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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?

Chris Tang
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Phil
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    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 08 '17 at 02:15
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    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 08 '17 at 03:19
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    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 08 '17 at 04:46

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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 $g(x) \le g(1)$.

Paul Sinclair
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    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. – Jack Pan 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 00:52
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    @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 02:28
  • @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
  • @PaulSinclair It's been a while since my topology, so I may not quite remember the right language and all, but I seem to remember that the restriction of the domain to a subset of a topological space, changes the definition of what "closed" and open" mean somewhat. So, for example, if you stay in the reals and you have a bounded and closed interval, D= [a,b], (which you are now taking to be the domain of your function), aren't intervals like [a,c) where c<b, technically "open" in D (a topological sense wrt the subspace topology induced by the closed interval)? – Matt Brenneman Oct 24 '18 at 12:45
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    @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
  • @Paul Sinclair. No need to be shocked. As the answer below points out, whether the frontier points of a compact subset of $\mathbb R$ can be classified as local extrema, depends on whether the definition of local extrema is given in terms of the standard metric topology on $\mathbb R$ or that of a subspace given by the domain of the function. In the first case $[a, c)$ is not an open set, in the second it is. My old calculus textbook agrees with MaxLi professor, and so do other textbooks. Whether this is suboptimal or not is a matter of what applications one has in mind. – Mr_3_7 Nov 06 '19 at 11:14
  • @Mr_3_7 - Did you double check that your old calc textbook actually defines local exrema without restriction to the domain? Or is that just the way you remember it? One of the many problems with this concept is that is cannot be extended sensibly to functions on more general sets. It depends on a "master space" ($\Bbb R$) of which all function domains must be subsets to even be applicable. When you start allowing domains to include, for example, complex numbers, then suddenly every real domain consists of nothing but boundary points, so no function with real domain ever has local extrema, – Paul Sinclair Nov 06 '19 at 17:39
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    @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? – Paul Sinclair Nov 06 '19 at 17:49
  • @Paul Sinclair. By establishing a tighter connection between the analytical properties of the function and the local extrema, you avoid the proliferation of new local extrema every time you look at the same analytical expression on different domains, which one might argue is not very principled. The "open" in $\mathbb R$ definition is looks to me as it avoids overclassification. I do not exactly understand the comment on "being less useful since they can be (absolute?) maxima/minima". We might disagree on this, but certainly such a definition is by no means "ridiculuous". – Mr_3_7 Nov 08 '19 at 08:52
  • This answer is bs. You can't just treat the function as if you can predict the values outside the interval in which its defined, the concept of a domain exists for a reason. Local minimums CANNOT and SHOULD NOT exist at the endpoints. Most textbooks and professors agree with me on this. – Allan Henriques Dec 11 '20 at 20:27
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    @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. – Paul Sinclair Dec 11 '20 at 20:35
  • @PaulSinclair When looking at the domain as if it sits inside a larger space you could imagine that the number line extends beyond where the function is defined, but you cannot plug in any of those other x values into the function, this way you could have something like boundary points. By saying that "the domain is the only thing that exist", you are just adding some unnecessary layer of abstraction on the problem when it is sufficient to say that there are no coordinates corresponding to x values outside the domain. Absolute extrema can lie at endpoints, but relative extrema cannot. – Allan Henriques Dec 11 '20 at 20:48
  • @AllanHenriques - I am not "adding some unnecessary layer of abstraction". Believe it or not, optimization is a concept for functions defined on many spaces. It is you are insisting on an unnatural idea that depends not only on the function, including the domain, but on some larger space where you've decided to insert it. And if someone else inserts it in a different space, suddenly what is a local extrema and what is not must change. This lowers the value of the terminology. – Paul Sinclair Dec 11 '20 at 21:45
  • @AllanHenriques - and why? You claim it is because "we can't predict the values of the function outside the interval in which its defined". That is false - It has no values outside of its domain. There is nothing to predict! You are trying to compare the functions behavior to your imagination. I am comparing only to actual values. – Paul Sinclair Dec 11 '20 at 21:47
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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.

Badam Baplan
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  • 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
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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.

Fadel Hossam Manour
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