# Non-unique direction of steepest ascent

The gradient $\nabla f$ of a differentiable function $f(x,y)$ points in the direction of steepest ascent at a given point $(x_0,y_0)$. The slope of this ascent is the magnitude of the gradient $\|\nabla f(x_0,y_0)\|$. It seems possible that there could be a non-parallel vector $u$ such that the (directional) derivative $\nabla_{u}f (x_0,y_0)$ in the direction of $u$ is also equal to $\|\nabla f(x_0,y_0)\|$. Thus the direction of steepest ascent would not be unique - the gradient just happens to equal one of directions of steepest ascent. Can this happen?

Is there a function $f(x,y)$ and a point $(x_0,y_0)$ such that

1. $f$ is differentiable at $(x_0,y_0)$,
2. There are two unit vectors $u\neq v$ such that $$\nabla_{u}f (x_0,y_0)=\nabla_{v}f (x_0,y_0)=\|\nabla f(x_0,y_0)\|\neq 0?$$

In the cases I've thought of where there are obviously two methods of steepest ascent (like a hyperbolic paraboloid) the gradient is zero.

If the gradient is not zero, then the steepest ascent direction is unique. This follows from the Cauchy-Schwarz inequality: $|(u,v)| \leq \| u \| \| v \|$ with equality if and only if $u=cv$ for some scalar $c$. Applying that here, the directional derivative $\nabla_u f$ is $(u,\nabla f)$, so $|\nabla_u f| \leq \| u \| \| \nabla f \|$ with equality if and only if $u=c \nabla f$. For $c>0$ you are following the steepest ascent direction, for $c<0$ you are following the steepest descent direction.