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Intermediate Value Theorem ... Browder, Theorem 3.16 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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:



Browder - 1 -  Theorem 3.16 ... ... PART 1 ... .png
Browder - 2 -  Theorem 3.16 ... ... PART 2 .png



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 one-sided 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 \(\displaystyle b - \delta \lt x \lt b \Longrightarrow \mid f(x) - f(b) \mid \lt \epsilon\)
 

MountEvariste

Well-known member
Jun 29, 2017
75
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 $|t-b|<\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 $|t-b|<\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 $|t-b|<\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 sign-preserving property of limits/continuous functions:

Sign-preserving 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 $|x-c|< \delta$ implies $f(x) > M$.

2. If $f(c)<0$ then there exists $N<0$ and $\delta >0$ s.t. $x \in I$ and $|x-c|< \delta$ implies $f(x) <N$.
 
Last edited:

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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 $|t-b|<\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 $|t-b|<\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 $|t-b|<\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 sign-preserving property of limits/continuous functions:

Sign-preserving 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 $|x-c|< \delta$ implies $f(x) > M$.

2. If $f(c)<0$ then there exists $N<0$ and $\delta >0$ s.t. $x \in I$ and $|x-c|< \delta$ implies $f(x) <N$.


Thanks MountEvariste for an informative and really helpful reply ...

Now working through what you have written ...

Thanks again ...

Peter