Differential of a function definition

Question

Consider $f: U \rightarrow \mathbb R$, $U \subset \mathbb R^n $ is an open set.

a. Show that if $f$ is differenciable on $a \in U$ then there is a unique vector $w$ such that

$\lim\limits_{h\mapsto 0}\dfrac{f(a + hv) - f(a)}{h} = \langle w,v \rangle$.

b. Is the reciprocal statement true ?

c. Find the parcials derivatives of $f$ on $a$ with respect to $w$.

I think part $a$ comes from the definition of a differentiable function, $w$ is unique because of the uniqueness of the limit. However, I'm not so sure about $b$ and $c$.

Any help will be appreciated !

Answer 1

If $f$ is differentiable at $a$, for $v\in \mathbb{R}^{n}$ such that $v+a \in U$, we can write

$df(a)v=\frac{\partial f}{\partial v}(a)= \sum^{n}_{i=1}\frac{\partial f}{\partial x_{i}}(a).v_{i} = <gradf(a),v>$.

But $\frac{\partial f}{\partial v}(a)=\lim\limits_{h\mapsto 0}\dfrac{f(a + hv) - f(a)}{h}\Rightarrow w=gradf(a)=(\frac{\partial f}{\partial x_{1}}(a),...,\frac{\partial f}{\partial x_{n}}(a))$.

So it exists $w$ and is unique. This solves (a) and (c).

The converse is false. Consider the function

$f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ defined by $f(0,0)=0$ and $f(x,y)=\frac{x^{3}y}{x^{6}+y^{2}}$ for (x,y)$\not=$ $(0,0)$.

This function has derived directional $\frac{\partial f}{\partial v}(a)=\lim\limits_{h\mapsto 0}\dfrac{f(a + hv) - f(a)}{h}$ for $a=(0,0)$. So $\lim\limits_{h\mapsto 0}\dfrac{f((0,0)+hv) - f(0,0)}{h}=<grad f(0,0),v>=<w,v>$, but it is not continuous at the origin, then it is not differentiable at the origin. This resolve (b).

Answer 2

b. No, take $n=2$ and let $f(x,y) = 1$ on $E=\{(x,x^2) : x\ne 0\}.$ Set $f=0$ everywhere else (including $(0,0)$). Then every line through $(0,0)$ intersects $E$ in at most one point. It follows that all directional derivatives of $f$ at $(0,0)$ are $0.$ Yet $f$ is not even continuous at $(0,0),$ much less differentiable there.