Counterexample to equivalence of definitions of differentiability

Question

i was wondering if the following (alleged) definitions of differentiability are equivalent.

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be continuous.

Definition 1: $f$ is called differentiable in $x \in \mathbb{R}^n$, if there exists a linear function $J:\mathbb{R}^n \rightarrow \mathbb{R}^m$, such that $$\operatorname{lim}_{v \rightarrow 0} \left|\left|\frac{f(x+v)-f(x)-Jv}{||v||} \right|\right|=0 $$

vs

Definition 2: $f$ is called differentiable in $x \in \mathbb{R}^n$, if there exists a linear function $J:\mathbb{R}^n \rightarrow \mathbb{R}^m$, such that $$\operatorname{lim}_{h \rightarrow 0} \left|\left|\frac{f(x+hv)-f(x)-Jhv}{h} \right|\right|=0 $$ $\forall v \in \mathbb{R}^n$ with $||v||=1$.

The first definition is the usual definition, the second definition basically states that all directional derivatives have to be given by a linear function.

If these definitions are not equivalent, there should be a continuous function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, such that all directional derivatives are given by a linear function, but $$\operatorname{lim}_{n \rightarrow \infty} \left|\left|\frac{f(x+v_n)-f(x)-Jv_n}{||v_n||} \right|\right| \neq 0 $$ for a suitable sequence $(v_n)_{n \in \mathbb{N}} \rightarrow 0$, $v_n \in \mathbb{R}^n$.

While writing this down i found a counterexample, so i made this a Q&A.

Answer 1

Let $$(x_n, y_n) = \left(\frac{sin(\frac{1}{n})}{n} , \frac{cos(\frac{1}{n})}{n}\right).$$ Let $\phi_n$ be a continuous bumpfunction, such that:

1. $0 \leq \phi_n(x,y) \leq \frac{1}{n}$ $\forall(x, y)$ $\forall n$
2. $\phi_n(x_n, y_n) = \frac{1}{n}$.
3. $\operatorname{support}(\phi_n)\subseteq B_{r_n}(x_n, y_n)$, where $B_{r_n}(x_n, y_n)$ is a ball with radius $r_n$ centered around $(x_n, y_n)$, such that $r_n = \operatorname{min}\left(\left|(x_n, y_n)-(x_{n-1}, y_{n-1})\right|, |(x_n, y_n)-(x_{n+1}, y_{n+1})|, \frac{1}{n} \right)/2.$

Let $$f(x, y)=\sum_{n=0}^{\infty}\phi_n(x, y).$$

$f$ is continuous and welldefined because the $\phi_n$ are continuous and their supports are mutually exclusive. Let $J=0.$ $f$ is constructed in a way, so that each line through the origin intersects at most one support of the bumpfcunctions $\phi_n$. So $$\operatorname{lim}_{h \rightarrow 0} \left|\left|\frac{f(hv)-f(0)-Jhv}{h} \right|\right|=0 $$ holds $\forall v \in \mathbb{R}^n$ with $||v||=1$ because for sufficiently small $h$, $hv$ will be out of the support of the $\phi_n$ which it intersects, if there is any.

But we have $$\operatorname{lim}_{n \rightarrow \infty} \left|\left|\frac{f(x_n, y_n)-f(0)- 0v_n}{||v_n||} \right|\right| = \operatorname{lim}_{n \rightarrow \infty} \left|\left|\frac{\frac{1}{n}}{\frac{1}{n}} \right|\right| = 1.$$

Answer 2

Let $P=\{(x,x^2):x\in \mathbb R, x\ne 0\}.$ Define $f:\mathbb R^2\to \mathbb R$ by setting $f = 1$ on $P,$ $f=0$ everywhere else. Then $f$ satisfies the second definition at $(0,0),$ with $J=0,$ but $f$ is not even continuous at $(0,0),$ much less differentiable, at $(0,0).$