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Todd Trimble
Todd Trimble
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Deleted Deleted Deleted Deleted Deleted DeletedThanks Willie's suggestion, I give my answer as follow:

Assume $p\in B\left(0,1\right)$ so that $\left|\nabla\left(p\right)\right|=\underset{\overline{B}\left(0,1\right)}{\max}\left|\nabla f\right|$. Set $g\left(t\right):=f\left(p+\nabla f\left(p\right)t\right)$ then $g:\left(-\delta,\beta\right)\rightarrow\mathbb{R}$ is convex. It's easy to see that there is $t_{0}>0$ such that $p+t_{0}\nabla f\left(p\right)\in\partial B\left(0,1\right)$. Therefore, $g'\left(0\right)\leq g'\left(t_{0}\right)$, which leads to $$\left|\nabla f\left(p\right)\right|\leq\left|\nabla f\left(p+t_{0}\nabla f\left(p\right)\right)\right|$$, we get the desired result.

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Thanks Willie's suggestion, I give my answer as follow:

Assume $p\in B\left(0,1\right)$ so that $\left|\nabla\left(p\right)\right|=\underset{\overline{B}\left(0,1\right)}{\max}\left|\nabla f\right|$. Set $g\left(t\right):=f\left(p+\nabla f\left(p\right)t\right)$ then $g:\left(-\delta,\beta\right)\rightarrow\mathbb{R}$ is convex. It's easy to see that there is $t_{0}>0$ such that $p+t_{0}\nabla f\left(p\right)\in\partial B\left(0,1\right)$. Therefore, $g'\left(0\right)\leq g'\left(t_{0}\right)$, which leads to $$\left|\nabla f\left(p\right)\right|\leq\left|\nabla f\left(p+t_{0}\nabla f\left(p\right)\right)\right|$$, we get the desired result.

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user97743
user97743
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Thanks Willie's suggestion, I give my answer as follow:

Assume $p\in B\left(0,1\right)$ so that $\left|\nabla\left(p\right)\right|=\underset{\overline{B}\left(0,1\right)}{\max}\left|\nabla f\right|$. Set $g\left(t\right):=f\left(p+\nabla f\left(p\right)t\right)$ then $g:\left(-\delta,\beta\right)\rightarrow\mathbb{R}$ is convex. It's easy to see that there is $t_{0}>0$ such that $p+t_{0}\nabla f\left(p\right)\in\partial B\left(0,1\right)$. Therefore, $g'\left(0\right)\leq g'\left(t_{0}\right)$, which leads to $$\left|\nabla f\left(p\right)\right|\leq\left|\nabla f\left(p+t_{0}\nabla f\left(p\right)\right)\right|$$, we get the desired result.Deleted Deleted Deleted Deleted Deleted Deleted

Thanks Willie's suggestion, I give my answer as follow:

Assume $p\in B\left(0,1\right)$ so that $\left|\nabla\left(p\right)\right|=\underset{\overline{B}\left(0,1\right)}{\max}\left|\nabla f\right|$. Set $g\left(t\right):=f\left(p+\nabla f\left(p\right)t\right)$ then $g:\left(-\delta,\beta\right)\rightarrow\mathbb{R}$ is convex. It's easy to see that there is $t_{0}>0$ such that $p+t_{0}\nabla f\left(p\right)\in\partial B\left(0,1\right)$. Therefore, $g'\left(0\right)\leq g'\left(t_{0}\right)$, which leads to $$\left|\nabla f\left(p\right)\right|\leq\left|\nabla f\left(p+t_{0}\nabla f\left(p\right)\right)\right|$$, we get the desired result.

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user97743
user97743
  • 31
  • 4

Thanks Willie's suggestion, I give my answer as follow:

Assume $p\in B\left(0,1\right)$ so that $\left|\nabla\left(p\right)\right|=\underset{\overline{B}\left(0,1\right)}{\max}\left|\nabla f\right|$. Set $g\left(t\right):=f\left(p+\nabla f\left(p\right)t\right)$ then $g:\left(-\delta,\beta\right)\rightarrow\mathbb{R}$ is convex. It's easy to see that there is $t_{0}>0$ such that $p+t_{0}\nabla f\left(p\right)\in\partial B\left(0,1\right)$. Therefore, $g'\left(0\right)\leq g'\left(t_{0}\right)$, which leads to $$\left|\nabla f\left(p\right)\right|\leq\left|\nabla f\left(p+t_{0}\nabla f\left(p\right)\right)\right|$$, we get the desired result.