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Has anybody got an idea how the equations

$\max\limits_{\partial B_r(x)}u = u\left(x+r \frac{Du(x)}{|Du(x)|}\right)$

$\min\limits_{\partial B_r(x)}u = u\left(x-r \frac{Du(x)}{|Du(x)|}\right)$

arise for linear and non-constant $u:\mathbb{R}^n\to\mathbb{R}$? It seems heuristically correct but I can't find a way to prove it. I tried to play with the linearity of $u$ around a little bit but I couldn't get near those equations unfortunately.

HelloEveryone
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    This just expresses the fact that the gradient is the direction with steepest increase. The point $x+r \frac{Du(x)}{|Du(x)|}$ is just the point of the spheric surface when you leave $x$ along the direction of the gradient. Of coarse, the key aspect here is the when $u$ is linear, the gradient is constant. – PierreCarre Jun 24 '22 at 09:16
  • Hi and thank you for your answer! This is heuristically completely logical and makes sense. Nevertheless is it not really satisfying as I am looking for a "formal" proof for it... – HelloEveryone Jun 24 '22 at 09:34

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As $u$ is linear, it's gradient $Du = d$ is constant, as pointed by @PierreCarre. You can then write $$u(x + h) = u(x) + \langle Du(x), h \rangle = u(x) + \langle d, h \rangle. $$ You can then use the Cauchy-Schwarz inegality to demonstrate you result.

Adrian Jarret
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  • Thanks for helping me. I wasn't able until now to really use your hint though.

    Of Course $u\left(x+r\frac{Du}{|Du|}\right)\leq \max\limits_{\partial B_r(x)}u$.

    Now there is an $s$ with $|s|=1$ as every linear function on an finite dimensional space is continiuous, such that

    $\max\limits_{\partial B_r(x)}u = u(x+rs) = u(x)+\langle d,rs\rangle \leq u(x)+|d|r$. But how can I continue? I don't see how $Du=const.$ helps me...also isn't there a term of $o(h)$ missing?

     – HelloEveryone Jun 24 '22 at 12:12
  • $o(h)$ is not missing as $u$ is a linear function. You have $u(x + rs) = \max_{|t| = 1} u(x + rt)$ which leads to $\langle d, s\rangle = \max_{|t| = 1} \langle d, t\rangle$. Then $s$ satisfies the optimality condition of the Cauchy-Schwarz inequality. – Adrian Jarret Jun 24 '22 at 12:29
  • Thank you! Think I've understood the rest of the argument as then $d$ and $s$ are l.d. and so on...but I can't see how you derive $\langle d,s\rangle = \max\limits_{|t|=1}\langle d,t\rangle$ from the steps before... – HelloEveryone Jun 24 '22 at 13:25
  • I think I understood it and proved it now. But one thing I didn't understand til now, is: where do we use that $d=Du(x)$ is constant? – HelloEveryone Jun 24 '22 at 17:05
  • Forget it, found my mistake! Cheers – HelloEveryone Jun 24 '22 at 17:25