A function $f:[a,b]\to \mathbb{R}$ is said to have a local maximum at $p$ if there exists a neighborhood $N_\delta(p)$ of $p$ such that $f(x) \leq f(p)$ for all $x \in N_\delta(p) \cap[a,b]$. The concept of local minimum at $p$ can be defined similarly.
Suppose that $f:[a,b]\to \mathbb{R}$ is continuous and has a local minimum at $p$ and at $q$, where $p < q$. Prove that $f$ has a local maximum at $r$, where $r \in (p,q)$.
Proof. Let $E + [p,q] \subseteq [a,b]$. Since $f$ is continuous on $[a,b]$, it is also continuous on $E$. By the Extreme Value Theorem, there exists a $r \in E$ such that $$f(r)=\sup_{x \in E}f(x). \label{7.48}\tag{7.48}$$ If $p < r < q$, then since the supremum \eqref{7.48} indicates that $f(r)$ is the maximum value of $f$ on $E$, it is definitely a local maximum of $f$. If $r= p$, then since $f(p)$ is a local minimum, there exists a $\delta > 0$ such that $p < x < p + \delta < q$ implies $$f(p) \leq f(x) \leq f(r),$$ i.e., $f(x) = f(p) = f(r)$ for all $x \in (p,p+\delta) \subseteq (p,q)$. Now we may pick any $r' \in (p,p+\delta)\subseteq (p,q)$ so that $$f(r')= f(r) = \sup_{x \in E}f(x),$$ i.e., $f$ has a local maximum at $r' \in (p,q)$. The case for $r = q$ is similar, so we omit the details here. Thus we have completed the proof of the problem.
Why does $f$ attain its local maximum at $r'\in (p,q)$. Is it because we have $f(x)\leq f(r') $ for all $ x \in (p,p+\delta)$?
The function $f$ is constant on $(p,p+\delta)$ !
Let $r' \in (p,p+\delta)$ then for any $x \in (p,p+\delta)$ we have
$$f(r') = f(x) \implies f(r') \geq f(x)$$
Now take an open neighbourhood of $r'$ which is a subset of $(p,p+\delta)$ to conclude that any $r' \in (p,p+\delta)$ is a local maximum.