i found following inequality in a textbook. Sadly there was no explanation or proof for it. I cant figure out one neither. Can someone help/has an idea?

$\Omega \subset \mathbb{R}^2$ be a bounded domain and $x_c$ is centroid with $$\int _\Omega (x-x_c) dx = 0$$

$C>0$ constant

a) for a smooth and bounded function (bounded derivatives) $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ it holds $$|\int _\Omega f(x)dx -|\Omega|f(x_c)|\leq C|\Omega|diam^2(\Omega)$$

b) for $x_a \in \Omega$ then $$|\int _\Omega f(x)dx -|\Omega|f(x_a)|\leq C|\Omega|diam(\Omega)$$

  • $\begingroup$ Seems like you could multiply $f$ with large constants and violate the inequality $\endgroup$ – zhw. May 10 '18 at 18:20

Here is a way to see b):

$$|\int _\Omega f(x)dx -|\Omega|f(x_a)|= |\int _\Omega f(x) - f(x_a)dx| = |\int _\Omega < grad(f)(\xi_x) , x - x_a>dx|$$ $$ \leq \int _\Omega |< grad(f)(\xi_x) , x - x_a>|dx \leq \int _\Omega C ||x-x_a||dx \leq C|\Omega|diam(\Omega)$$

  • $\begingroup$ does that hold true if C is independed of $\Omega$? $\endgroup$ – Uralmok May 11 '18 at 1:55
  • $\begingroup$ Yes. Since according to your assumptions, $grad(f)$ is globally bounded: $||grad(f)||\leq C$ on each $\Omega$. $\endgroup$ – trancelocation May 11 '18 at 4:23

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