Analysis (left and right-hand limits, monotonicity)

[raphile]

Homework Statement

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.

Homework Equations

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!

 

[Sajet]

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.