4

Assume $f$ is continuous over $[a,b]$ and $f$ has a local maximum value at every point in $(a,b)$. Show that $f$ is constant.

Intuitively it makes sense that the constant function should be the constant function, but maybe one strategy is to prove that it is a necessary and sufficient condition on $f$. In other words, $f$ is continuous over $[a,b]$ and $f$ has a local maximum value at every point in $(a,b)$ if and only if $f$ is constant. Alternatively we can think of it as such: at every point $A_i$ over $[a,b]$, there exists some interval $J_i$ such that $A_i$ is greater than or equal to every value in of $f$ in $J_i$.

I am unsure which route to take to solve this.

user19405892
  • 15,870
  • It's not a local maximum--that's not possible with a constant function since the sign of the derivative never changes (it's always $0$)--there are no local extrema. – Jared Mar 31 '16 at 01:32
  • @Jared Every point for a constant function is a local maximum, though. – user19405892 Mar 31 '16 at 01:34
  • @Jared I think this is the following definition they're using for $f$ having local maxima at $x=x^$: $$(\exists \epsilon > 0)(\forall x \in \Bbb{R})(\lvert x-x^ \rvert < \epsilon \implies f(x^*) \geq f(x))$$ It's the same definiton as used in this Math StackExchange question. – Noble Mushtak Mar 31 '16 at 01:34
  • @NobleMushtak I'm not sure...but it would seem to me that this would label inflection points where $f'(x) =0$ as local extrema...is that wrong (I mean I'm fairly certain that inflection points are not local extrema)? – Jared Mar 31 '16 at 01:38
  • @Jared Inflection points are in no way local extrema. That definition of local maximum would not include inflection points as local extrema. – Noble Mushtak Mar 31 '16 at 01:39
  • @user19405892 Every point for a constant function is an absolute maximum (and minimum)--not local (at least in my opinion--that may be wrong it sounds like though). – Jared Mar 31 '16 at 01:40
  • @Jared Aren't all absolute maximums also local maximums, though? Also, this paper seems to deal with this issue, but it seems really complicated. I'll try to post an easier explanation if I can understand the paper. – Noble Mushtak Mar 31 '16 at 01:41
  • @NobleMushtak No, absolute extrema are definitely not necessarily local extrema. Usually when we talk about $\mathbb{R}\mapsto\mathbb{R}$ functions, and the function is differentiable (therefore continuous) then the absolute extrema over the domain $(-\infty, \infty)$, if they exist, will be a local extrema. – Jared Mar 31 '16 at 01:45
  • @Jared I never said that absolute extreme are local extrema. – user19405892 Mar 31 '16 at 01:48
  • ahh, sorry, I meant to address @NobleMushtak. – Jared Mar 31 '16 at 01:49
  • I'm wondering what can be said if $f$ is not assumed to be continuous. So far I have (1)When $a<a'<b<b'$, the set$ f[a',b']$ is countable and well-ordered (by the usual order on $R$), and (2) The image of $(a,b)$ is countable. – DanielWainfleet Mar 31 '16 at 03:14
  • @Jared Can you give an example of an absolute maximum that would not fit the definition of local maximum I have above? I don't see how an absolute maximum would not fit that definition since an absolute maximum has $f(x^*) \geq f(x)$ for all $x$ in the domain. – Noble Mushtak Mar 31 '16 at 11:47
  • 1
    @NobleMushtak Consider the function $f(x) = x^3$ over the domain $x \in [-3, 3]$. The absolute maximum is at $x = 3$ and has a value of $27$ and the absolute minimum occurs at $x = -3$ and has a value of $-27$. Neither of these are local extrema since the function $f(x) = x^3$ has no local extrema. If you're saying that we should only consider $x$ inside the given domain (which your definition does not) then yes, those two points satisfy your definition of local extrema--but I've never seen any text book or Calculus course (at any level) that would consider those local extrema. – Jared Mar 31 '16 at 23:02
  • @Jared Oh...That makes sense! Thanks for the example! If I changed my definition to $\forall x \in D$ where $D$ was the domain of $f$, then that example would also say that those absolute extrema are not local extrema since there's no positive interval around $-3$ or $3$ as they're on the boundary of the domain. However, that doesn't apply to this because all of the points said to be local maximums are in $(a, b)$ while the domain is $[a, b]$, so the hypothesis never says that a local maximum is on the boundary of the domain. – Noble Mushtak Mar 31 '16 at 23:26

3 Answers3

2

Let $c\in [a,b]$ be a point where $f$ attains its global minimum value $m$.

Let $A = \{ x \in [a,c] : f(y)=m \text{ for all } y\in [x,c] \}$ and let $a'=\inf A$.

Since $f$ is continuous, $f(a')=m$ and so $a'\in A$ and $A=[a',c]$.

Since $a'$ is a local maximum, there is an interval $I$ around $a'$ such that $f(x)\le f(a')=m$ for all $x \in I$. This implies that $f(x)=m$ for all $x \in I$. Therefore, $a'=a$, because otherwise $I$ would contain a point of $A$ less than $a'$. Thus, $A=[a,c]$.

Analogously, by considering $B = \{ x \in [c,b] : f(y)=m \text{ for all } y\in [c,x] \}$, we get that $b'=\sup B=b$ and $B=[c,b]$.

Therefore, $f$ is constant in $[a,b]$, because $[a,b]=[a,c]\cup[c,b]=A\cup B$ and $f$ is constant, equal to $m$, in $A\cup B$.

lhf
  • 221,500
1

Suppose $f$ has local maximum at every $x \in (a,b)$. Let $U \subset (a,b)$ be the open set for which $x$ is a local maximum. Let $y$ be different from $x$ and in $U$. WLOG suppose $y>x$. Suppose $f(x) \not = f(y)$. Let $V$ be the open set for which $y$ is a local maximum. Then on $V \cap U$ both $x,y$ are local maximums. However, $x<y$ and $y \in U \cap V \subset U$ implies $f(x)\geq f(y)$ but $U \cap V \subset V$ and $x<y$ implies $f(x)\leq f(y)$; hence $f(x) = f(y)$. Since this we can do this for any $x \in (a,b)$ we are done i.e $f$ is constant on $(a,b)$.

Remark: A more general statement is that if $X$ is a connected topological space, $f: X \to \mathbb{R}$ is continuous (and $\mathbb{R}$ has the usual topology) and $f$ is local constant, then $f$ is constant.

Faraad Armwood
  • 9,156
0

Hint: you need to break the continuity of $f$.

Also, do not assume that $f$ is differentiable -- you can't use that $f'(x) = 0$.

User001
  • 1