On continuity and existence of all directional derivatives at a point of a scalar field whose gradient at that point is $\vec 0$

Question

Let $f:\mathbb R^2 \to \mathbb R$ be a function such that for some $a \in \mathbb R^2$ , $\nabla f(a)$ exists and equals $\vec 0$.

1. Is $f$ necessarily continuous at $a$ ?
2. Do all directional derivatives $f'(a;y)$ exist at $a$?
3. Suppose $f$ is continuous at $a$ and $\nabla f(a)=0$. Do the directional derivatives at $a$ exist in this case?

Answer 1

The answer to all your questions is "no".

For 1, there are discontinuous functions whose partial derivatives exist and are equal to zero in all directions.

As for 2 and 3., the answer is "no" again. Consider $f(x,y)=x^{1/3}y^{1/3}$. $f(x,y)$ is continuous at the origin, and the gradient is zero at $(0,0)$ because $f$ is zero along the axes, but it doesn't have directional derivatives in other directions: Indeed: $$\frac{f(hy_1,hy_1)-f(0,0)}{h}=\frac{h^{2/3}y_1^{1/3}y_2^{1/3}}{h}\approx h^{-1/3}$$ and so the limit does isn't finite as $h\to 0$.

Answer 2

It's not necessarily even continuous at the point since the derivatives only tell you about continuity in the directions of the coordinate axes. As a counterexample consider $$f(x,y)= x^2+y^2, x \neq y$$ $$=1, x=y \neq 0$$ $$=0, x=y=0$$

At (0,0), $\nabla f = <0,0>,$ but the function is not continuous at the origin.

Answer 3

$\newcommand{\Reals}{\mathbf{R}}$Let $f:\Reals^{2} \to \Reals$ be a function, and assume $\nabla f(0) = 0$.

1. Is $f$ necessarily continuous at $0$?

   • No: Define $$ f(x, y) = \begin{cases} 1 & \text{if $y = x^{2}$ and $x \neq 0$,} \\ 0 & \text{otherwise.} \end{cases} $$ In this example, all directional derivatives of $f$ exist and are $0$ at the origin, but $f$ is discontinuous at the origin. (The file linked by uniquesolution gives a rational function having the same properties.)

2. Do all the directional derivatives of $f$ exist at $0$?

   • No: Define $$ f(x, y) = \begin{cases} 0 & \text{if $xy = 0$,} \\ 1 & \text{otherwise.} \end{cases} $$ (As Paul notes, partial derivatives at the origin only measure behavior along the coordinate axes.)

3. What if $f$ is continuous?

   • No: Define $f(x, y) = \min\bigl(|x|, |y|\bigr)$, which vanishes along both coordinate axes (so both partials exist at the origin) and is continuous (as the minimum of two continuous functions), but fails to have directional derivatives at the origin along every non-coordinate direction.