# Is $|D(u)(p)|\leq\left|\frac{\partial u}{\partial \nu}(p)\right|$?

Let $\nu$ be the unit normal of $\partial U$, let $p\in \partial U$ and $u=0$ on $\partial U$.

Is it true that: $$|D(u)(p)|\leq\left|\frac{\partial u}{\partial \nu}(p)\right|$$

I.e. is it the case that:

$$|D(u)(p)|\leq\left|\nu \cdot D(u)(p)\right|$$

It is a fact that was used in my lecture, and I wasn't sure why it is true. I feel as though we could have that normal pointing at say $(1,0,\cdots,0)$ and what then if $D(u)(p)$ was concentrated in the other coordinates?

This is true because $u=0$ on $\partial U$ (otherwise there is no reason for it to be true). I suppose $\partial U$ is at least $C^1$, so it has a tangent space at the point $p$. Let $v_1,\dots,v_{n-1},v_n$ be a basis for $\mathbb{R}^n$ where $v_n=\nu$ and $v_1,\dots,v_{n-1}$ forms a basis for the tangent space to $\partial U$ at $p$. Then $Du(p)\cdot v_i=0$ for $i=1,\dots,n-1$, since $u=0$ on $\partial U$. Thus
$$Du(p) = \sum_{i=1}^n (Du(p)\cdot v_i)v_i = (Du(p)\cdot \nu)\nu,$$
and we have $|Du(p)|=|Du(p)\cdot \nu|$.