# Can endpoints be local minimum?

My textbook defines local maximum as follows:

A function $$f$$ has local maximum value at point $$c$$ within its domain $$D$$ if $$f(x)\leq f(c)$$ for all $$x$$ in its domain lying in some open interval containing $$c$$.

The question asks to find any local maximum or minimum values in the function

$$g(x)=x^2-4x+4$$ in the domain $$1\leq x<+\infty$$.

The answer at the back has the point $$(1,1)$$, which is the endpoint.

According to the definition given in the textbook, I would think endpoints cannot be local minimum or maximum given that they cannot be in an open interval containing themselves. (ex: the open interval $$(1,3)$$ does not contain $$1$$). Where am I wrong?

• Your question is an excellent and important one. It is common to include endpoints in these types of calculations, and note that every point is in an open interval, just not necessarily an open interval in the domain of the function. Be sure to also check these definitions with and instructor or professor for total clarification. Commented Feb 8, 2017 at 2:15
• This is mainly just a matter of convention - lacking a deep mathematical basis for preferring inclusion or exclusion of these points. I have seen professors and texts that disagree on this point. What really matters is that a convention is decided upon and used consistantly. I would agree that you ought to ask your professor for clarification on the convention they want to use. Commented Feb 8, 2017 at 3:19
• Yes, it is a matter of convention. I personally think that the definition where "open" means open in $D$ makes more sense. Apparently, so does your textbook. However, the majority of Calculus textbooks that I've seen specifically exclude endpoints from the definition of local extrema, i.e. they treat "open" as being open in $\mathbb{R}$. Commented Feb 8, 2017 at 4:46

Actually, the question is settled by reading the definition you provided carefully:

A function $f$ has local maximum value at point $c$ within its domain $D$ if $f(x)\leq f(c)$ for all $x$ in its domain lying in some open interval containing $c$.

I.e., the points $x$ for which the condition must hold are required to both be in the open interval and in $D$.

To see that $(1,1)$ is a local maximum, consider the open interval $(0, 2)$. If $x \in (0, 2)$ and $x$ is also in the domain $[1,\infty)$, then $1 \le x < 2$. Now $g(x) = x^2-4x + 4 = (2 - x)^2$. So $g(1) = (2 - 1)^2 = 1^2 = 1$, but if $x > 1$, then $0 < 2 - x < 1$, so $0 < (2-x)^2 = g(x) < 1$. So for $x$ in the open interval $(0,2)$ and also in the domain $[1,\infty)$, we have that $g(x) \le g(1)$.

• For a different perspective, this is not how my calculus professor taught it recently. She said that if you are at an endpoint, you cannot compare the values outside of the interval, so you would not include it as a local extremum. Commented Nov 18, 2017 at 20:58
• @PaulSinclair, it's actually common for textbooks to take this perspective. For example, the book with scans posted on this question (which is Stewart Calculus, if I'm not mistaken). Commented Feb 21, 2018 at 2:28
• @MattBrenneman - $[a,c)$ is indeed an open neighborhood of $a$ in $D$. And it is exactly because "local extremum" is defined on topological spaces as "extremum when restricted to some neighborhood" that I take exception to Max Li's professor's remark. Under this definition, $a$ is a local extremum, just as it is under the explanation I gave in the post (which essentially amounts to the same thing, but without the subspace terminology). Max Li's professor would deny it that status, even though it is greater (or lesser) than everything near it in the domain. Commented Oct 24, 2018 at 16:22
• @Mr_3_7 It is simply a ridiculous definition, for which I can think of not a single reason it would ever be preferable to the sensible definition given in the OP. The best I can come up with is so that one can say "all local extrema are critical points", instead of saying "all local extremum are critical or boundary points" (given differentiability). But that is actually less useful, since the boundary points may be the maximum or minimum. Can you give even one example where this non-domain definition is more useful than the one here? Commented Nov 6, 2019 at 17:49
• @AllanHenriques - You've reviewed "most textbooks" and interviewed "most professors"? It is your concept that is trying to change the answer based on what is outside the domain. Mine says the domain is the only thing that exists. When you look only at the domain, there is no such thing as a "boundary point" or anything beyond it. So concepts depend on the domain sitting inside some larger space. So if you want a definition that does not depend on things outside the domain, you cannot dis-allow boundary points as extrema, because there is no such thing. Commented Dec 11, 2020 at 20:35

I think fundamentally the comments are right, and you should speak with your teacher to confirm definitions and expectations. But there's also a point to make about topology here, which could justify the book's definition and answer as consistent.

The definition of local maximum you gave is:

A function $f$ has a local maximum at point $c$ within its domain $D$ if $f(x) \leq f(c)$ for all $x$ in its domain lying in some ** open ** interval containing $c$.

If you interpret this as saying that the interval can come from $\mathbb{R}$, and is not restricted to $D$, then you have no problem, as others have pointed out. But like you I am thinking about being restricted to $D$ and my instinct is to think only about intervals in $D$. This can still be ok, if we just alter our interpretation of "open" a little bit (in a natural way)...

Now, whenever we say "open" we're really saying "open with respect to ** insert topology here ** ." A lot of the time it's obvious from context or the textbook has established a practice of contextual implication, but in this case (without knowing your book) I'd argue there are two reasonable interpretations:

1. We might be talking open intervals with respect to the standard topology on $\mathbb{R}$ (which is what you've probably been using in your class), but
2. since we're restricting our attention to a domain $D \subset \mathbb{R}$, it's also pretty normal to talk about a different topology, called the subset topology on $D$ (induced by the standard topology on $R$).

In the subset topology on $D \subset \mathbb{R}$ (induced by the standard topology), a set $S$ is open if and only if $S$ is the intersection $D \cap X$, with $X$ open in $\mathbb{R}$ with respect to the standard topology on $\mathbb{R}$.

We're often more interested in the subset topology than the usual topology on the whole space just because of situations like the one you're in, in which a definition doesn't work quite like you expect when $D \not= \mathbb{R}$.

So let's work with a slightly different definition of local maximum:

A function $f$ has a local maximum at point $c$ within its domain $D$ if $f(x) \leq f(c)$ for all $x$ in its domain lying in some interval $I$ containing $c$ such that $I$ is open with respect to the subset topology on $D$.

Now back to your case. Let $D = [1, \infty)$. For any $a > 1$, we have that $$[1,a) = D \cap (-a,a)$$ Since $(-a,a)$ is open in $\mathbb{R}$ with respect to the standard topology, $[1,a)$ is open in $D$ with respect to the subset topology on $D$. This intuitively makes sense, because if you were an ant walking on $f(D)$, when you came to $f(1)$ you'd have nowhere to go but down.

• I am going to have to spend some time trying to understand your answer, but thank you nonetheless.
– Phil
Commented Feb 11, 2017 at 23:13

No you guys all missed the correct answer, simply look at the function X^2 how many local maximum or minimum does it have? The answer is it has only one local minimum which is also an absolute minimum. Now if you modify that definition in your textbook to work for closed intervals then you would simply have an endless number of local minimum and maximum values which is not possible.