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Analysis (left and right-hand limits, monotonicity)

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Assume that [itex]f[/itex] is a monotone increasing function defined on [itex]\mathbb{R}[/itex] and that for some [itex]x_0\in \mathbb{R}[/itex] the left and right limit coincide. Can you prove that [itex]f[/itex] is continuous at [itex]x_0[/itex]? Either give a complete proof or a counterexample.

    2. Relevant equations



    3. The attempt at a solution

    I believe the function should be continuous at [itex]x_0\text{ }[/itex] (otherwise, my whole answer is incorrect!). This is the proof I've come up with:

    Given that the left and right limits coincide at [itex]x_0[/itex], call this limit [itex]L[/itex]. Then we know (by definitions of left and right limits):

    [itex]\forall \epsilon>0 \text{ }\exists \delta_1>0: \forall x\in \mathbb{R}\text{ with } -\delta_1<x-x_0<0\text{ we have }|f(x)-L|<\epsilon[/itex]
    [itex]\forall \epsilon>0 \text{ }\exists \delta_2>0: \forall x\in \mathbb{R}\text{ with } 0<x-x_0<\delta_2\text{ we have }|f(x)-L|<\epsilon[/itex]

    Suppose (for a contradiction) that the function is not continuous at [itex]x_0[/itex]. Then we can say that [itex]|f(x_0)-L|=\tau>0[/itex], where [itex]\tau[/itex] is some positive number. If [itex]f(x_0)>L[/itex] then we have [itex]f(x_0)-L=\tau[/itex], and if [itex]f(x_0)<L[/itex] then we have [itex]L-f(x_0)=\tau[/itex].

    In the case where [itex]f(x_0)-L=\tau[/itex], take [itex]\epsilon=\tau\text{ }[/itex] in the right-hand limit definition. Then we can say that for [itex]x\in (x_0,x_0+\delta_2)[/itex] (in particular, for [itex]x=x_0+\delta_2/2[/itex]) we have [itex]|f(x)-L|<\tau[/itex] and hence [itex]f(x_0+\delta_2/2)<L+\tau[/itex]. This means we have [itex]f(x_0)=L+\tau[/itex] but [itex]f(x_0+\delta_2/2)<L+\tau[/itex], which contradicts the statement that [itex]f[/itex] is monotonically increasing.

    In the case where [itex]L-f(x_0)=\tau[/itex], take [itex]\epsilon=\tau\text{ }[/itex] in the left-hand limit definition. Then we can say that for [itex]x\in (x_0-\delta_1,0)[/itex] (in particular, for [itex]x=x_0-\delta_1/2[/itex]) we have [itex]|f(x)-L|<\tau[/itex] and hence [itex]f(x_0-\delta_1/2)>L-\tau[/itex]. This means we have [itex]f(x_0)=L-\tau[/itex] but [itex]f(x_0-\delta_1/2)>L-\tau[/itex], which contradicts the statement that [itex]f[/itex] is monotonically increasing.

    I'd really like someone to check whether this proof makes sense or possibly if there is a simpler way to do it, or even if I have the wrong answer to start with. Thanks!
     
  2. jcsd
  3. May 7, 2012 #2
    Yes, this works. Sure, you could shorten the second part if you like. For instance you can use that by monotony f(x0) is stuck between two converging sequences with the same limit point, i.e.

    [itex]f(x_0-1/n) \leq f(x_0) \leq f(x_0+1/n)[/itex] for all n

    [itex]\Rightarrow \lim_{n \in \mathbb N} f(x_0-1/n) \leq f(x_0) \leq \lim_{n \in \mathbb N} f(x_0+1/n)[/itex]

    [itex]\Rightarrow L \leq f(x_0) \leq L \Rightarrow f(x_0) = L[/itex]

    but yours is fine, too.

    Also, depending on your definition of continuity (epsilon-delta-criterion?), you might want to point out how all of this fits together in proving the continuity of f.
     
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