If $f(x) \leq f(a)$ for all $x$ in an open ball $B(a)$ , then prove that $\nabla f(a) = 0$

Question

Question statement : Assume $f$ is a scalar field function differentiable at each point of an n-ball $B(a)$ . If $f(x) \leq f(a) \ \ \forall \ x $ in an open ball $B(a)$ , then prove that $\nabla f(a) = 0$
My try : $f(a+he_i)-f(a)\leq0$ where $e_i$ is a unit coordinate vector. Now I divide both sides by $h>0$ and take limit $\lim_{h\to 0} \frac{f(a+he_i)-f(a)}{h}\leq0$ which gives $\frac{\partial f}{\partial x_i}\leq0$ at $a$. Doing the same thing, but now dividing both sides by $h<0$ and take limit $\lim_{h\to 0} \frac{f(a+he_i)-f(a)}{h}\geq0$ which gives $\frac{\partial f}{\partial x_i}\geq0$ at $a$. Thus $\frac{\partial f}{\partial x_i}=0$ at $a$ for all $x_i, \ i \in {1,2,...n}$. Thus $\nabla f(a) = 0$.

I have no clue as to whether this "weird" method of mine is correct or not. Someone please guide.

Answer 1

Your proof is fine. Alternatively, if you know the connection between the directional derivative and the gradient $$ \nabla f(a) \cdot v = \nabla_v f(a) = \lim_{h \to 0} \frac{f(a+hv)-f(a)}{h} $$ then you can set $v = \nabla f(a)$ and conclude that $$ \Vert \nabla f(a) \Vert^2 = \lim_{h \to 0+} \frac{f(a+h \nabla f(a))-f(a)}{h} \le 0 \, . $$

Answer 2

You can also think of that like that.

Let $a=(a_{1},a_{2},...,a_{n})$.

Freeze $n-1$ of them - then $f$ is a function of one variable $x_{i}$ which has local maximum in $a_{i}$ so $df/dx_{i}=0$ at $a$.