Prove Minkowski's Inequality for Integrals

Question

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 $$\left|\left|\int f(\cdot, y)d\nu(y)\right|\right|_p \le \int||f(\cdot, y)||_pd\nu(y).$$

Answer 1

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.

Answer 2

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.

Answer 3

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.