Showing $f'(a;v_1+v_2)=f'(a;v_1)+f'(a;v_2)$ where $f$ may not be differentiable but directional derivatives exist

Question

Show $f'(a;v_1+v_2)=f'(a;v_1)+f'(a;v_2)$ where $f$ may not be differentiable but all directional derivatives of $f$ exist. $f:\mathbb R^n\to \mathbb R$

It will be enough to show that $$\lim_{h\to 0}{f(a+hv_1+hv_2)-f(a+hv_1)\over h}=f'(a;v_2)$$

Any hints? I tried to apply Taylor polynomial on LHS but that gave $f(a;v_1+v_2)-f(a;v_1)$ which is essentially useless for me

Answer 1

Here's a classic counterexample. The partial derivatives (at $a=0$) of $$f(x,y) = \begin{cases} \frac{xy^2}{x^2+y^2}, & (x,y)\ne (0,0) \\ 0, & (x,y)=(0,0)\end{cases}$$ are clearly both $0$. But what is $f'(0,v)$ for $v=(1,1)$?

Answer 2

It's not true. "Saw" the counterexample immediately, took a bit to see how to write it down coherently:

Using complex notation for vectors in $\Bbb R^2$:

Say $\phi:\Bbb R\to\Bbb R$ is smooth, $2\pi$-periodic, and satisfies $$\theta(t+\pi)=\theta(t)+\pi.$$Define $f:\Bbb R^2\to\Bbb R$ by $$f(re^{it})=r\cos(\theta(t))\quad(r\ge0, t\in\Bbb R).$$

Now $f$ is smooth away from the origin and continuous at the origin. It has directional derivatives at the origin: The point to the condition on $\theta(t)$ is that we actually have $f(re^{it})=r\cos(\theta(t))$ for $r\in\Bbb R$, which shows that $$f'(0,e^{it})=\cos(\theta(t)).$$So $f'(0,v)$ is certainly not linear.