Use monotonic property of subsequences to prove one sided limits at a point of a monotonic function
We were given a two parter homework question:
- Prove that if every subsequence of a sequence $x_n$ is converging to L then $x_n$ converges to L.
- Use what you proved above to prove that if $f(x)$ is a monotonic increasing function at $(a,b)$ then $\forall x_0 \in (a,b):$
$i. \lim_{x \to x_0^+} f(x) = inf_{x>x_0} f(x) $
$ii. \lim_{x \to x_0^-} f(x) = sup_{x<x_0} f(x) $
$iii. \lim_{x \to x_0^-} f(x) \le f(x_0) \le \lim_{x \to x_0^+} f(x) $
Now part 1. was easy to prove (used something similiar to Monotonic subsequences and convergence)
As for part 2, I had no idea how to use the first part in my proof. What I attempted was take a series $x_n$ of increasing values whose limit is $x_0$, therefore the series of $f(x_n) \to f(x_0)$ (from the monotonic property of $f(x)$), now since this series is monotonic and bounded it converges to it's supremum, same for the series of $x_0$ and $f(x_0)$ values from the right of $x_0$ (only this time they are decreasing).
Still not sure how to prove $iii$ but in my proof I haven't used the property I proved at part 1, could anyone point me in the right direction?
$\endgroup$1 Answer
$\begingroup$You cannot conclude that $f(x_n)\to f(x_0)$ from $x_n\to x_0$ just using monotonicity--you need continuity, as well.
Instead, note that for all $x>x_0$ with $x\in(a,b),$ we have $f(x)>f(x_0).$ Since $x_0\in(a,b),$ there exists some $x\in(a,b)$ with $x>x_0,$ and so $\{f(x):x\in(a,b),x>x_0\}$ is a non-empty set of reals that is bounded below by $f(x_0),$ so has a greatest lower bound. Put $y_0=\inf_{x>x_0}f(x).$ Now suppose that $x_n\searrow x_0$ ($x_n$ converges monotonically to $x_0$ from above). We know that $f(x_{n+1})<f(x_n)$ for all $n$ since $f$ is monotone increasing and $x_{n+1}<x_n$ for all $n,$ so $\bigl\{f(x_n)\bigr\}_{n=1}^\infty$ is a monotone decreasing sequence. Moreover, by definition of $y_0,$ we have that $f(x_n)\ge y_0$ for all $n,$ meaning that $\bigl\{f(x_n)\bigr\}_{n=1}^\infty$ is bounded below by $y_0$, and so $f(x_n)\searrow y$ for some $y\ge y_0$.
Now, if we had $y>y_0,$ then by definition of infimum, there would be some $x\in(a,b)$ with $x>x_0$ such that $f(x)<y.$ Since $x\in(a,b)$ with $x>x_0$ and since $x_n\searrow x_0,$ then we know that $x>x_n>x_0$ for all but finitely many $n$. But then $f(x_n)<f(x)<y$ for all but finitely-many $n,$ which is impossible, since $f(x_n)\searrow y.$ Therefore, we must have $y\le y_0,$ and so $y=y_0.$ Thus, for all sequences $\{x_n\}_{n=1}^\infty$ of points of $(a,b)$ with $x_n\searrow x_0,$ we have that $f(x_n)\searrow y_0=\inf_{x>x_0}f(x).$ Thus, $$\lim_{x\searrow x_0}f(x)=\inf_{x>x_0}f(x),$$ as desired.
Similar arguments can be used for 2(ii). Part (iii) is all about monotonicity of $f.$
P.S.: Part 1 can be done even more easily when you observe that $y_n=x_{n+1}$ gives a subsequence of $\{x_n\}_{n=1}^\infty.$
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