1
$\begingroup$

Give a proof or counterexample of the following statement: Let $f$ be a real-valued function, that is defined and continuous on all of $\mathbb{R}^2$ except at the origin. It has a removable discontinuity at the origin provided that the limit $\lim_{ (x,y)\to(0,0)} f(x, y)$ exists along all parabolas that contain the origin.

  • 0
    @mrf: Thank you, I completely missed that requirement. I could multiply $f$ by something continuous on x>0 that goes from$0$to $1$ to $0$ on each vertical line segment from $(x,0)$ to $(x,x^3)$, but at that point it is probably getting more complicated than David Mitra's examples.2012-02-17

1 Answers 1

4

Here's an example that satisfies the criteria for parabolas of the form $y=ax^2+bx$ or $x=ay^2+by$:

Let $f(x,y)={\cases{xy^3\over x^2+y^6,&$(x,y)\ne(0,0)$ \cr 0,\phantom{\biggl|}& otherwise}}.$

First we show that the limit as $(x,y)$ approaches the origin along one of the parabolic paths given above is 0:

Along the parabola $y=ax^2+bx$, $a\ne 0$:

$f(x,y)={x(ax^2+bx)^3\over x^2+(ax^2+bx)^6} \quad\buildrel{x\rightarrow0}\over\longrightarrow\quad 0, $

as two applications of L'Hopital's rule will verify (or observe that the dominant term upstairs is $ax^7$ and the dominant term downstairs is $x^2$).

Along the parabola $x=ay^2+by$, $a\ne 0$:

$\eqalign{f(x,y)={(ay^2+by)y^3\over (ay^2+by)^2 +y^6} &={ay^5+by^4\over a^2y^4+2aby^3+b^2y^2+y^6}\cr &={ay^3+by^2\over a^2y^2+2aby +b^2 +y^4}\cr & \buildrel{y\rightarrow0}\over\longrightarrow\quad 0,}$ as easily seen when $b\ne 0$. For $b=0$, we have $ {ay^3+by^2\over a^2y^2+2aby +b^2 +y^4} ={ay^3 \over a^2y^2 +y^4}={ay \over a^2 +y^2} \quad \buildrel{y\rightarrow0}\over\longrightarrow\quad 0, $ as well.

Now we show that $\lim\limits_{(x,y)\rightarrow(0,0)} f(x,y)$ does not exist (and thus, $f$ is discontinuous at the origin, but the discontinuity is not removable):

Just observe that along the path $x=y^3$: $ f(x,y)={y^6\over 2y^6}\quad\buildrel{y\rightarrow0}\over\longrightarrow\quad {1\over2}. $


I'm not sure what happens for a general parabola that passes through the origin...


Incidentally, in, Counterexamples in Analysis, by Bernard R. Gelbaum and John M. H. Olmsted, page 116, an example is given of a function which has no limit at $(0,0)$, but such that for any path of the form $x^m=(y/c)^n$, where $c\ne 0$ and $m,n$ are relatively prime positive integers, the limit as $(x,y)$ approaches the origin along the path is zero. The function with the stated properties is: $ f(x,y)=\cases{ {e^{-1/x^2}y\over e^{-2/x^2}+y^2 },& $x\ne0$\cr 0\phantom{\biggl|} ,&$x=0$}. $

  • 0
    @Jonas Thanks again.. "contains the origin".2012-02-16