5
$\begingroup$

Assume $f : I \rightarrow \mathbb{R}$ is a non-decreasing on an open interval $I$ and that $f$ satisfies the Intermediate value property or Darboux's property on $I$ (that is, for any $a < b$ in $I$ and any $L$ between $f(a)$ and $f(b)$, there exists $c \in [a, b]$ such that $ f (c) = L)$.

Then, prove that $f$ is continuous.

However, I know that a function can be discontinuous and also satisfy the IVT at the same time. Could someone point me in the right direction?

$\endgroup$
3
  • $\begingroup$ Obviously, you must use the fact that $f$ is non-decreasing. $\endgroup$
    – 5xum
    May 27, 2014 at 12:10
  • $\begingroup$ That is, if $f$ is non-decreasing, it can only have jump discontinuities. $\endgroup$
    – PA6OTA
    May 27, 2014 at 12:37
  • $\begingroup$ $f(x) = \begin{cases}\sin(\frac{1}{x}) & ,\text{if}\ x\neq 0 \\0& ,\text{if}\ x = 0 \end{cases}$ satisfies Intermediate Value Property, but is not continuous at $0$ $\endgroup$
    – OBDA
    May 27, 2014 at 13:25

2 Answers 2

2
$\begingroup$

A monotonic function can have only discontinuities of first kind or simple discontinuity.

The idea of the proof is that if $p$ is a point of discontinuity, then $f(p-)$ and $f(p+)$ exists and $f(p-)<f(p+)$. We can easily show that $$\color{red}l=\sup\limits_{a<t<p}f(t)=f(p-)<f(p+)=\inf\limits_{p<t<b}f(t)=\color{red}L.$$

Now using definition of $\sup$ and $\inf$, choose some $\varepsilon>0$ and there exists $r$ and $s$ such that $$ r<x_1<p\Rightarrow l-\varepsilon<f(x_1)\le l\\ p<x_2<s \Rightarrow L\le f(x_2)<L+\varepsilon$$

Observe that $f(x_1)<\dfrac{l+L}{2}<f(x_2)$ then by intermediate value property, there should be some $c\in (x_1,x_2)$ such that $f(c)=\dfrac{l+L}{2}$. A contradiction!

$\endgroup$
2
  • $\begingroup$ @Jack D'Aurizio:How this is a contradiction?? $\endgroup$
    – P.Styles
    Nov 10, 2017 at 15:26
  • $\begingroup$ @Jack D'Aurizio:Actually yours name is first that clicked to my mind so, i tagged you here.Sorry!!,for the discomfort $\endgroup$
    – P.Styles
    Nov 10, 2017 at 15:30
1
$\begingroup$

$f$ is non decreasing on $[a,b]$ then for any $x_1\leq x_2 \implies f(x_1)\leq f(x_2)$.

for $c\in [a,b]$, we want to prove that $f$ is continuous at $c$, so consider $\epsilon>0$. We want to find a delta such that $x\in (c-\delta,c+\delta) \cap [a,b] \implies |f(x)-f(a)|<\epsilon$.

So look in the interval $(f(a)-\epsilon,f(a)+\epsilon)$. We know that there exists $f(y)$ and $f(z)$ in $(f(a)-\epsilon,f(a)+\epsilon)$ such that $f(y)\leq f(a) \leq f(z)$ [IVT gurantees this]

So again by applying IVT, we get for all $x \in (y,z) \implies f(x) \in (f(a)-\epsilon,f(a)+\epsilon)$. choose $\delta = \min [\dfrac{z-a}{2},\dfrac{a-y}{2}]$

$\endgroup$
1
  • 1
    $\begingroup$ You should mention the two cases where $f(a)$ is not an interior point. $\endgroup$
    – Philipp
    Feb 21, 2021 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.