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Tricky continuity problem!

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Given that:

    [itex]
    f(x,y) =
    \begin{cases}
    xy/(x² + y²), & \text{if }(x,y) \neq (0,0) \\
    3n+1, & \text{if }(x,y) = (0,0)
    \end{cases}
    [/itex]

    Discuss the continuity of that function from E² to ℝ.

    2. Relevant equations

    • Definition of continuity
    • Definition of uniform continuity

    3. The attempt at a solution

    First, I said that f(x,y) is continuous everywhere except at (0,0) since ½ and 0 occur when (x,y) = (0,0) [can't put them in formal proof].

    I want to test if f(x,y) is uniformly continuous. Using the definition of uniform continuity, [itex]\forall \epsilon > 0 \exists \delta > 0[/itex] such that:

    If [itex]|(x,y) - (a,b)| < \delta[/itex] then [itex]|f(x,y) - f(a,b)| < \epsilon[/itex]

    Given these intervals, I have:

    [itex]|(x,y) - (a,b)| < \delta[/itex]
    [itex]√((x - a)² + (y - b)²) < \delta[/itex] ← I'm not quite sure if this is correct.

    [itex]|f(x,y) - f(a,b)| < \epsilon[/itex]
    [itex]|xy/(x² + y²) - ab/(a² + b²)| < \epsilon[/itex] ← I'm stuck here. I don't know how to get from delta interval to here and figure out the appropriate substitution of [itex]\delta[/itex] and [itex]\epsilon[/itex]
     
  2. jcsd
  3. Apr 24, 2013 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The statement that "½ and 0 occur when (x,y) = (0,0)" is incorrect- or badly stated. When (x, y)= (0, 0), the function value is 0. I don't know what you mean by "½ and 0 occur".
    What you mean, I believe, is that as (x,y) approaches (0, 0) along the line y= 0 we have [itex]\lim_{x\to 0}\frac{x(0)}{x^2}= 0[/itex] but approaching (0, 0) along the line y= x, we have [itex]\lim_{x\to 0}\frac{x(x)}{x^2+ x^2}= 1/2[/itex].

    Since f maps R2 to R, the "| |" in [itex]|f(x,y)- f(a,b)|[/itex] is the usual absolute value. But what does it mean in [itex]|(x,y)- (a,b)|[/itex]?

    Isn't it sufficient to note that f(x,y) is uniformly continuous on any closed and bounded set[/tex] on which it is continuous? So this f is uniformly continuous on any set that does not include (0, 0) or have (0, 0) in its boundary.
     
  4. Apr 24, 2013 #3

    mfb

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    2016 Award

    Staff: Mentor

    "uniformly continuous" implies "continuous". If the function is not continuous, why do you test if it is uniformly continuous?

    It is. I would expect that the square of this inequality is more useful.

    What does that 3n+1 do there? What is n?
     
  5. Apr 24, 2013 #4


    Yes, that is what I mean that xy/(x² + y²) is ½ and 0 at the origin.

    But how do you know that f(x,y) is uniformly continuous?
     
    Last edited: Apr 24, 2013
  6. Apr 24, 2013 #5
    Yes, that is true. But continuous function doesn't really imply uniform continuous function. For instance, f(x) = x² is not uniformly continuous, though it is continuous.

    Well, if we take x = 3n + 1 and y = n and work out the computation, then the inequalities hold.
     
  7. Apr 24, 2013 #6
    Never mind. I checked out the theorem online that my instructor hasn't gone over yet. It's true that the domain consisting of all real values except the origin is closed, so if f is continuous on the closed and bounded set, then it's uniformly continuous on that same set.
     
  8. Apr 24, 2013 #7

    mathwonk

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    Science Advisor
    Homework Helper

    The complement of the origin is not closed because it is possible for sequence of non zero values to converge to zero.

    It is also not bounded. So there is a valid question here of whether the given function is uniformly continuous on the complement of the origin, i.e. the same set where it is continuous.

    But it seems unlikely, since if e = 1/4 >0, then for any d>0, you can always find two points "d-near" (0,0) hence "d-near" each other , with on on the line y=0 and the other on the line y=x, such that at one of them f is close to 0, and at the other f is close to 1/2.

    If the values of f at the two points are close enough to 0 and 1/2 respectively, then the values of f at the two points will be further than 1/4 apart.

    A typical bounded and closed set in the complement of the origin, on which the function will be uniformly continuous, is a closed disc centered at the origin, with a smaller such closed disc removed, i.e. a closed annulus centered at the origin.
     
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