Wikipedia says that:

A real-valued function $f$ defined on a real line is said to have a local (or relative) maximum point at the point $x^*$, if there exists some $\varepsilon > 0$ such that $f(x^*) \ge f(x)$ whenever $\lvert x − x^*\rvert < \varepsilon$. The value of the function at this point is called maximum of the function. Similarly, a function has a local minimum point at $x^*$, if $f(x^*) \le f(x)$ whenever $\lvert x − x^*\rvert < \varepsilon$. The value of the function at this point is called minimum of the function.

Now I don't understand one thing. Why, in the definition, is the use of the "liberal-inequality" - $f(x^*) \ge f(x),\ \:\ f(x^*) \le f(x)$ - instead of strict inequality? Agreed that this is a definition, and it doesn't make sense to question a definition - but I'd like to know what's the motivation behind using the liberal inequality.

  • $\begingroup$ Take the following function: $$ f(x)=\begin{cases} -x+1&x\in(-\infty,-1]\\ 0& x\in[-1,1]\\ x-1& x\in[1,\infty) \end{cases} $$ Then $f$ has no local extrema if we take the strict inequality definition. But all $x\in(-1,1)$ satisfy $f'(x)=0$ and the 'loose' inequality definition. This is not so much an answer as a problematic situation. $\endgroup$ – Ian Coley Apr 12 '13 at 16:09

One way to think about this is in the real world sense. Suppose you're walking on a plateau, that is, a bump rising out the ground with a flattened out top (here we assume the ideal situation in which anywhere on the top is at the same level as another). If you're standing on top of it, at any point, you would be at the highest height you can be walking on the plateau, correct? However, suppose we apply the math definition to our situation in which $\geq$ is replaced by $>$. Then it would be impossible to reach the highest height because at every point on top of the plateau, you are at the same height, meaning there isn't a point on it that is strictly higher than all others. This is course absurd as we know we're standing at the highest height on the plateau! So it makes sense to define the maximum with a 'liberal' inequality in order to more properly match what we would think in real life. Does this make sense?

  • $\begingroup$ I do get your point, but the thing is, we often say maximum is attained at this point, minimum is attained at that point. So that gives a feeling (to me, that is!) that no other point in the neighborhood of the point attains a greater or smaller value. $\endgroup$ – Parth Thakkar Apr 13 '13 at 17:07
  • $\begingroup$ Probably, it is similar to limits - when something (x) tends to some point (a), it just gets as close as possible to a but never becomes a (The question I get here is, how can you say that the limit is 'a' and not something else in the neighborhood of 'a'). Btw, I do know the exact definition of limits, but still, it seems odd! $\endgroup$ – Parth Thakkar Apr 13 '13 at 17:12
  • $\begingroup$ Well, it is not exactly as you say... if a sequence $(x_n)_n$ has a limit point $a$, you can very well have $x_n=a$ for some $n$. The point is that it may happen but it is not necessary that the $x_n$ never coincide with $a$. The idea is, as you say, that the $x_n$ go closer and closer to $a$. More precisely, for any open ball $B$ around $a$, $x_n\in B$ for all but finite $n\in\mathbb N$. You can easily see by the definition I gave you that the sequence $(y_n)_n$ with $y_n=a$ for all $n\in\mathbb N$ has $a$ as limit point. $\endgroup$ – Simone Aug 13 '13 at 17:13
  • $\begingroup$ But the liberal inequality has also its problems. In a plateau situation, you cannot determine if you are on a local maximum or minimum by varying x a little and checking f(x). Is this unavoidable? $\endgroup$ – Adam Nov 18 '13 at 15:22

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