Intermediate Value Theorem

Note that because of continuity,

[!math|{“type”:“lemma”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:0}] Lemma 1. If function $f$ is continuous at $c$ and $f(c)\ne 0$, there is an interval $(c-\delta ,c+\delta )$ around $c$ where $f$ has the same size as $c$.

\begin{proof} For $ϵ=\frac{|f(c)|}{2}$, $\exists \delta >0$ s.t. $|x-c|<\delta ⟹|f(x)-f(c)|<\frac{|f(c)|}{2}$. $\therefore f(x)$ has the same sign as $f(c)$. \end{proof}

Due to the above lemma, we have

Bolzano’s Theorem §

[!math|{“type”:“theorem”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Bolzano’s Theorem”,“label”:“bolzano-theorem”,“_index”:1}] Theorem 2 (Bolzano’s Theorem). Suppose $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs. Then, $\exists c\in (a,b):f(c)=0$.

\begin{proof} Suppose WLOG that $f(a)<0,f(b)>0$. Let $B$ be the set $\{x\in [a,b]\mid f(x)\le 0\}$. This is bounded and nonempty so $c=\text{sup}B$ exists. We claim $f(c)=0$.

If $f(c)>0$, then by ^54e2e4 , it is positive in some neighborhood of $c$, so there is $c^{\prime }<c$ which is an upper bound for $B$, contradicting the definition of $c$.

If $f(c)<0$, then again by ^54e2e4, there is $c^{\prime \prime }>c$ s.t. $f(c^{\prime \prime })<0$, again a contradiction \end{proof}

Generalizing Bolzano’s, we get

Intermediate Value Theorem §

[!math|{“type”:“theorem”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Intermediate Value Theorem”,“label”:“intermediate-value-theorem”,“_index”:2}] Theorem 3 (Intermediate Value Theorem). Let $f\in C([a,b])$. If $x_1$ and $x_2$ are 2 points in $[a,b]$ with $x_1<x_2$ and $f(x_1)\ne f(x_2)$, then $f$ takes every value between $f(x_1)$ and $f(x_2)$ in the interval $(x_1,x_2)$.

\begin{proof} Suppose WLOG that $f(x_1)<f(x_2)$. For $t\in (f(x_1),f(x_2))$, consider $g(x)=f(x)-t$. Then, $g(x_1)<0$, $g(x_2)>0$, and $g$ is continuous, so we can apply ^1959fb to find $c\mid g(c)=0$. Then, $f(c)-t=0$, or $f(c)=t$. \end{proof}