# Prove Minkowski's Inequality for Integrals

I am interested in proving the following claims :

Suppose that ($$X$$, $$\mathcal{M}$$, $$\mu$$) and ($$Y$$, $$\mathcal{N}$$, $$\nu$$) are $$\sigma$$-finite measure spaces, and let $$f$$ be an ($$\mathcal{M} \otimes \mathcal{N}$$)-measurable function on $$X \times Y.$$

a) If $$f \ge 0$$ and $$1 \le p < \infty$$, then

$$\left[\int \left(\int f(x,y) d\nu(y) \right)^pd\mu(x)\right]^\frac{1}{p} \le \int \left[\int f(x,y)^p d\mu(x)\right]^\frac{1}{p}d\nu(y)$$

b) If $$1 \le p \le \infty$$, $$f(\cdot, y) \in L^p(\mu)$$ for a.e. $$y$$, and the function $$y \to ||f(\cdot, y)||_p$$ is in $$L^1(\nu)$$, then $$f(x, \cdot) \in L^1(\nu)$$ for a.e. $$x$$, the function $$x \to \int f(x,y) d\nu(y)$$ is in $$L^p(\mu)$$, and $$\left|\left|\int f(\cdot, y)d\nu(y)\right|\right|_p \le \int||f(\cdot, y)||_pd\nu(y).$$

• Commented Dec 20, 2019 at 16:06
• As I understand, $f(.,y)$ is a function of one variable for each $y$. So does $\int f(.,y)dy$ make sense? Commented Jan 6, 2020 at 17:53
• Hey I'm looking for references of generalizations of the Minkowski inequality. I'm not an expert in functional analysis. Does b) is true for any f and any norm that satisfies such conditions? On b it not necessary to assume that f is positive? Commented Nov 23, 2020 at 12:20
• How would one use Theorem 6.14 as claimed by Folland? Commented Dec 6, 2022 at 23:52

I am writing this answer just because it took me a while to understand CQNKZX's answer. I think that expanding it can help other people who, like me, were lost and really want to understand what is going on. So I decided to write a more detailed answer based entirely on the arguments of CQNZKX and Folland.

Let $$p\in (1,\infty)$$ and $$q$$ such that $$1/p + 1/q = 1$$. First of all, remember that the map \begin{align*} \Lambda: L^p(\mu) &\to L^q(\mu)^*\\ f &\mapsto \left(\Lambda(f)(g):= \int f(x)g(x)\ \mathrm{d} \mu\right) \end{align*} is an isometry with its image. Let $$g\in L^q(\mu)$$

\begin{align} \left|\Lambda\left(\int f(\cdot,y)\mathrm{d}\nu(y) \right)(g)\right|&= \int \left(\int f(x,y)\mathrm{d}\nu(y)\right)g(x) \mathrm{d}\mu(x)\\ &=\int \int f(x,y)g(x) \mathrm{d}\mu(x)\mathrm{d}\nu(y) \ \ \ \ \ \ \ \ \ \quad (1) \\ &\leq \int \left \Vert f(\cdot,y) \right\Vert_p \|g\|_q\ \mathbb{d} \nu(y) \ \ \ \ \ \ \ \ \quad \quad \ \ \quad (2)\\ &\leq \|g\|_q \int\|f(\cdot,y)\|_p \ \mathrm{d} \nu(y). \end{align}

In (1), we have used Fubinni's theorem, and in (2), Holder's inequality.

Since $$\Lambda$$ is an isometry and the above equality holds for every $$g\in L^q(\mu)$$

$$\left\|\int f(\cdot,y)\ \mathrm{d}\nu(y) \right\|_p = \left\| \Lambda\left( \int f(\cdot,y)\ \mathrm{d}\nu(y) \right)\right\|_{\mathrm{operator}} \leq \int\|f(\cdot,y)\|_p \ \mathrm{d} \nu(y).$$

If $$p=1$$, the inequality is trivial.

• Is this the proof of b)? And a)? Commented Jan 26, 2021 at 23:33
• I guess the condition $f \ge 0$ is to guarantee those integrals are well defined (including $+\infty$)? Commented Sep 30, 2022 at 21:38

This is several year late, but here is another proof also based on Holder's inequality:

Without loss of generality we can assume that $$f\geq0$$. The case $$p=1$$ is a restatement of Fubini's theorem. Suppose that $$p>1$$ and let $$H(x)=\int_Y f(x,y)\,\nu(dy)$$. From Fubini's theorem and then H"older's inequality we obtain \begin{aligned} \|H\|^p_{L_p(\mu)} &=\int_X \int_Y f(x,y)\,\nu(dy) H^{p-1}(x)\,\mu(dx) =\int_Y \int_X f(x,y) H^{p-1}(x)\,\mu(dx)\,\nu(dy)\\ &\quad \leq \int_Y \Big(\int_X|f(x,y)|^p\,\mu(dx)\Big)^{\tfrac{1}{p}}\|H\|^{p-1}_{L_p(\mu)} \,\nu(dy). \end{aligned} The conclusion follows immediately if $$\|H\|_p<\infty$$. If $$\|H\|_p=\infty$$, choose monotone sequences of sets $$A_n\subset X$$ and $$B_n\subset Y$$ such that $$\mu(A_n)\vee\nu(B_n)<\infty$$, and for any $$k\in\mathbb{N}$$ define $$f_k=f\vee k$$. Then $$\left(\int_{A_n}\Big(\int_{B_m}f_k(x,y)\,\nu(dy)\Big)^p\,\mu(dx) \right)^{1/p}\leq \int_{B_m}\Big(\int_{A_n}|f_k(x,y)|^p\,\mu(dx)\Big)^{1/p}\,\nu(dy).$$ Letting first $$k\rightarrow\infty$$, then $$n\rightarrow\infty$$ and finally $$m\rightarrow\infty$$ we obtain the desired result.

• I like this proof more than the standard duality method because it's very analogous to the way the Minkowski triangle inequality en.wikipedia.org/wiki/Minkowski_inequality is proven!
– D.R.
Commented Dec 6, 2022 at 5:08
• Do we have the reversed inequality in case $p \in (0, 1)$? Commented Jun 28 at 9:42
• @Akira: I think id does for $f\geq0$. The proof I gave relays on Holder's inequality. You may try to use this version. Commented Jun 28 at 14:05

Following Folland's proof (the inequality after applying Tonelli and Holder), consider $\int f(x,y) \,dν(y)$ as a linear functional(not necessarily bounded) on $L_q(\mu)$. If it's bounded, then $\int f(x,y) \,dν(y)$ must be in $L_p(\mu)$ and the result is immediate. Otherwise the RHS must be infinity.