differentiable equiavlence If a function is differentiable that implies that the partial derivatives exists (from every direction) and that the derivative is linear. But a function having these two properties need not to be differentiable (any example?). My prof claimed that there is no equivalence to (total) differentiable rather than the definition, really? can one prove that?
 A: It is a well known fact (see T. Apostol's book on Analysis) that if a function $f$  has a partial derivative, say $\partial_1f$  at a point $c$, and the remaining partial derivatives exist and are continuous at  every element in a neighborhood of $c$, then $f$ is differentiable at $c$.

*

*If this does not occur, $f$ may not be differentiable. For example
$$f(x,y)=(x+y)\mathbb{1}_{\{xy=0\}}(x,y) +\mathbb{1}_{\{xy\neq0\}}(x,y)$$ has partial derivatives $\partial_1f$  and $\partial_2f$ at $(0,0)$, but it is  not differentiable


*Another classical example (counter example rather) is
$$g(x,y)=\frac{xy^2}{x^2+y^4}\mathbb{1}_{\{x\neq0\}}(x,y)$$
For this function, all directional derivatives exists at $(0,0)$ and in spite of this, $g$ fails to be continuous at $(0,0)$ which it turn, implies that $g$ is not differentiable at $(0,0)$


*Consider the function $F:\mathbb{R}^3\rightarrow\mathbb{R}$ defined by
\begin{equation*}
F(x,y)=\left\{\begin{array}{lcr}
x+y-\frac{x^3y}{x^4+y^2} &\quad &\text{if}\quad (x,y)\neq(0,0)\\
0 &\quad& \text{otherwise}
\end{array}\right.
\end{equation*}
For any $\mathbf{v}=(h,k)$, the directional derivative $D_vF(\mathbf{0})=h+k$, however, $F$ is not differentiable at $\mathbf{0}$ since
$$
\frac{F(h,k)-h-k}{\sqrt{h^2+k^2}}=\frac{1}{\sqrt{h^2+k^2}}\frac{h^3k}{h^4+k^2}
$$
fails to converge to $0$ as  $(h,k)\rightarrow(0,0)$.

The common definition of (Fréchet) differentiability says that $f:U\rightarrow\mathbb{R}^m$, $U\subset \mathbb{R}^n$ open,  is differentiable at $c\in U$ if there exists a linear map $A:\mathbb{R}^n\rightarrow\mathbb{R}^m$ such that for all $h$ small enough
$$f(x+h)=f(x)+Ah_r(h)$$
where $\lim_{h\rightarrow0}\frac{\|r(h)\|}{\|h\|}=0$.
There is another notion of differentiability (Gâteaux) that deals with directional derivatives. It says that $f$ is (Gâteaux) differentiable at $c$ if there is a map $L:\mathbb{R}^n\rightarrow\mathbb{R}^m$ such that for any $u\in\mathbb{R}^n$,
$$\lim_{\lambda\rightarrow0}\frac{f(x+\lambda u)-f(x)}{\lambda}=Lu$$
(it is implicit that the limit exists)


*

*Fréchet implies Gâteaux and in such case $A=L$.


*Gâteaux implies Fréchet under special conditions (such as the ones I mentioned above), and in such case $L$ is linear and $A=L$. There are other set of alternative conditions but that is beyond the point here.

A: A standard example goes something like this: define a function $f:\mathbb{R}^2 \to \mathbb{R}$ with the following properties:

*

*$f(x,y)=0$ whenever $y \geq 1.5x^2$ or $y \leq 0.5x^2$. In other words, we only allow $f$ to take nonzero values in the open region between the parabolas $y=0.5x^2$ and $y=1.5x^2$.

*$f(x,y)=1$ whenever $x \neq 0$ and $y=x^2$. In other words $f$ takes the value $1$ everywhere on the parabola $y=x^2$, except for the origin, where we already have required $f(0,0)=0$.

It is not too hard to convince yourself such a function exists. It is even possible to find such functions which are smooth (i.e. $C^\infty$) at every point other than $(0,0)$.
On the other hand, it is clear that such a function of is not even continuous at $(0,0)$, since the value there is $0$, even though we can approach along the parabola $y=x^2$ where the value is $1$.
Finally, in spite of the discontinuity, we can check that the derivative of this function at $(0,0)$ is $0$ from every direction (in particular the derivative is "linear", which seems to be a property you want). The reason for this is that, if $L$ is any line through the $(0,0)$, then $f$ only takes the value $0$ on a sufficiently small interval of the line centered at $(0,0)$. The intuition here is that parabolas vanish faster than lines so, for instance, the parabola $y=1.5x^2$ must, for sufficiently small $x > 0$, scoot underneath of any line $y=mx$ with $m > 0$.
