A detailed proof of Minkowski's inequality for integrals

Question

I'm reading Theorem 6.19 in textbook Real Analysis: Modern Techniques and Their Applications by Gerald B. Folland. The proof given by the author is very sketchy. I also change the original statement of part (ii) to make it more concise.

• Could you verify if my proof and my formulation of part (ii) are correct?

• In part (ii), the case $\infty$ is obtained by the monotonicity of integral. Why don't we do the same for part (i)?

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Let

• $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be $\sigma$-finite measure spaces, and $(E, \| \cdot \|)$ a Banach space.

• $\| \cdot \|_p$ is the $L_p$-norm.

• $\lambda = \mu \otimes \nu$ the product measure of $\mu$ and $\nu$.

• $f: X \times Y \to E$ is $\lambda$-measurable.

(i) If $1 \le p < \infty$ and $E = \mathbb R_+$, then $$\left[ \int_X \left( \int_Y f(x, y) \mathrm d \nu(y) \right )^{p} \mathrm d \mu(x) \right]^{1 / p} \leq \int_Y \left [ \int_X f^p (x, y) \mathrm d \mu(x) \right]^{1 / p} \mathrm d \nu(y).$$

(ii) If $1 \le p \le \infty$ and $f (x, \cdot)$ is $\nu$-integrable for $\mu$-a.e. $x \in X$, then $$\left [ \int_X \left \| \int_Y f(x, y) \mathrm d \nu(y)\right\|^{p} \mathrm d \mu(x)\right]^{1 / p} \leq \int_Y \left [ \int_X \|f(x, y)\|^{p} \mathrm d \mu(x) \right]^{1 / p} \mathrm d \nu(y).$$

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Proof: We start with (i). Let $R$ be the RHS and $L'_q(X, \mu, \mathbb R_+) \triangleq \{g \in L_q(X, \mu, \mathbb R_+) \mid \|g\|_q = 1\}$. The case $p=1$ is true by Tonelli's theorem. For $p >1$, let $q$ be its Hölder's conjugate and $h:X \to \mathbb R, x \mapsto \int_Y f(x, y) \mathrm d \nu(y)$.

First, we consider a function $\varphi: L'_q(X, \mu, \mathbb R_+) \to \overline{\mathbb R}_+, \, g \mapsto \int_X h g \mathrm d \mu$. By Hölder's inequality, $$\sup_{g \in L'_q(X, \mu, \mathbb R_+)} \varphi(g) \le \sup_{g \in L'_q(X, \mu, \mathbb R_+)} \|h\|_p \cdot \|g\|_q = \|h\|_p.$$

Because $h$ is non-negative and measurable, there is a non-decreasing sequence $(h_n)$ of non-negative simple functions converging $\mu$-a.e. to $h$. We define a sequence $(g_n)$ by $$g_n \triangleq \frac{h^{p-1}_n}{\|h_n\|_p^{p-1}}, \quad n \in \mathbb N.$$

It follows from $(p-1)q = p$ that $\|g_n\|_q =1$ and $\|h_n\|_p = \int_X h_n g_n \mathrm d \mu$. This means $g_n \in L'_q(X, \mu, \mathbb R_+)$ for all $n$. As such, \begin{align} \|h\|_p &= \lim_n \|h_n\|_p &&= \lim_n \int_X h_n g_n \mathrm d \mu \\ &\le \lim_n \varphi (g_n) && \le \sup_{g \in L'_q(X, \mu, \mathbb R_+)} \varphi(g). \end{align}

It follows that $\sup_{g \in L'_q(X, \mu, \mathbb R_+)} \varphi(g) = \|h\|_p$. For all $g \in L'_0(X, \mu, \mathbb R_+)$, \begin{align} \varphi (g) =& \int_Y \left ( \int_X f(x, y) g(x) \mathrm d \mu (x) \right ) \mathrm d \nu (y) \quad \text{by Tonelli's theorem} \\ \le& \int_Y \left ( \int_X f^p(x, y) \mathrm d \mu (x) \right )^{1/p} \left ( \int_X g^q(x) \mathrm d \mu (x) \right )^{1/q} \mathrm d \nu (y) \quad \text{by Hölder's inequality} \\ =& R. \end{align} Hence $R$ is an upper bound of $\varphi$ and thus $\|h\|_p \le R$. This completes the proof of (i). Now we're going to prove (ii). First, we consider the case $1 \le p < \infty$. Let's denote the LHS by $L$. Notice that $\|f\|$ is non-negative and measurable. We have \begin{align} L &\le \left [ \int_X \left ( \int_Y \| f(x, y) \| \mathrm d \nu(y)\right )^{p} \mathrm d \mu(x)\right]^{1 / p} \quad \text{because} \quad \left \|\int f \right \|^p \le \left | \int \|f\| \right |^p = \left ( \int \|f\| \right )^p\\ &\le \int_Y \left [ \int_X \|f(x, y)\|^{p} \mathrm d \mu(x) \right]^{1 / p} \mathrm d \nu(y) \quad \text{by (i)}. \end{align}

The case $p =\infty$ is then obtained by the monotonicity of integral.

Answer 1

Note that $\mathbb{R}_+$ is not a Banach space.

One place where your proof breaks down is in defining $g_n$. It is possible that $\lVert h_n \rVert_{L^p} = \infty$ for some $n$, so $g_n$ is not well defined. You need your $h_n$ to be in $L^p(X, \mu)$. The way Folland achieves this in his proof of theorem 6.14 is by using the $\sigma$-finite hypothesis on $X$: Take $E_n \subset X$ of finite measure such that $E_n \nearrow X$, and set $h_n' = h_n\chi_{E_n}$. Then $h_n'$ are simple, $h_n' \nearrow h$, and $h_n' \in L^{p}(X, \mu)$. I think the rest of your proof for (i) goes through.

Basically for $\sigma$-finite $(X, F, \mu)$, for measurable $h \geq 0$ we have $\lVert h \rVert_{L^p} = \sup\{\int_{X}hg\,d\mu : g \geq 0, \lVert g \rVert_{L^q} = 1\}$. You proved this for $p \in [1, \infty)$. A simple argument (which also uses $\sigma$-finiteness of $\mu$) proves it for $p = \infty$. Then the proof of the actual estimate is easy, and it works for all $p \in [1, \infty]$ since Holder's inequality holds for all such $p$. But this only works when $y \mapsto \lVert f(., y) \rVert_{L^p}$ is measurable. For $p < \infty$, this is a consequence of Tonelli's theorem, but I don't think it necessarily holds for $p = \infty$. So your statement for (ii) needs to be careful here.

I agree with you that trying to use monotonicity for $p = \infty$ is not as obvious as it looks because we only have $f(x, y) \leq \lVert f(., y) \rVert_{L^{\infty}}$ for a.e. $x \in X$, when we need it for a.e. $y \in Y$ for a.e. $x \in X$.