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Does there exist a function from $R^2$ to $R$ that is not continuous at origin, but whose restriction to every single polynomial curve through the origin is continuous.

I am not too sure how to approach this problem. I tried just creating some functions that I thought would have said property and then trying to find paths for them (such as $sin(x)$ or $e^x - 1$) that would contradict it. I know this isn't a good approach because there is no way to test every single polynomial, so there must be something I am missing. My initial guess was that it is impossible since any function can at least be approximated to some degree using a polynomial, but once again I know that isn't a valid proof

Any help would be appreciated, I am really stumped

Thanks

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2 Answers 2

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Yes, there exists such a function. Consider, for example, $$ f(x,y) := \begin{cases} 1, & \text{if}\ y = |x|^{3/2}, \ x\neq 0,\\ 0, & \text{otherwise}. \end{cases} $$

A. Rosenthal proved that (see here):

If a function $f\colon \mathbb{R}^2\to\mathbb{R}$ is continuous at a point $P$ along every convex curve through $P$ which is (at least) once differentiable, then it is continuous at $P$.

Yet, $f$ can be continuous at $P$ along every curve which is (at least) twice differentiable without being continuous at $P$.

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You have not defined a "polynomial curve". Im considering curves of the form $$\gamma:\quad t\mapsto{\bf z}(t)=\bigl(p(t),q(t)\bigr)\qquad(-\infty<t<\infty)\tag{1}$$ with $p$ and $q$ polynomials, $\>{\bf z}(0)={\bf 0}$ and $\>{\bf z}'(0)\ne{\bf 0}$. (This includes graphs $y=q(x)$, $q$ a polynomial.)

The function I propose is $$f(x,y):=\left\{\eqalign{1\quad&\bigl(x\ne 0\ \wedge\ 0<|y|<e^{-1/x^2}\bigr)\cr 0\quad&({\rm otherwise})\ .\cr}\right.$$ The reasons this works is the following: Any curve $\gamma$ considered in $(1)$ has a well defined tangent at ${\bf 0}$. If this tangent is not horizontal then $f\bigl({\bf z}(t)\bigr)\equiv0$ for small $|t|$. If this tangent is horizontal then either $y(t)\equiv0$, or there is an $r\geq2$ with $|y(t)|\geq Cx^r$ for small $|t|$. One then can check that $f\bigl({\bf z}(t)\bigr)\equiv0$ for small $|t|$ as well.

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