Given function $f(\vec x) \in \mathbb R, \vec x \in \mathbb R^n $, We know that for all unit direction $\hat u$:
$$
\partial_{\hat u} f = \hat u \cdot \nabla f
$$
Consider function $f$:
$$
f(\left<x,y,z\right>) = z – \min(\lvert x \rvert, \lvert y \rvert)\times\frac{\lvert y \rvert}{y}
$$
At the origin, along direction $\hat u=\left<\frac{1}{\sqrt 3}, \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3} \right>$:
$$
\partial_{\hat u} f(\vec 0) = \lim_{t \to 0} \frac{f(t \hat u) – f(\vec 0)}{t} = \lim_{t \to 0} \frac{\left(\frac{1}{\sqrt 3} – \min\left(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\right)\times 1\right) – 0}{t} = \lim_{t \to 0} \frac{0 – 0}{t} = 0\\
\hat u \cdot \nabla f(\vec 0) = \hat u \cdot \left<f_x,f_y,f_z \right> = \left<\frac{1}{\sqrt 3}, \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3} \right> \cdot \left<0,0,1\right> = \frac{1}{\sqrt 3}\\
\partial_{\hat u} f(\vec 0) \ne \hat u \cdot \nabla f(\vec 0)
$$
So there must be other requirements for:
$$
\partial_{\hat u} f = \hat u \cdot \nabla f
$$
From the graph of the function $f$, we can see that even the directional derivative is defined for all direction at the origin, the tangent surface at the origin is not a plane.
Here are my questions:
- does $\nabla f(\vec 0)$ exist ?
- what's the requirement for $\partial_{\hat u} f = \hat u \cdot \nabla f$
Edit: Proof that $f$ is not differentiable at the origin
For $f$ to be differentiable, there is a linear transformation $T$ such that:
$$
\lim_{\lVert \mathbf h\rVert \to 0}
\frac{f(\mathbf a+\mathbf h)-f(\mathbf a)-T_a(\mathbf h)}{\lVert \mathbf h\rVert} = \mathbf 0.
$$
Since $\mathbf h$ can approach to $\mathbf 0$ follow any trajectory, assume $\mathbf h$ is following direction $\hat u$, $\mathbf h = t \hat u$ and $\lVert \hat u \rVert = 1$, then we have:
$$
\lim_{t \to 0}
\frac{f(\mathbf a+t \hat u)-f(\mathbf a)-T_a(t \hat u)}{t} = 0\\
\lim_{t \to 0}\frac{f(\mathbf a+t \hat u)-f(\mathbf a)}{t} = \lim_{t \to 0}\frac{T_a(t \hat u)}{t}\\
$$
To the left we have:
$$
\lim_{t \to 0}\frac{f(\mathbf a+t \hat u)-f(\mathbf a)}{t} = \partial_{\hat u} f(a)
$$
To the right, since $T$ is linear, we have:
$$
\lim_{t \to 0}\frac{T_a(t \hat u)}{t} = \lim_{t \to 0}\frac{t T_a(\hat u)}{t} = T_a(\hat u)
$$
So
$$
\partial_{\hat u} f(a) = T_a(\hat u)
$$
$T$ is linear and I have already proofed that $\partial_{\hat u} f(0)$ is not linear (above in the question). So $f$ is not differentiable at the origin.
What have surprised me is that I thought linearity of $\partial_{\hat u} f$ is a property of differentiable function, but in reality, it is a requirement. Although, technically there is no difference between property and requirement, they are both $\Rightarrow$ in math world.
Best Answer
In order to ensure that $\partial_{\hat u} f = \hat u \cdot \nabla f$ at a given point $\mathbf a \in \mathbb R^n,$ you can simply require that the function $f$ is differentiable at $\mathbf a.$
There may be weaker conditions that ensure that $\partial_{\hat u} f = \hat u \cdot \nabla f$, but I do not know what they are. Differentiability seems to be the one usually used.
Of course you must also define what it means for a function over $\mathbb R^n$ to be differentiable at $\mathbf a.$ Here is a typical definition:
A good definition of the gradient $\nabla f$, in turn, would require that $f$ be differentiable at every point where $\nabla f$ is defined.
Your function does not have such a derivative at $\mathbf 0.$
There is more that can be said about this, but that is the fundamental issue.