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Problem of the Week #360 - July 21, 2020

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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,909
Here is this week's POTW:

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Let $f : \Bbb R^n \to \Bbb R$ be differentiable at $\mathbf{x}_0$. If $f(\mathbf{x}_0) = 0$ and $\mathbf{x}_0$ has an open neighborhood $V \subset \Bbb R^n$ such that $f(\mathbf{x}) \ge 0$ for all $\mathbf{x}\in V$, prove that $\mathbf{x}_0$ is a critical point of $f$, i.e., $\nabla f(\mathbf{x}_0) = \mathbf{0}$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,909
No one answered this POTW. You can read my solution below.

Fix $i\in \{1,\ldots, n\}$. If $h$ is a sufficiently small positive number such that $\mathbf{x}_0 \pm h\mathbf{e}_i\in V$, then $0 \le f(\mathbf{x}_0 \pm h\mathbf{e}_i) = f(\mathbf{x}_0) \pm h\partial_if(\mathbf{x}_0) + o(h) = \pm h\partial_i f(\mathbf{x}_0) + o(h)$ so that $-o(h) \le \partial_i f(\mathbf{x}_0) h \le o(h)$ or $-\frac{o(h)}{h} \le \partial_i f(\mathbf{x}_0) \le \frac{o(h)}{h}$. Letting $h \to 0$ results in $\partial_i f(\mathbf{x}_0) = 0$.