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When I read the Normal derivative of Wiki and this question, I am confuse with them.

Assume $\Omega$ is bounded open subset of $R^n$ and $\partial\Omega$ is smooth, and some smooth function $f$ reach its max in $x_0\in\partial\Omega$.

Then why $\frac{\partial f}{\partial \overrightarrow n}(x_0)\le 0$ ? $n$ is out normal.

If the define of direct derivative is $$ \frac{\partial f}{\partial n}(x_0)=\lim_{t\rightarrow 0}\frac{f(x_0+t\overrightarrow n)-f(x_0)}{t} $$

$x_0+t\overrightarrow n$ is not in $\Omega$, so, $f(x_0+t\overrightarrow n)$ is meaningless. So, I don't know how to define the direct derivative along the out normal when $x_0\in \partial\Omega$.

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Enhao Lan
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2 Answers2

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It's even worse: since $\Omega$ is open, $\partial \Omega \cap \Omega = \emptyset$, so it's not even clear if $\lim_{x \to x_0}f(x)$ itself exists. Take for example $f(x) = \frac{1}{x}$ on the open interval $(0,1)$.

So, in order to define the directional derivative of $f$ at $\partial \Omega$, $f$ needs to be defined (and continuously differentiable) on an open set $U$ containing $\Omega$, i.e. $\Omega \subset U$.

However, if $f$ is bounded (on $\Omega$), you can say a lot more -- in particular, you can extend $f$ to an open subset containing $\Omega$, which is exactly what you need; I believe this is an application of the Whitney extension theorem, see also this question.

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Frits Veerman
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If $f$ extends continuously to the boundary, you could just restrict $t$ to negative real numbers in your definition, and everything should be fine.

Steven Gubkin
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