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Showing that the directional derivatives exist but f is not continuous

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data
    It says:

    [tex] \displaystyle f:{{\mathbb{R}}^{2}}\to \mathbb{R}[/tex]

    [tex] \displaystyle f\left( x,y \right)=\left\{ \begin{align}
    & 1\text{ if 0<y<}{{\text{x}}^{2}} \\
    & 0\text{ in other cases} \\
    \end{align} \right.[/tex]

    Show that all the directional derivatives about (0,0) exist but f is not continuous in (0,0).

    2. Relevant equations
    Directional derivative:
    [tex] \displaystyle \frac{\partial f}{\partial v}\left( 0,0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( \left( 0,0 \right)+\left( ha,hb \right) \right)-f\left( 0,0 \right)}{h}[/tex]


    3. The attempt at a solution

    I write the equation for the directionar derivative respect to a generic v=(a,b) about the origin:

    [tex] \displaystyle \frac{\partial f}{\partial v}\left( 0,0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( \left( 0,0 \right)+\left( ha,hb \right) \right)-f\left( 0,0 \right)}{h}[/tex]

    That is:

    [tex] \displaystyle =\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( ha,hb \right)}{h}[/tex]

    I should arrive that the limit does not depend on v=(a,b). But I'm stuck here. Which is the value of [tex] \displaystyle f\left( ha,hb \right)[/tex] 1 or 0? It clearly depends on the vector v. If I get close to the origin by the y-axis then [tex] \displaystyle f\left( 0,hb \right)=0[/tex].

    Thanks!!

    PD: I hope I'm not breaking the rules. I've recieved several infractions which I'm not going to discuss here. It's not my intention.
     
    Last edited: Aug 29, 2012
  2. jcsd
  3. Aug 29, 2012 #2

    Mark44

    Staff: Mentor

    Are you sure you have posted the exact wording of the problem? The function isn't continuous at (0, 0). In fact it's discontinuous all along the parabola y = x2 in the x-y plane.
     
  4. Aug 29, 2012 #3
    Yes, I can imagine what the graph looks like and its discontinuity is pretty obvious but what I'm asked to show is that the directional derivatives do exist around the origin but the function is still discontinuous.

    Thanks Mark!
     
  5. Aug 29, 2012 #4

    Mark44

    Staff: Mentor

    I don't see how the directional derivative in the direction of -j, at (0, 0), could exist.

    Look at the difference quotient using these values:

    v = (0, -1), h = .1
    v = (0, -1), h = .01
    v = (0, -1), h = .001
    and so on...
     
  6. Aug 29, 2012 #5
    I think you're right, that's why I made this thdread.

    The directional derivative respecto to v=(0,-1) about (0,0) is:

    [tex] \displaystyle \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 0,-h \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{h}[/tex]

    which is infinite, isn't it?

    Thanks!

    PD: I am looking at the wording of the problem right now and it says that. Maybe just a mistake from the professors.
     
  7. Aug 29, 2012 #6
    Sorry, that's wrong. The limit is zero in all directions because f(0,-h)=0. The numerator is zero, then the limit is zero.
     
  8. Aug 29, 2012 #7

    vela

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    You should be able to prove that given (a,b), if h is small enough, f(ha,hb) will equal 0.
     
  9. Aug 29, 2012 #8
    Yes, I see. Do I have to do any formal proof with deltas and epsilon? Or can I just write it directly?

    Thanks vela!
     
  10. Aug 29, 2012 #9

    vela

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    Depends on what the grader wants. You'd know better than us.
     
  11. Aug 30, 2012 #10

    Mark44

    Staff: Mentor

    What about my example, where (a, b) = (0, -1). No matter how small h is (but is still positive), f(ha, hb) = 1. Is there a mistake in my thinking here?
     
  12. Aug 30, 2012 #11

    vela

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    f(x,y) = 0 when y≤0. It's only equal to 1 between the x-axis and the parabola y=x2.
     
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