
MHB Master
#1
February 1st, 2020,
03:14
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...
I am currently reading Chapter 3: Continuous Functions on Intervals and am currently focused on Section 3.1 Limits and Continuity ... ...
I need some help in understanding the proof of Theorem 3.16 ...
Theorem 3.16 and its proof read as follows:
In the above proof by Andrew Browder we read the following:
" ... ... But $ \displaystyle f(b) \gt y$ implies (since $ \displaystyle f$ is continuous at $ \displaystyle b$) that there exists $ \displaystyle \delta \gt 0$ such that $ \displaystyle f(t) \gt y$ for all $ \displaystyle t$ with $ \displaystyle b  \delta \lt t \leq b$. ... ... "
My question is as follows:
How do we demonstrate explicitly and rigorously that since $ \displaystyle f$ is continuous at $ \displaystyle b$ and $ \displaystyle f(b) \gt y $ therefore we have that there exists $ \displaystyle \delta \gt 0$ such that $ \displaystyle f(t) \gt y$ for all $ \displaystyle t$ with $ \displaystyle b  \delta \lt t \leq b$. ... ...
Help will be much appreciated ...
Peter
***NOTE***
The relevant definition of onesided continuity for the above is as follows:
$ \displaystyle f$ is continuous from the left at $ \displaystyle b$ implies that for every $ \displaystyle \epsilon \gt 0$ there exists $ \displaystyle \delta \gt 0$ such that for all $ \displaystyle x \in [a, b]$ ...
we have that $ \displaystyle b  \delta \lt x \lt b \Longrightarrow \mid f(x)  f(b) \mid \lt \epsilon$

February 1st, 2020 03:14
# ADS
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MHB Apprentice
#2
February 1st, 2020,
08:11
We have $f(b) > y$ or $f(b)y >0$. Define $g(b) = f(b)y>0$.
Since $f$ is continuous at $b$, $g$ is continuous at $b$.
Thus for all $\epsilon_0 >0$, there exist $\delta_0 >0$ s.t. $t \in [a,b]$ and $tb<\delta_0 $ imply $g(t)g(b) < \epsilon_0$.
In particular, for $ \epsilon := g(b)/2>0$ there exists $\delta >0$ s.t. $t \in [a,b]$ and $tb<\delta $ imply $g(t)g(b) < g(b)/2$.
Therefore $g(b)/2 <g(t)g(b) < g(b)/2$. Hence $g(t)> g(b)g(b)/2 = g(b)/2$.
Rewriting $g(t) > g(b)/2$ in terms of $f$ we have $f(t) > \frac{1}{2}(f(b)+y)> \frac{1}{2}(y+y) = y$ as $f(b)>y$.
Thus $f(t)>y$ whenever $t \in [a,b]$ and $tb<\delta $; that's $t \in [a,b] \cap (b\delta, b+\delta)$, i.e. whenever $b\delta < t \le b.$
This calculation can be skipped by appealing to signpreserving property of limits/continuous functions:
Signpreserving property (for continuous functions): Let $f: I \to \mathbb{R}$ be continuous at $ c \in I \subseteq \mathbb{R}$.
1. If $f(c)>0$ then there exists $M>0$ and $\delta >0$ s.t. $x \in I$ and $xc< \delta$ implies $f(x) > M$.
2. If $f(c)<0$ then there exists $N<0$ and $\delta >0$ s.t. $x \in I$ and $xc< \delta$ implies $f(x) <N$.
Last edited by MountEvariste; February 1st, 2020 at 11:04.

MHB Master
#3
February 1st, 2020,
19:07
Thread Author
Originally Posted by
MountEvariste
We have $f(b) > y$ or $f(b)y >0$. Define $g(b) = f(b)y>0$.
Since $f$ is continuous at $b$, $g$ is continuous at $b$.
Thus for all $\epsilon_0 >0$, there exist $\delta_0 >0$ s.t. $t \in [a,b]$ and $tb<\delta_0 $ imply $g(t)g(b) < \epsilon_0$.
In particular, for $ \epsilon := g(b)/2>0$ there exists $\delta >0$ s.t. $t \in [a,b]$ and $tb<\delta $ imply $g(t)g(b) < g(b)/2$.
Therefore $g(b)/2 <g(t)g(b) < g(b)/2$. Hence $g(t)> g(b)g(b)/2 = g(b)/2$.
Rewriting $g(t) > g(b)/2$ in terms of $f$ we have $f(t) > \frac{1}{2}(f(b)+y)> \frac{1}{2}(y+y) = y$ as $f(b)>y$.
Thus $f(t)>y$ whenever $t \in [a,b]$ and $tb<\delta $; that's $t \in [a,b] \cap (b\delta, b+\delta)$, i.e. whenever $b\delta < t \le b.$
This calculation can be skipped by appealing to signpreserving property of limits/continuous functions:
Signpreserving property (for continuous functions): Let $f: I \to \mathbb{R}$ be continuous at $ c \in I \subseteq \mathbb{R}$.
1. If $f(c)>0$ then there exists $M>0$ and $\delta >0$ s.t. $x \in I$ and $xc< \delta$ implies $f(x) > M$.
2. If $f(c)<0$ then there exists $N<0$ and $\delta >0$ s.t. $x \in I$ and $xc< \delta$ implies $f(x) <N$.
Thanks MountEvariste for an informative and really helpful reply ...
Now working through what you have written ...
Thanks again ...
Peter