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Let $f:\mathbb{R}\to \mathbb{R}$ be defined as $$f(x) = \begin{cases} e^{\frac{1}{x^2-1}}, & \text{if }|x|<1 \\ 0, & \text{if }|x|\geq 1. \end{cases}$$ It is not difficult to check that $f\in C^1(\mathbb{R})$ with $f(x)=0$ if $|x|>1$ and $f(x)>0$ if $|x|\leq 1$. Now I'd like to construct the similar function on the plane. Let $f:\mathbb{R}^2\to \mathbb{R}$ be defined as $$f(x,y) = \begin{cases} e^{\frac{1}{x^2+y^2-1}}, & \text{if } x^2+y^2<1 \\ 0, & \text{if }x^2+y^2\geq 1. \end{cases}$$ I want to prove that $f\in C^1(\mathbb{R}^2; \mathbb{R})$, i.e., partial derivatives of $f$ exists and continuous. It is not difficult to verify that $$\frac{\partial f}{\partial x}(x,y) = \begin{cases} -\frac{2x e^{\frac{1}{x^2+y^2-1}}}{(x^2+y^2-1)^2}, & \text{if } x^2+y^2<1 \\ 0, & \text{if }x^2+y^2\geq 1. \end{cases}$$ It shows that $\frac{\partial f}{\partial x}$ exists everywhere on $\mathbb{R}^2$. But what about continuity of $\frac{\partial f}{\partial x}$?

From the definition of $\frac{\partial f}{\partial x}$ it follows that it is continuous on $\{(x,y)\in \mathbb{R}^2: x^2+y^2<1\}\cup \{(x,y)\in \mathbb{R}^2: x^2+y^2>1\}$. But is it continuous on the boundary? I mean suppose that $(x_0,y_0)\in \mathbb{R}^2$ such that $x_0^2+y_0^2=1$. How to show that $$\lim_{(x,y)\to (x_0,y_0)}\frac{\partial f}{\partial x}(x,y)=0.$$

I have some issues to show this rigorously. Can anyone show the proof please?

RFZ
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  • There seems to be a mistake in your derivative, I think it should be something like $-\frac{2xe^{\frac1{x^2+y^2-1}}}{(x^2+y^2-1)^2}$, at least the exp should have a $y^2$ extra. I did not check the rest of this derivative. I think then you will find why things go wrong? – student91 Feb 27 '23 at 14:57
  • @student91, thank you for your remark! I had the right answer on the paper but made a typo when typing. No it still unclear how to to compute that limit. – RFZ Feb 27 '23 at 17:04
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    No need to calulate the derivatives! If $\varphi:\mathbb R\to \mathbb R$ is ($n$-times) continuously differentiable then so is $f:\mathbb R^2\to\mathbb R$, $(x,y)\mapsto \varphi(x^2+y^2)$. – Jochen Feb 27 '23 at 17:20
  • @Jochen, indeed, it is much easier than I thought! So it follows basically from the fact that composition of smooth functions is smooth, right? – RFZ Feb 27 '23 at 17:41
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    Yes, of course. – Jochen Feb 27 '23 at 17:44
  • @Jochen, just a small correction: I guess you meant $f:\mathbb{R}^2\to \mathbb{R}$, $(x,y)\mapsto f(\sqrt{x^2+y^2})$ – RFZ Feb 28 '23 at 02:58
  • This depends on the definition of $\varphi$. It is better to avoid the non-differentiable square root. – Jochen Feb 28 '23 at 08:11
  • Express $f$ and $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ in polar coordinates. Note that all three go to $0$ as $r\to 1$. As $r$ is continuous and the boundary is $r = 1$, you cannot approach the boundary without $r \to 1$. – Paul Sinclair Feb 28 '23 at 17:52
  • In this answer to a question I did an user gives a beautiful example plot of the function you are exploring. – Joako Nov 06 '23 at 02:41

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