Is this condition necessary for a function $f : \mathbb{R}^m \to \mathbb{R}^n$ to be differentiable?

Question

Consider the following definitions:

A function $f : U \subseteq \mathbb{R}^m \to \mathbb{R}^n$ (where $U$ is an open set) is differentiable at a point $a \in U$ if there exists a linear map $T : \mathbb{R}^m \to \mathbb{R}^n$ such that

$$\lim_{v \to 0} \dfrac{f(a+v) - f(a) - T(v)}{\|v\|} = 0$$

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The directional derivative of $f$ at $a$ in the direction of $v \in \mathbb{R}^m$ is

$$\dfrac{\partial f}{\partial v}(a) = \lim_{t \to 0} \dfrac{f(a+tv) - f(a)}{t}$$

Now, it is easy to show that $\dfrac{\partial f}{\partial (\alpha v)}(a) = \alpha\dfrac{\partial f}{\partial v}(a)$ for any $v$ and $\alpha \neq 0$ (if it exists).

It is also easy to show that if $f$ is differentiable at $a$, then $\dfrac{\partial f}{\partial v}(a)$ exists for all $v$, with the special property that

$$\forall \, v,w \in \mathbb{R^m} \qquad \dfrac{\partial f}{\partial (v + w)}(a) = \dfrac{\partial f}{\partial v}(a) + \dfrac{\partial f}{\partial w}(a)$$

So far so good.

Now, there is an equivalent definition of differentiability at a point, as follows:

(Alternative Definition) A function $f : U \subseteq \mathbb{R}^m \to \mathbb{R}^n$ (where $U$ is an open set) is differentiable at a point $a \in U$ if all the following conditions hold:

• For all $v \in \mathbb{R}^m$ the directional derivative $\dfrac{\partial f}{\partial v}(a)$ exists.

• For all $v,w \in \mathbb{R^m}$, we have $\dfrac{\partial f}{\partial (v + w)}(a) = \dfrac{\partial f}{\partial v}(a) + \dfrac{\partial f}{\partial w}(a)$.

• $\lim_\limits{v \to 0} \dfrac{f(a+v) - f(a) - \frac{\partial f}{\partial v}(a)}{\|v\|} = 0$

I also understand that this definition is equivalent to the first one. So far so good.

But is the second bullet on the Alternative Definition necessary? Or does it follow from the first and third bullets?

Answer 1

The second bullet doesn't follow from the first and the third. Let $n = m = 1$ and consider the function $f(x) = |x|$ at $x = 0$. The directional derivatives are

$$ \frac{\partial f}{\partial v}(0) = \lim_{t \to 0^{+}} \frac{f(tv) - f(0)}{t} = \lim_{t \to 0} \frac{|tv|}{t} = |v| $$

and we have

$$ \lim_{v \to 0} \frac{f(0 + v) - f(0) - \frac{\partial f}{\partial v}(0)}{|v|} = \lim_{v \to 0} \frac{|v| - |v| }{|v|} = 0$$

but the map $v \mapsto \frac{\partial f}{\partial v}(0)$ is not additive.

BTW, if $v \mapsto \frac{\partial f}{\partial v}(a)$ happens to be differentiable at $v = 0$ then this (together with the homogeneity of the map) forces $v \mapsto \frac{\partial f}{\partial v}(a)$ to be linear.