Function totally differentiable in $(0,0)$

Question

I asked this question here but didn't have an account yet and I can't comment yet on another question. So please forgive me for asking again.

Consider the following function:

$$f:\mathbb R^2\rightarrow\mathbb R,(x,y)\mapsto\begin{cases} x,&y=0 \\ y,&x=0\\0,&\text{else}\end{cases}$$

So far I have, that this function is continuous in $(0,0)$, as for every sequence $(x_n,y_n)\rightarrow (0,0)$ we have $$f(x_n,y_n)=\begin{cases} x_n,& y_n=0 \\ y_n,&x_n=0 \\ 0,&\text{else}\end{cases} \rightarrow 0.$$

Further I have calculated all directional derivatives:

If $(v_1,v_2)\in\mathbb R^2\setminus\{0\}$ with $v_1\cdot v_2\neq 0$, we have $\frac{1}{t}f(tv_1,tv_2)=0.$ For $v_1=0,v_2\neq 0$ we have $\frac{1}{t}f(tv_1,tv_2)=\frac{1}{t}\cdot tv_2=v_2$ and for $v_1\neq 0, v_2=0$ we get $\frac{1}{t}f(tv_1,tv_2)=v_1$. Thus we get for the partial derivatives: $$\frac{\partial f}{\partial x}(0,0)=1,\quad \frac{\partial f}{\partial y}(0,0)=1.$$

I now want to know, if this function is totally differentiable in $(0,0)$. The partial derivatives are not continuous in $(0,0)$, so I can't use that to say that the function is totally differentiable. But as $f$ is continuous in $(0,0)$ I can't rule out that the function is not totally differentiable. I know that by definition $f$ is totally differentiable in $(0,0)$ if there exists a Matrix $T$ and a function $\varphi$ with $$f(x,y)=f(0,0)+T\cdot \left((x,y)-(0,0)\right)+\Vert (x,y)-(0,0)\Vert\cdot\varphi(x,y)$$ but I don't know how to calculate $T$ and $\varphi$.

Regarding the comment if I'm sure that the partial derivatives are not continuous:

For $(x,y)\neq (0,0)$ I get $\frac{\partial f}{\partial x}(x,y)=\begin{cases} 1, & y=0\\0,&\text{else}\end{cases}$ which seems to me not continuous in $(0,0)$. Or am I wrong here?

Regarding the answer by H. H. Rugh: what notation do you mean? And how do you get to that equation?

I can't react to my own (other) question, so what should I do instead? — ichsens, Aug 20 '16 at 22:56
If $f$ is differentiable at $a$, then necessarily, $df_a(h_1,h_2) = \frac{\partial f}{\partial x}(a)h_1+\frac{\partial f}{\partial y}(a)h_2$. This gives you the only possible candidate for $T$. — paf, Aug 20 '16 at 22:57
really good excercise to train the understanding of partial, directional and total derivative! — Philipp, Aug 30 '19 at 17:31

score 1 · Accepted Answer · answered Aug 20 '16 at 22:59

1

You are definitely right about the partial derivatives not being continuous at the origin. Regarding notation, I think you are missing some factors of $t$ in the arguments when calculating directional derivatives. Returning to the derivative itself: A necessary condition (not sufficient in general) is that the directional derivative is linear as a function of the directional vector. So one should have: $$ Df_{(0,0)} (v_1,v_2) = \lim_{t\rightarrow 0} \frac{1}{t} (f(t v_1,t v_2)-f(0,0))= v_1 \partial_x f(0,0) + v_2 \partial_y f(0,0) = v_1 + v_2$$ which is not the case since e.g. $Df_{(0,0} (1,1)=0$

answered Aug 20 '16 at 22:59

H. H. Rugh

35,236

You are right, I forgot the $t$ in the argument! I still don't fully understand your equation, I will take some time to think about that. Does it have something to do with $D_vF(a)=DF(a)\cdot v$ if $F$ is totally differentiable? That the directional derivative is the product of the differential with the directional vector? – ichsens Aug 20 '16 at 23:02
Indeed! we just have different notation for the same thing. I personally wouldn't call it the product of the differential with the directional vector, but rather say that the differential is a linear map that maps directional vectors into ${\Bbb R}$. But this is mostly a matter of taste. – H. H. Rugh Aug 20 '16 at 23:09
Alright, thank you, I will think about that! – ichsens Aug 20 '16 at 23:10

Function totally differentiable in $(0,0)$

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