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Trying to prove Minkowski's inequality

$$\int dy \bigg[ \int f(x,y)^p dx\bigg]^{1/p} \geq \bigg[ \int dx \bigg(\int f(x,y)dy\bigg)^p\bigg]^{1/p}$$

1 Answers1

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Here is an approach using Fubini-Tonelli and Hölder's inequality. It does not use the triangle inequality.

We define $h(x) := \int_{\Omega_1} f(x,y) \mathrm{d}y$ for all $x \in \Omega_2$. We have to estimate $\|h\|_p$. We take an arbitrary $g \in L^{q}(\Omega_2)$ and get \begin{align*} \int_{\Omega_2} h(x) g(x) \mathrm{d}x &= \int_{\Omega_2} \int_{\Omega_1} f(x,y) g(x) \mathrm{d} y \mathrm{d} x \\ &= \int_{\Omega_1} \int_{\Omega_2} f(x,y) g(x) \mathrm{d} x \mathrm{d} y \\ &\le \int_{\Omega_1} \left(\int_{\Omega_2} f(x,y)^p \mathrm{d}x\right)^{1/p} \left(\int_{\Omega_2} g(x)^{q} \mathrm{d} x\right)^{1/q} \mathrm{d} y \\ &= \int_{\Omega_1} \left(\int_{\Omega_2} f(x,y)^p \mathrm{d} x\right)^{1/p} \mathrm{d} y \|g\|_{q} . \end{align*} Here, we used the Fubini-Tonelli theorem and Hölders inequality. Now, \begin{equation*} \|h\|_p = \sup_{\|g\|_{q} \le 1} \int_{\Omega_2} h(x) g(x) \mathrm{d}x \end{equation*} yields $$ \left( \int_{\Omega_2}\left(\int_{\Omega_1} f(x,y) \mathrm{d}y\right)^p \mathrm dx \right)^{1/p} = \|h\|_p \le \int_{\Omega_1} \left(\int_{\Omega_2} f(x,y)^p \mathrm{d} x\right)^{1/p} \mathrm{d} y $$

gerw
  • 31,359
  • sorry how does the last step give us the estimate? I think my idea was to write the LHS as $\int dy [\int f(x,y)^p dx]^{1/p} = \int dy \sup_{||g(x,y)||_Lq \leq 1} \int f(x,y)g(x,y) dx$ but im a little stuck with how to do that for the RHS –  Dec 08 '22 at 17:44
  • I have added some more terms. Does it help? – gerw Dec 08 '22 at 19:50
  • I get it up to the very last step how do we know that LHS = $||h||_p$? –  Dec 10 '22 at 03:28
  • Also is there a typo in the LHS should it be $\bigg( \int \bigg(\int f(x,y)dy\bigg)^p dx\bigg)^{1/p}$? –  Dec 10 '22 at 03:35
  • The last step is just the definition of $|h|_p$. And yes, an integral was missing on the left-hand side. – gerw Dec 10 '22 at 19:42