# 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. – The Count Feb 8 '17 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. – David Feb 8 '17 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 $\mathbb{R}$. – zipirovich Feb 8 '17 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. – Max Li Nov 18 '17 at 20:58
• @MaxLi - your calculus professor said WHAT!? Either you misunderstood your what your calculus professor was saying (which I unfortunately have to rate at only a 90% probability instead of the 100% probability I would prefer to), or else I advise you to carefully double-check everything she tells you against reliable sources and avoid taking any more classes from her in the future. Some things I can shrug off as differences in taste (is $0 \in \Bbb N$ or not?), and some things as pointless but not really harmful (is $0^0 = 1$ or undefined?). But this violates the basic meaning of the words. – Paul Sinclair Nov 19 '17 at 0:52
• @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). – PersonX Feb 21 '18 at 2:28
• @PersonX - Just because my comment has a wider application than one person does not at all make me recant. – Paul Sinclair Feb 21 '18 at 13:33
• @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. – Paul Sinclair Oct 24 '18 at 16:22

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 Feb 11 '17 at 23:13