Analysis (left and right-hand limits, monotonicity)

[raphile]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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

\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
\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

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

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

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

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

\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)

\Rightarrow L \leq f(x_0) \leq L \Rightarrow f(x_0) = L

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.