# does there exist a function of two variables that is not continuous at origin, but whose restriction to every polynomial is,

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

Yes, there exists such a function. Consider, for example, $$f(x,y) := \begin{cases} 1, & \text{if}\ y = |x|^{3/2}, \ x\neq 0,\\ 0, & \text{otherwise}. \end{cases}$$
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 $$P$$.
You have not defined a "polynomial curve". Im considering curves of the form $$\gamma:\quad t\mapsto{\bf z}(t)=\bigl(p(t),q(t)\bigr)\qquad(-\infty with $$p$$ and $$q$$ polynomials, $$\>{\bf z}(0)={\bf 0}$$ and $$\>{\bf z}'(0)\ne{\bf 0}$$. (This includes graphs $$y=q(x)$$, $$q$$ a polynomial.)
The function I propose is f(x,y):=\left\{\eqalign{1\quad&\bigl(x\ne 0\ \wedge\ 0<|y| The reasons this works is the following: Any curve $$\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.