Prove that
$$f(x, y):=\left\{\begin{array}{ll}
\left(x^{2}+y^{2}\right) \sin \frac{1}{\sqrt{ x^{2}+y^{2}}}, & (x, y) \neq 0 \text { } \\
0, & (x, y)=0 \text { }
\end{array}\right.$$
is differentiable on $\mathbf{R}^{2}$ but not continuously differentiable at $(0,0)$.
My Proof-trying. Initially, we will show that $f$ is differentiable on $\mathbb{R}^2$.
If $f(x,y)\neq (0,0)$ then
$$\frac{\partial f}{\partial x} =f_{x}(x, y)=\frac{-x}{\sqrt{x^{2}+y^{2}}} \cos \frac{1}{\sqrt{x^{2}+y^{2}}}+2 x \sin \frac{1}{\sqrt{x^{2}+y^{2}}}$$
and we can see easily that $f_{x}(x, y)$ is continuous at $(x,y)$ as $(x,y)\neq (0,0).$ Similarly $\frac{\partial f}{\partial y} =f_{y}(x, y)$ is exist and it is continuous. Hence $f$ is differentiable at $(x,y)$ as $(x,y)\neq (0,0).$
Now we will check if $(x,y)=(0,0).$ By definition
$$f_{x}(0,0)=\lim _{h \rightarrow 0} \frac{f(h, 0)-f(0,0)}{h}=\lim _{h \rightarrow 0} h \sin \frac{1}{|h|}=0.$$ Similarly $f_y(0,0)=0.$ Then by the definiton of differentiability and for $B=\left[\begin{array}{ll}
0 & 0
\end{array}\right]$, $h=(h,k)$
$$\lim_{h\to 0} \frac{f(h, k)-f(0,0)-B \cdot(h, k)}{\|(h, k)\|}=\lim_{h\to 0}\sqrt{h^{2}+k^{2}} \sin \frac{1}{\sqrt{h^{2}+k^{2}}} = 0$$ So $f$ is differentiable at $(0, 0).$
Therefore we get $f$ is differentiable on $\mathbb{R^2}$.
My questions: $1)$ What is the '' continuously differentiable at (0,0) '' mean?
$2)$ We know that if $f$ is differentiable at $a$, then we can say that $f$ is continuos at $a$ so since $f$ is differentiable on $\mathbb{R^2}$, can we say $f$ is continuous on $\mathbb{R^2}$?
$3)$ How can I show that $f$ is not continuously differentiable at $(0,0)$
Best Answer
It means that the (Fréchet) derivative at $(0,0)$ is continuous.
Yes. As saying that a map is continuous on a subset is equivalent to say that the map is continuous at each point of the subset.
For this, it is enough to prove that one of the partial derivative is not continuous at $(0,0)$. Which is indeed the case for $\frac{\partial f}{\partial x}(0,0)$.