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From the result proven in How prove this, we can easily prove that if $\Omega$ is an open, bounded smooth subset of $\mathbb{R}$ then: $\int_{\Omega} f^2 dx \le (\sup_{x,y\in \Omega} |x-y|)^{2} \int_{\Omega} |Df|^2 dx$ when $f \in C^{1}(\bar{\Omega})$ and $f=0$ on the boundary of $\Omega$. Now how to use that to prove the generalized result in $\mathbb{R}^{n}$ ? I don't seem able to get anywhere with it. Any help is appreciated.

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It seems to me this question will not have an easy generalisation to the $\mathbb{R}^n$ case with out at least more assumptions on the domain $\Omega$. At the very least, I'd imagine you'd need to assume that $\Omega$ is convex so that it contains line segments. For motivation you could try the proof assuming the domain is a cube, or else convex and zero outside of some cube. It would be much easier to apply the fundamental theorem of calculus similar to the post you linked.

To see a full proof with the necessary assumptions (and a sharper constant), you could look here https://ems.press/content/serial-article-files/1064#:~:text=The%20classical%20proof%20for%20the,general%20assumptions%20on%20%CE%A9%20(cf. or here https://link.springer.com/article/10.1007/BF00252910 In these proofs they break the convex domain in to smaller convex domains which can be approximated sensibly by cubes. There's also more discussion here Constructive proof of Poincaré's inequality on convex domains . If you're not aware, this type of inequality is called a Poincare inequality and is used frequently in the study of PDEs. If you want to drop the convexity assumption you can, but you will lose the constant that you want in the inequality for a different one. Hope that helps!

Edit: For showing the inequality on a cube, (assuming still zero on the boundary) let $\Omega = [p,q]^n$ for $p<q$. Then $$ \begin{aligned} \int_{\Omega}f(x)^2 dx &= \int_p^q ... \int_p^q f(x_1, \dots, x_n)^2dx_1\dots dx_n \\ &= \int_{\Omega} \left( \int_p^{x_1} \frac{\partial f}{\partial x_1} (y, \dots, x_n) dy \right)^2 dx_1\dots dx_n \\ &\leq \int_{\Omega} (x_1 - p)\left( \int_p^{x_1} \frac{\partial f}{\partial x_1} (y, \dots, x_n) ^2 dy \right)dx_1\dots dx_n \\ &\leq (q-p)\int_{\Omega} \left( \int_p^{q} \frac{\partial f}{\partial x_1} (y, \dots, x_n) ^2 dy \right)dx_1\dots dx_n \\ &= (q-p)^2\int_{\Omega} \frac{\partial f}{\partial x_1} (y, \dots, x_n) ^2 dy dx_2\dots dx_n \\ &\leq \text{diam}({\Omega})^2\int_{\Omega} \frac{\partial f}{\partial x_1} (y, x_2, \dots, x_n) ^2 + \dots \frac{\partial f}{\partial x_n} (y, x_2 \dots, x_n) ^2 dy dx_2 \dots dx_n \end{aligned} $$

which is the desired inequality. Basically use the exact same argument as the one you linked but for a partial derivative, and follow the calculation through unless I'm missing something.

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  • $\begingroup$ Thanks for your response ! Pretty interesting reading. Isn't it that since $\Omega$ is an open bounded subset of $\mathbb{R}^{n}$, then it is a countable union of disjoint open cubes, so it suffices to prove the result over an open cube ? I don't seem able to prove it for a cube though...any hints on how to generalize the result I linked to that case ? $\endgroup$
    – elaoumam
    Nov 8, 2023 at 1:45
  • $\begingroup$ I've added an edit with a calculation for a cube. Interesting point, I guess you could do that and drop the convexity condition to get what you need. But you can see how conexity allows you to use a similar argument to the one for the cube but projecting in the direction of a coordinate axis to the boundary of the domain, as in the linked articles. Using convexity this way I guess gives sharper inequalities than the one you're after. $\endgroup$
    – brighton
    Nov 8, 2023 at 11:49
  • $\begingroup$ Thanks a lot ! It does work and it's quite easier than I thought, like for quite some time I didn't see that f vanished on the sides of the cube not only on its vertices... $\endgroup$
    – elaoumam
    Nov 8, 2023 at 17:07

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