commented Apr 21, 2020 at 1:35

I'll add some details on the Minkowski inequality (this question is the canonical Math.SE reference for the equality cases, but almost all of it concerns Hölder's inequality).

The standard proof of the Minkowski inequality begins with $$ \begin{align*} \int |f+g|^p &\le \int |f||f+g|^{p-1} + \int |g||f+g|^{p-1} \\ &\le \|f\|_p \| |f+g|^{p-1}\|_q + \|g\|_p \| |f+g|^{p-1}\|_q \end{align*} $$ where $q$ is the conjugate exponent to $p$. This simplifies to $\|f+g\|_p^p \le (\|f\|_p+\|g\|_p) \|f+g\|_p^{p-1} $ as wanted. So, if equality holds, it also holds in the two instances of Hölder's inequality above. Hence $|g|^p$ and $|f|^p$ are both constant multiples of $(|f+g|^{p-1})^q$, which makes them collinear vectors in $L^1$.

Additionally, the equality case requires $|f+g| = |f|+|g|$, which means the signs (or arguments, in the complex case) of $f$ and $g$ must agree a.e. where the functions are not zero. Conclusion: $f$ and $g$ are collinear vectors in $L^p$.


I want to add some details around the answer by "community wiki". This is essentially an answer to my own question.

Disclaimer : I did not found these details anywhere. What you are going to read is the result of my own (lack of) understanding of the subject. Feel free to comment if something seems unclear/wrong.

I drop the "almost everywhere" precaution.

There are reals (not both vanishing) $s$ and $t$ such that

$s|f(x)|^p=t|g(x)|^p$ (1)

And there are functions $a$ and $b$ taking values in the positive reals

$a(x)f(x)=b(x)g(x)$ (2)

I have to prove that there exist constants $\alpha$ and $\beta$ non both vanishing such that $\alpha f(x)=\beta g(x)$ for every $x$.

Case $s=0$

In that case $g(x)=0$ and we have $\alpha f(x)=\beta g(x)$ with $\alpha=0$ and $\beta=1$.

Case $s\neq 0$

Using the $p$-th root, we have

$|f(x)|=\lambda |g(x)|$

with $\lambda=\sqrt[p]{t/s}$. In particular, $a(x)$ and $b(x)$ are both non vanishing.

Case $s\neq 0$, $g(x)\neq 0$

Taking the module in (2), $a(x)\lambda|g(x)|=b(x)|g(x)|$, so that

$a(x)f(x)=\lambda a(x)g(x)$

and thus $f(x)=\lambda g(x)$.

I conclude that, if there are points in which $g(x)\neq 0$ we have to choose the constants $(\alpha,\beta)$ under the form (\alpha, \lambda\alpha).

Now we pass to the last case:

Case $s\neq 0$, $g(x)=0$

Equation (1) reads


with $s\neq 0$. Thus $f(x)=0$, and any $\alpha, \beta$ works.


If you read French, I wrote much more details here.


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