# Why tangent surface is a plane

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 $$f$$:

$$f(\left) = z - \min(\lvert x \rvert, \lvert y \rvert)\times\frac{\lvert y \rvert}{y}$$ At the origin, along direction $$\hat u=\left<\frac{1}{\sqrt 3}, \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3} \right>$$: $$\partial_{\hat u} f(\vec 0) = \lim_{t \to 0} \frac{f(t \hat u) - f(\vec 0)}{t} = \lim_{t \to 0} \frac{\left(\frac{1}{\sqrt 3} - \min\left(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\right)\times 1\right) - 0}{t} = \lim_{t \to 0} \frac{0 - 0}{t} = 0\\ \hat u \cdot \nabla f(\vec 0) = \hat u \cdot \left = \left<\frac{1}{\sqrt 3}, \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3} \right> \cdot \left<0,0,1\right> = \frac{1}{\sqrt 3}\\ \partial_{\hat u} f(\vec 0) \ne \hat u \cdot \nabla f(\vec 0)$$

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:

1. does $$\nabla f(\vec 0)$$ exist ?
2. 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.

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 function $$f: A \to \mathbb{R}^n$$, $$A \subseteq \mathbb{R}^m$$ is differentiable at a point $$\mathbf a \in \mathbb R^n,$$ if there is a linear transformation $$T$$ such that $$\lim_{\lVert \mathbf h\rVert \to 0} \frac{f(\mathbf a+\mathbf h)-f(\mathbf a)-T(\mathbf h)}{\lVert \mathbf h\rVert} = \mathbf 0.$$

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 $$\mathbf 0.$$

• Thanks, I have written a proof that $f$ is not differentiable at the origin in my question. And for linearity of partial derivative, I have an already answered question here math.stackexchange.com/questions/3079820 – Zang MingJie Jan 19 '19 at 21:35
You are taking a derivative where your function doesn't have one. $$f(x) = |x|$$ does not have a derivative at 0. However it is convex and has the subdifferential $$[-1,1]$$ at $$0$$.
Edit: Your partial derivative wrt $$x$$ is not smooth. Consider $$|y|>0$$, then the function $$g(x) = \max(|x|,|y|)\frac{|y|}{y}$$ does not have a derivative at $$x=0$$. So even if your function has partial derivatives at $$(0,0,0)$$, the partial derivative wrt $$x$$ is not continuous.
• My $f$ is already carefully chosen that the derivative exist for all directions. you can try derivative it using limitation. – Zang MingJie Jan 18 '19 at 12:02
• @Zang MingJie It is discontinuous along $x=0$ and also along $y=0$. Verify that the left and right derivatives there do not match. – lightxbulb Jan 18 '19 at 12:15
• @lightxbulb $f \vert_{x=0} = z - \min(0, \lvert y \rvert)\times \frac{\lvert y \rvert}{y} = z$ can you explain why it is discontinuous. – Zang MingJie Jan 18 '19 at 12:33
• @Randall this maybe the answer I need, I want to know exactly the requirements for a function be differentiable. Any why my $f$ is not differentiable. – Zang MingJie Jan 18 '19 at 12:35