Prove that there exists no scalar field which has derivative $> 0$ at a fixed point wrt every direction

Question

The problem statement : Prove that there is no scalar field $f$ such that $\nabla_\vec{y}f(\vec{a})>0$ for a fixed vector $\vec{a}$ and every non-zero vector $\vec{y}$.
Here $\nabla_\vec{y}f(\vec{a})$ represents the derivative of $f$ at $\vec{a}$ with respect to $\vec{y}$.
$$ \nabla_\vec{y}f(\vec{a})=\lim_{h\rightarrow 0}\frac{f(\vec{a}+h\vec{y})-f(\vec{a})}{h}$$ I proved it using that if $\vec{y}\neq0$ then $$\nabla_\vec{-y}f(\vec{a})=-\nabla_\vec{y}f(\vec{a})<0$$ This gives a contradiction. Is there any other way to prove this ?

Answer 1

I think that every proof you find is going to be equivalent to the one you already gave, but here is a proof by meme. If such a function existed, it would be possible to walk from one point on the graph to another, uphill both ways.

Answer 2

Note that the directional derivative is the inner product of the gradient vector at a given point and the unit vector in some given direction. Now, if we replace our unit vector with the negative of the vector, the directional derivative changes its sign, and it is due to the definition of the inner product.

Answer 3

Let's call $\nabla^{\vec{a}} f : \vec{y} \rightarrow \nabla_{\vec{y}} f (\vec{a})$.

$\nabla^{\vec{a}}$ is a linear application, which can only be either identically zero (which is forbidden by hypothesis) or take both positive and negative values.

(Of course this ends up being much the same as your proof, especially if you want to give explicit illustration of the statement)