Non differentiability of $|x| + |y|$
I have to show non differentiability of $f(x,y) = |x| + |y|$ at $(0,0)$.
Now we know this is continuous at $(0,0)$, so I tried finding $f_x$ and $f_y$ using the theorem, which says that if $f(x,y)$ is differentiable at $(0,0)$, then
$\Delta f = f(x+h, y+k)- f(x,y) = hf_x(0,0) + kf_y(0,0) + \epsilon_1 h + \epsilon_2 k$
where $\epsilon_i$ are function of $h,k$ and Both $\to 0$ as $h,k \to 0$. Now finding it here, we have:
$$\Delta f = |h| + |k|\tag{1}$$
comparing, i took $\epsilon_i = 0$ and $f_x(0,0)$ and $f_y(0,0)$ depend on path whether $h,k > 0$ or $< 0$. That is $f_x(0,0)$ is $1$ for $h>0$ and $-1$ for $h< 0$.
Can we now say that function is non-differentiable as $f_x$ and $f_y$ dont exist at $(0,0)?
Actually by checking $h$ and different value of $f_x$ have I checked existence of $f_x(0,0)$ or continuity of $f_{x}(x,y)$ at $(0,0)$ ? I think I have done the former.
How do we know that in $(1)$ we have to take $\epsilon_1$ as coefficient of $|h|$ or as $0$ ?
1 Answer
$\begingroup$You should check that the partial derivatives at the origin exist before looking at $\Delta f$.
In this case, since $f(x,0)=|x|$ is not differentiable at $x=0$ (in the one-variable sense), the partial derivative $f'_x(0,0)$ doesn't exist. Therefore $f$ isn't differentiable (in the two-variable sense) at the origin.
$\endgroup$ 1
