Exercise 4.5.5

Let $f$ be continuous, and let $L\in (f(a),f(b))$, we must find $c\in (a,b)$ with $f(c)=L$. (If $f(a)>f(b)$ then instead consider $f^{\prime }(x)=-f(x)$ and $L^{\prime }=-L$)

(a)

Let $c=\sup <!--\; FUNCTION\; APPLICATION\; -->\{x:f(x)\le L\}$. $f(c)<L$ is not possible since we could find $\delta$ small enough that $f(c+\delta )<L$ contradicting $c$ being an upper bound. And $f(c)>L$ is impossible since we could find $\delta$ small enough that $f(c-\delta )>L$ contradicting $c$ being the least upper bound (since we found a smaller upper bound). Thus we must have $f(c)=L$ completing the proof.

A detail we glossed over is $c\in (a,b)$, which can be seen since $f(b)>L$ and $f(a)<L$ has $f(a+\delta )<L$ meaning $a$ cannot be the least upper bound.

(b)

Let $I_0=[a,b]$ and bisect into two intervals, let $I_1$ be the interval where $L$ is still between $f$ at the endpoints. Continue like this to get a sequence $I_n\subseteq I_{n-1}$, $I_n\ne \varnothing$, and let $c\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{I}_n$.

Suppose for contradiction that $f(c)<L$; then there must be some $𝜖>0$ so that $V_{𝜖}(f(c))<L$. Since $f$ is continuous, there must also be some $\delta$ so that $f(V_{\delta }(c))<L$ - but this contradicts our construction in that one endpoint of $I_n$ is mapped to a number greater than $L$, no matter how large $n$ gets and how small $I_n$ is as a result. A similar argument shows $f(c)$ cannot be larger than $L$, and therefore $f(c)=L$.