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We know that continuity along all directions does not imply that the function is continuous in multivariate space. Intuitively is it right to think that a function can be discontinuous along a particular path even if it is continuous in all directions?

Can we say a function $f(x_1,\ldots,x_n)$ is continuous at a given point $p\in\mathbb R^n$ if there exists an open set containing $p$ such that $f(x)$ is continuous along each direction $x_1,....,x_n$ at each point in p?

Curious
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  • Just for clarification, are you asking that: Given $f(x_1, \ldots, x_n)$ that is continuous at a point $p$ along any straight path through $p$, can there be a path through $p$ such that $f$ restricted to the path is discontinuous? – chriseur Jan 05 '15 at 05:42
  • @chriseur I know that a function can be discontinuous along a path even if it is continuous along all lines passing through it. Bit i want to know whether the converse is true( Can we prove continuity at a point p by proving that the function is continuous along each path passing through p ) – Curious Jan 05 '15 at 06:03

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No. $$ \frac{x^2 y}{x^4 + y^2}, $$ evaluated along any line $y=mx$ through the origin, does approach $0.$ Indeed, for nonzero $m,x$ we get $$ \left| \frac{m x^3}{x^4 + m^2 x^2} \right| \leq \left| \frac{x}{m} \right| $$ with constant $m.$ We also take the function as $0$ along the $x$ and $y$ axes. If you wish, define the function to be, for example, $0$ at the origin. That seems sensible, at least it gives substance to the idea that the function is continuous along lines through the origin.

However, along the curved path $$ y = x^2 $$ the function stays exactly $1/2,$ and along the curved path $$ y = -x^2 $$ the function stays exactly $-1/2.$

That is, there is no value for the function at the origin that would make the thing continuous.

Will Jagy
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  • @Will Jagy: The function given does not approach 0 along any line $y=mx$; your are saying that $\displaystyle \frac{mx^3}{x^4+ m^2x^2} \to 0$ as $x\to 0$. – chriseur Jan 05 '15 at 05:45
  • @chriseur It does. You can write that as $\dfrac{mx}{x^2 + m^2}$ (for the purpose of finding the limit), which is zero divided by non-zero. – M. Vinay Jan 05 '15 at 05:51
  • @Vinay: yes, when $m\neq 0$, but the function you've written is clearly not continuous at the origin ($x,m = 0$) (see the answer below) – chriseur Jan 05 '15 at 06:26
  • @chriseur Perhaps I'm misunderstanding the example, but the answer says the function "approaches $0$" as $x \to 0$, meaning the limit is $0$, which is correct. Of course the value of the function is not defined at the origin, so the function is not continuous. – M. Vinay Jan 05 '15 at 08:18
  • The example should be completed by saying that the value of the function at $(0,0)$ is $0$. On any line through $(0,0)$ the limit is $0$, but the function is not continuous at $(0,0)$. – egreg Jan 05 '15 at 10:41
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    @Vinay: yikes, I stand corrected (as I've found some egregious error in my thinking). I apologize for disruption I may have caused. – chriseur Jan 05 '15 at 18:09
  • @chriseur, I put in a few extra notes in the answer. I was asleep for most of the comments above. – Will Jagy Jan 05 '15 at 18:18