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

#### Peter

##### Well-known member
MHB Site Helper
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 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
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
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 $f(x) <N$.