if $f$ with continuous partial derivatives at $(x_0,y_0)$. Then there exists a unit vector $\vec{u}$ which $D_\vec{u} (f)(x_0,y_0)=0$

Question

Prove or disprove:

let $f$ be a function with continuous partial derivatives at $(x_0,y_0)$. Then there exists a unit vector $\vec{u}$ for which $D_\vec{u} (f)(x_0,y_0)=0$

It feels like it's not true, but I'm new to this material and don't really understand it.

score 1 · Accepted Answer · answered Jul 11 '17 at 17:17

If the range of $f(x, y)$ is $\Bbb R$, which I assume here, then:

For any vector $\vec w = (w_x, w_y)$ at $(x_0, y_0)$, we have

$D_{\vec w}f(x_0, y_0) = \lim_{t \to 0}\dfrac{f(x_0 + t w_x, y_0 + tw_y) - f(x_0, y_0)}{t} = \nabla f(x_0, y_0) \cdot \vec w. \tag{1}$

Now if

$\nabla f(x_0, y_0) = 0, \tag{2}$

then clearly

$D_{\vec w} f(x_0, y_0) = \nabla f(x_0, y_0) \cdot \vec w = 0 \tag{3}$

for all $\vec w$; if

$\nabla f(x_0, y_0) \ne 0, \tag{4}$

then we may choose $\vec w$ normal to $\nabla f(x_0, y_0)$, e.g.,

$\vec w = (-f_y(x_0, y_0), f_x(x_0, y_0)) \ne0, \tag{5}$

where $\nabla f = (f_x, f_y)$. Since $\vec w \ne 0$, we may take

$\vec u = \dfrac{\vec w}{\Vert \vec w \Vert}; \tag{6}$

we then have

$\Vert u \Vert = 1, \tag{7}$

and

$D_{\vec u} f(x_0, y_0) = 0, \tag{8}$

as desired.

Geometrically: Just take a vector tangent to a level curve of $f$ (more generally, level hypersurface) — Ted Shifrin, Jul 11 '17 at 18:25
@Ted Shifrin: indeed, and such tangents are $\bot$ to $\nabla f$. — Robert Lewis, Jul 11 '17 at 18:46

if $f$ with continuous partial derivatives at $(x_0,y_0)$. Then there exists a unit vector $\vec{u}$ which $D_\vec{u} (f)(x_0,y_0)=0$

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