The problem from Munkres' Analysis on Manifold is that
Show that the function $f(x,y) = |xy|$ is differentiable at 0, but is not of class $C^1$ in any neighborhood of $0$.
My thought on the first part is that since $Df(0)= [0 0]$, $f$ is differentiable at $0$.
However, I have no idea of how to approach to the second part.
This was my initial thought :
Let $x=(x_1, x_2)\in\Bbb R^2$.
Then,
$D_1f(x) =\begin{cases}x_2&\text{ if }\quad x_1x_2 &> 0\\x_1&\text{ if }\quad x_1x_2&<0\end{cases}$
Therefore, $D_1f(x) =0\quad as \quad x \to 0$.
Since $D_1f(0) = 0$, this implies that $D_1f$ is continuous at $0$. Similar argument goes for $D_2f$.
Is this result wrong or compatible to what the problem requires?
And how can I prove that $f$ is not continuously differentiable in neighborhood of $0$?
Thanks!
Your prove for differentiability is okay.
Not $C^1$:
Notice that $D_1 f$ does not exist at $(0,y)$ for any $y\ne 0$.
Differentiable:
The function $f$ is constant zero on the axes. Thus, the partial derivatives at $(0,0)$ are $0$. Hence, the candidate for $Df(0)$ is $(0,0)$. Now, for $(x,y) \ne (0,0)$ we have $$ \begin{align} |f(x,y) - f(0,0) - 0\cdot x - 0\cdot y| &= |x||y| \\ &= \min(|x|,|y|)\max(|x|,|y|) \\ &\le \min(|x|,|y|) \sqrt{x^2 + y^2}. \end{align}$$ As $\min(|x|,|y|)$ tends to $0$ for $(x,y)\to (0,0)$, we have shown that $f$ is differentiable at $(0,0)$ with $Df(0,0) = (0,0)$.