# How to prove the generalized integral inequality?

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.

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.
Edit: For showing the inequality on a cube, (assuming still zero on the boundary) let $$\Omega = [p,q]^n$$ for $$p. 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}
• 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 ? Commented Nov 8, 2023 at 1:45