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.

  • 2
    Do you have any thoughts on the problem? I would also try to phrase this more as a question than a command. That ruffles some users.2012-02-16
  • 0
    It looks like you've transcribed a math problem, but without putting any of your own thoughts or words around it: this is rather off-putting, more in questions written like this than your earlier ones, because people like to know an OP is open for engagement with the community but this sends all the wrong signals.2012-02-16
  • 2
    What about $f(x,y)=1$ if $x>0$ and $0$f(x,y)=0$ otherwise? – 2012-02-16
  • 0
    @JonasMeyer: Your function is not continuous outside the origin.2012-02-16
  • 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
    You seem to have only considered parabolas with axes parallel to one of the coordinate axes and with vertex at the origin.2012-02-16
  • 0
    @Jonas Yes, thanks for pointing this out. I will think about what happens for a general parabola with vertex at the origin...2012-02-16
  • 0
    I don't see any assumption in the problem that the vertex is at the origin.2012-02-16
  • 0
    @Jonas Thanks again.. "contains the origin".2012-02-16