Let $f:(a,b)\to \mathbb{R}$ be a strictly increasing function. Does the limit $\lim_{x\to a^+}f(x)$ necessarily exist and is a real number or $-\infty$? If so, is it true that $\ell=\lim_{x\to a}f(x)\le f(x) \ \ \forall x\in (a,b)$? Please provide proofs.
One sided limit of an increasing function defined on an open interval
-
0Is this a homework problem? What have you tried so far? – 2012-05-13
-
0No. So far I am convinced that the answer to both questions is yes and tried to prove the second one. Suppose that $\exists x_0\in (a,b):f(x_0)<\ell$. Then, $\exists \xi>0:a
and so $f(x_0) . I tried $\epsilon=f(x_0+\xi)-f(x_0)$ to get a contradiction but I can't show that. – 2012-05-13 -
0I have reached this point: $\exists \delta>0:0
. If is show that $\exists x\in (a,b): 0 – 2012-05-13x_0-\xi$ I am done.
2 Answers
Both statements are true.
There are two cases to consider. Let $S=\{f(x):x\in(a,b)\}$. Suppose first that $S$ is bounded below, i.e., that there is some $y\in\Bbb R$ such that $y
To prove this, let $\epsilon>0$. Then $u+\epsilon>u$, so $u+\epsilon$ is not a lower bound of $S$, and there is therefore some $a+\delta\in(a,b)$ such that $f(a+\delta)$$u\le f(x)
The second is an immediate consequence of the argument just given: $u\le f(x)$ for every $x\in(a,b)$.
Now suppose that $S$ is not bounded below. Then for every $u\in\Bbb R$ there is an $x_u\in(a,b)$ such that $f(x_u)$f$ is strictly increasing, we have $f(x)
-
0You say that $u=\inf S$ and then that $u$ is not a lower bound of $S$. Do you mean $u+\epsilon$? – 2012-05-13
-
0@SomeoneContinuous: Yes, that was a typo. I also still have to add the other case. – 2012-05-13
-
0At the second line, didn't you mean $u\in \mathbb{R}$? – 2014-07-15
-
0Actually it follows that $l=\lim_{x\to a} f(x)
provided $f$ is strictly increasing on the open interval $(a,b).$ – 2015-12-05 -
0@azc: True, but I was addressing the actual question. – 2015-12-05
As for the second statement, I think there is a more precise result. Below is my trial, please check if it is correct.
Lemma Let $f\colon D\subset\mathbb{R}\to\mathbb{R}.$ Suppose that $a, b\in D$ with $a and the point $a$ is a right-sided limit point of $D.$ Let $f$ is strictly increasing on $(a,b)\cap D.$ If $f$ has a right limit $f(a+0)$ at $a,$ then \begin{gather*} f(x)>f(a+0),\qquad\forall x\in (a,b)\cap D. \end{gather*}
Proof: Let $A=f(a+0):=\lim\limits_{x\to a+0}f(x).$
If $A=-\infty,$ then there is nothing to show. So we assume that $A\in\mathbb{R}.$
We prove the statement by contradiction. If otherwise there exists $x_1\in (a,b)\cap D$ such that $f(x_1) then, for $\epsilon_1=\frac{A-f(x_1)}{2},$ there exists $\delta_1>0$ with $\delta