# Can a function have a strict local extremum at each point?

A problem given in Spivak's Calculus text is to show that a function $f:[a,b]\to \mathbb{R}$ cannot have a strict local maximum at each point. I will sketch the proof below the fold.

My question is: can a function $f:[a,b]\to \mathbb{R}$ have a strict local extremum at each point, perhaps a combination of strict local minima and maxima?

If there did exist such a function, the sets $\{x: x\text{ a strict local maximum}\}$ and $\{x: x\text{ a strict local minimum}\}$ must both be dense, since if not, once we get our hands on an interval which contains only one type of extremum, the proof goes forward as before.

Addendum: Such a function, if it were to exist, could not be continuous on any nonempty open set, because I claim a continuous function with a local (non-strict) extremum at each point must be a constant, which is a contradiction, because a constant function has no strict extrema.

Proof that $f:[a,b]$ cannot have a strict local max at each point: Suppose $f$ is such a function. take any point $x_1$ in $[a,b]$ and surround it with an interval $[a_1,b_1]$ such that $0<b_1-a_1< \frac12 (b-a)$ such that $f(x)<f(x_1)$ for all $x\in [a_1,b_1]$. Now take any point in $[a_1,b_1]$ and get an interval $[a_2,b_2]$, again reducing the length by at least half, and so on. These intervals must have a common point $p$, which cannot be a local maximum, since it has points $x_i$ arbitrarily close to it such that $f(x)>p$.

• Thomae's function has a strict local maximum at every rational number and a not-strict local minimum at each irrational number. That doesn't quite meet the criteria for being a counterexample, but it might be a good starting point. Aug 11 '14 at 23:03

No.

It follows from the fact that a function can have only countably many strict local maxima and minima.

Part 1. Although continuousness is mentioned in the question, it isn't needed.

I think your inquiry comes from Chapter 11 Question 71b) of Spivak's Calculus Edition IV, which reads as follows:

Suppose that every point is a local strict maximum point for $$f$$. Let $$x_1$$ be any number and choose $$a_1 \lt x_1 \lt b_1$$ with $$b_1-a_1 \lt 1$$ such that $$f(x_1) \gt f(x)$$ for all $$x \in [a_1,b_1]$$. Let $$x_2 \neq x_1$$ be any any point in $$(a_1,b_1)$$ and choose $$a_1 \leq a_2 \lt x_2 \lt b_2 \leq b_1$$ with $$b_2-a_2 \lt \frac{1}{2}$$ such that $$f(x_2) \gt f(x)$$ for all $$x$$ in $$[a_2,b_2]$$. Continue in this way, and use the Nested Interval Theorem (Problem 8-14) to obtain a contradiction.

It seems as though there are several typos in Spivak's proposed approach; however, the theme of the method is valid, and we can proceed as follows:

Suppose that every point in $$\text{dom}(f)$$ is a local strict maximum point of $$f$$. Consider an arbitrary $$x_1$$. By definition, there is a $$\delta \gt 0$$ such that for any $$x \in (x_1 - \delta, x_1 + \delta) \setminus \{x_1\}$$: $$f(x_1) \gt f(x)$$. Now, let $$\delta_1 \lt\min\left(\delta,\frac{1}{1\cdot2}\right)$$ and choose an $$a_1,b_1 \in (x_1-\delta_1,x_1+\delta_1)\setminus \{x_1\}$$ such that $$a_1 \lt x_1 \lt b_1$$. We see that for any $$x \in [a_1,b_1] \setminus \{x_1\}$$: $$f(x_1) \gt f(x)$$. Importantly, note that $$x_1 \in [a_1,b_1]$$ and $$|a_1 - b_1| \lt \frac{1}{1}$$.

Next, consider any $$x_2 \in (a_1,x_1)$$. By definition, there is a $$\delta$$ such that for any $$x \in (x_2 - \delta, x_2+\delta) \setminus \{x_2\}$$: $$f(x_2) \gt f(x)$$. Let $$\delta_2 \lt\min\left(\delta, \frac{1}{2 \cdot 2},|x_1-x_2|,|a_1-x_2|\right)$$ and choose an $$a_2,b_2 \in (x_2-\delta_2,x_2 + \delta_2)\setminus\{x_2\}$$ such that $$a_2 \lt x_2 \lt b_2$$. Note that $$x_1 \notin [a_2,b_2]$$ and $$x_2 \in [a_2,b_2]$$. Therefore, we can say that for all $$x \in [a_2,b_2] \setminus \{x_2\}$$: $$f(x_2) \gt f(x)$$. Further, from our definition of $$\delta_2$$, we have that: $$\delta_2 \lt x_1 - x_2$$. Because $$x_1 \lt b_1$$, we have $$x_1 - x_2 \lt b_1 -x_2$$. Together, we have $$\delta_2+x_2 \lt b_1$$. But $$b_2 \lt x_2 + \delta_2$$, so we conclude that $$b_2 \lt b_1$$. Similarly, we know that $$-\delta_2 \gt a_1 - x_2$$. This implies that $$-\delta_2 + x_2 \gt a_1$$, but $$a_2 \gt x_2 -\delta_2$$. Therefore, we must have $$a_2 \gt a_1$$, which means that $$a_1 \lt a_2 \lt b_2 \lt b_1$$, where $$|a_2 - b_2| \lt \frac{1}{2}$$.

We can continue in this fashion for any $$n \in \mathbb N$$, which allows us to make two intermediate conclusions:

1. Noting that for any $$n$$ we have $$0 \lt |a_n -b_n| \lt \frac{1}{n}$$, we must have that $$\displaystyle \lim_{n \to \infty} |a_n-b_n|=0$$.

2. Letting $$I_n=[a_n,b_n]$$, we have that for any $$n$$: $$I_{n+1} \subset I_n$$.

Under these conditions, we have that $$\bigcap _{i=0}^\infty I_n \neq \emptyset$$. In particular, we have that $$\bigcap _{i=0}^\infty I_n \neq \emptyset$$ is a singleton. Call this element $$x^*$$. By assumption, $$x^*$$ is a local strict maximum point of $$f$$, which means there is a $$\delta^*$$ such that for all $$x \in (x^*-\delta^*,x^*+\delta^*) \setminus\{x^*\}: f(x^*) \gt f(x)$$ $$\quad(\dagger_1)$$.

However, note that by the Archimedean Property, there is an $$n$$ such that $$\frac{1}{n} \lt \delta^*$$. Consider the $$I_n$$ whose total length $$|a_n-b_n| \lt \frac{1}{n}$$. Because $$x^* \in [a_n,b_n]$$, it must be the case that all of $$[a_n,b_n]$$ fits inside $$(x^*-\delta^*,x^*+\delta^*)$$. i.e. $$[a_n,b_n] \subset (x^*-\delta^*,x^*+\delta^*)$$. By construction, we know that all $$x \in [a_n,b_n]\setminus {x_n}$$ satisfy $$f(x_n) \gt f(x) \quad (\dagger_2)$$. More importantly, by construction, we also know that $$x_n \notin [a_{n+1},b_{n+1}]$$. This implies that $$x_n \neq x^*$$. $$(\dagger_2)$$ would therefore imply that $$f(x_n) \gt f(x^*)$$, which would contradict $$(\dagger_1)$$ because $$x_n \in (x^*-\delta^*,x^*+\delta^*) \setminus\{x^*\}$$.

So it must be the case that not every point in $$\text{dom}(f)$$ is a local strict maximum point of $$f$$.