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Continuity of piecewise function of two variables

  1. Sep 22, 2014 #1
    The question looks like this.
    Let ##f(x, y)## = 0 if [itex]y\leq 0[/itex] or [itex]y\geq x^4[/itex], and [itex]f(x, y)[/itex] = 1 if [itex]0 < y < x^4 [/itex].
    (a) show that [itex]f(x, y) \rightarrow 0[/itex] as [itex](x, y) \rightarrow (0, 0)[/itex] along any path through (0, 0) of the form [itex] y = mx^a [/itex] with [itex]a < 4[/itex].
    (b) Despite part (a), show that [itex]f[/itex] is discontinuous at (0, 0)
    (c) Show that [itex]f[/itex] is discontinuous on two entire curves.
    What I've came to conclusion is that when [itex] x<0, m>0 [/itex], and [itex]a[/itex] being an odd number, [itex]y[/itex] becomes smaller then zero, so [itex]f(x, y)[/itex] can't be any larger than zero. But I don't think that's not enough. I think I need to find a way to generalize that [itex] mx^a (a<4) [/itex]is larger than [itex]x^4[/itex] or smaller than 0 when [itex]x[/itex] and [itex]y[/itex] is close enough to zero, where I cant' quite get to.
    In regarding (b), I know [itex]f(x, y)[/itex] is discontinuous on certain directions, but can't elaborate it in decent form.
    In regarding (C), How can I show it?
     
    Last edited: Sep 23, 2014
  2. jcsd
  3. Sep 23, 2014 #2

    pasmith

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    For (a): The only way [itex]f(x,mx^a)[/itex] can equal 1 is if [itex]0 < mx^a < x^4[/itex]. If both [itex]m[/itex] and [itex]x^a[/itex] are positive then this requires [itex]|x| > m^{1/(4 - a)}[/itex]. Is that true as [itex]|x| \to 0[/itex], or is there a point beyond which this constraint is violated? What can you say about the case when both [itex]m[/itex] and [itex]x^a[/itex] are negative?

    For (b): Can you show that there are points (x,y) arbitrarily close to (0,0) for which f(x,y) = 1, and yet f(0,0) = 0?

    For (c): Use the above idea: Let (X,Y) be a point on such a curve; you need to show that there are points (x,y) arbitrarily close to (X,Y) such that [itex]|f(x,y) -f(X,Y)| = 1[/itex].
     
  4. Sep 23, 2014 #3
    I know why the curve is incontinuous except for points on x-axis or y-axis, but I want to know how to prove it elaborately. How can I do that? can it be done by epsilon-delta? If so, then how? If not, what is the 'other way'?
     
  5. Sep 23, 2014 #4

    pasmith

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    The definition of continuity at a point is:

    "[itex]f: \mathbb{R}^2 \to \mathbb{R}[/itex] is continuous at [itex]\mathbf{a} \in \mathbb{R}^2[/itex] if and only if for every [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that for all [itex]\mathbb{x} \in \mathbb{R}^2[/itex], if [itex]\|\mathbf{x} - \mathbf{a}\| < \delta[/itex] then [itex]|f(\mathbf{x}) - f(\mathbf{a})| < \epsilon[/itex]".

    The negation of that is:

    "[itex]f: \mathbb{R}^2 \to \mathbb{R}[/itex] is not continuous at [itex]\mathbf{a} \in \mathbb{R}^2[/itex] if and only if there exists an [itex]\epsilon > 0[/itex] such that for all [itex]\delta > 0[/itex] there exists an [itex]\mathbf{x} \in \mathbb{R}^2[/itex] such that [itex]\|\mathbf{x} - \mathbf{a}\| < \delta[/itex] and [itex]|f(\mathbf{x}) - f(\mathbf{a})| \geq \epsilon[/itex]."
     
  6. Sep 23, 2014 #5
    Understood. But, if we approach (0, 0) along the y-axis, regardless of if y >0 or y<0, isn't it continuous? Is it considered discontinuous because there is no 'width' of the contacting point?
     
  7. Sep 23, 2014 #6

    pasmith

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    Any non-constant function from the plane to {0,1} is of necessity discontinuous. If you compose such a function with a path on which the function is constant, then the result is a constant function from (an interval of) the reals to {0,1}, which is continuous. Here the y-axis lies entirely within the region in which f = 0. There are, of course, points arbitrarily close to (0,0) which lie in the region in which f = 1, which is why f is not continuous at (0,0).

    If you want an epsilon-delta proof of that, take [itex]\epsilon = \frac12[/itex] and [itex]\delta > 0[/itex] arbitrary. Let [itex]\delta' = \min\{\delta, 2^{4/3}\}[/itex] and consider the point [itex](\frac12 \delta', \frac{1}{32}\delta'^4)[/itex]. This by construction satisfies [tex]
    \|(\tfrac12 \delta', \tfrac{1}{32}\delta'^4)\| = \left( \frac14 \delta'^2 + \frac{1}{2^{10}}\delta'^8\right)^{1/2} < \sqrt{\tfrac12 \delta'^2} < \delta
    [/tex] and [tex]f(\tfrac12 \delta', \tfrac{1}{32}\delta'^4) = 1.[/tex]
     
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