Math for Non-Geeks/ Extreme value theorem

{{#invoke:Math for Non-Geeks/Seite|oben}} In the following we are going to deal with continuous functions on compact intervals. These are intervals that are closed and bounded, i.e. have the form $[a,b]$. We will see that such functions are always bounded and attain a minimum and maximum. This theorem is called the Extreme Value Theorem. It is used in mathematics to prove the existence of relative extrema, i.e. points of a function that are "at the extreme" of being the lowest point in the graph (the minimum) or the highest point in the graph (the maximum).

Motivation

File:Satz vom Minimum und Maximum – Intuition und Motivation.webm Let's take a continuous function $f$ which is defined on a compact interval $[a,b]$. I.e. we are considering a function $f:[a,b]\to ℝ$. This function has the value $f(a)$ at the point $a$ and the value $f(b)$ at the point $b$.

The function values f(a) and f(b)

Now $f$ is defined for every intermediate point between $a$ and $b$. Intuitively, for functions without gaps in the domain of definition, continuity means that we are able to draw the graph without lifting the pen from the paper. Hence, the graph of $f$ connects the points $(a,f(a))$ and $(b,f(b))$ by a continuous path without jumps. The following graph provides an example for such a function $f$:

The fraph of the function f

We note that the function $f$ above is bounded. And it attains a maximum and minimum value:

The function f ist bounded and attans a maximum and a minimum

Is it always that way? Try for yourself to connect the points $(a,f(a))$ and $(b,f(b))$ by different graphs without lifting the pen. Could you imagine to draw the graph of an unbounded function - even if your paper was infinitely large?

Intuitively, the answer is no. No matter how far your graph goes up or down, you need to return to the end points at $(a,f(a))$ or $(b,f(b))$. Going to infinity "forces you to lift the pen" and is therefore not allowed. However, the function $f$ can attain very large values (like $10^{100}$ or more) while staying bounded, as long as you return in order to reach the end point. This situation is illustrated in the following figure:

Another bounded function with values f(a) and f(b), attaining a maximum and a minimum

So our first intuition tells us that when connecting the two points $(a,f(a))$ and $(b,f(b))$ without lifting the pen, our function stays bounded. And it attains a maximum and a minimum. Now let's think about what could go wrong when phrasing this intuition in a mathematical way. The end points $a$ and $b$ of the domain of definition $[a,b]$ could become problematic: For an open domain of definition $(a,b)$, the function could run towards $\pm \mathrm{\infty }$ at $a$ or $b$, or it could converge towards a value without attaining it. Including the boundary points $\{a,b\}$ in the domain of definition excludes these cases as it "catches the function" at the end points $(a,f(a))$ and $(b,f(b))$. If we move a boundary to infinity, let's say by considering the domain of definition $[a,\mathrm{\infty })$, the function could have "infinitely much time" and might run towards infinity while being continuous. This happens for instance for $f(x)=x^2$. So we also expect problems with unbounded domains of definition. A statement like "the maximum and minimum are attained" can only be expected to hold true on a compact interval $[a,b]$. Now, in between $a$ and $b$, the function could also "break out" and tend towards $\pm \mathrm{\infty }$ (like $f(x)=\frac{1}{x}$ near $x=0$). This scenario will be prevented by assuming continuity of the function $f$.

In the following, we will mathematically verify that our intuition is true. That means we prove a statement like "the maximum and minimum are attained" for continuous functions defined on a compact interval $[a,b]$ and discuss what may go wrong if we choose other domains of definition.

Assumptions of the theorem

File:Voraussetzungen des Satzes vom Maximum und Minimum.webm Let's take a look at the assumptions made within the extreme value theorem:

• $f$ is a continuous function
• $f$ is defined on a compact interval $[a,b]$

Are those assumptions really necessary or can we relax them without losing validity of the extreme value theorem?

Assumption of continuity

Function $f(x)=\frac{1}{x}$ on the interval $[-1,1]$ with $f(0)=0$

First, we note that continuity prevents the function $f$ from "breaking out" to $+\mathrm{\infty }$ or $-\mathrm{\infty }$ within its domain of definition. Ij we just allow any function $f:[a,b]\to ℝ$ , no matter whether it is continuous or not, we will find non-continuous functions which are violating the extreme value theorem. The following function is unbounded (so it does not attain any extrema) and non-continuous at $x=0$ :

So we cannot simply drop the assumption that $f$ is continuous.

Interval-assumption

The function $f(x)=\frac{1}{x}$ on $(0,1]$

The domain of definition is also important. It must include its boundary (i.e. be closed). This way we "catch" the function at the interval boundary and make sure it does not "run away" towards infinity. The function $g:(0,1]\to ℝ:x\mapsto \frac{1}{x}$ is an example which "runs away" as we approach $x=0$.

Unboundedly large domains are also problematic, since the function has "infinitely much time" to run away. An easy example is the function $h:[0,\mathrm{\infty })\to ℝ:x\mapsto x^2$. And there are functions, which are defined on a bounded domain of definition, continuous and do not "run away" towards infinity, but do not attain an extremum. This happens if there are open boundaries or gaps within the domain of definition. The extremum would then be attained at the boundary (or the gap) - but this argument has been removed from the domain of definition.

The same may happen when removing a maximum or minimum from the interior of the domain of definition instead of the boundary - which creates a gap. Of course, the function may also "run away" at such a gap. An example for this effect is the continuous but unbounded function $j:[-1,1]\setminus \{0\}\to ℝ:x\mapsto \frac{1}{x}$. The argument $x=0$ is excluded from the domain of definition $[-1,1]\setminus \{0\}=[-1,0)\cup (0,1]$ . So this function is well defined and continuous, but it "runs away" at the gap. In a mathematical language, the function is unbounded and hence violates the conclusion of the extreme value theorem.

Outlook: Generalization of the theorem

So far, we only considered intervals (possibly with gaps) as candidates for the domain of definition. Is this restriction really necessary? This time, the answer is no. For instance, we can take the union of two intervals $D=[a,b]\cup [c,d]$ with $a<b<c<d$ and define some continuous and real-valued function $j$ on $D$. If we restrict $j$ to only $[a,b]$ or $[c,d]$ , we can apply the extreme value theorem. Both the functions $j_1=j{|}_{[a,b]}$ and $j_2=j{|}_{[c,d]}$ with restricted domain of definition are bounded and hence attain a maximum and a minimum. The function $j$ must therefore also be bounded. Its maximum is the larger of the both maxima of $j_1$ and $j_2$ , so $j$ also attains a maximum (the same holds for the minimum). Therefore, every continuous function defined on the union of two closed intervals $[a,b]\cup [c,d]$ fulfills the conclusion of the extreme value theorem. The same holds if we consider three or more closed intervals - or an even larger class of domains of definition. In fact, we can precisely state what this larger class of domains of definition is:

If we take a second look at the proof, we note that the domain of definition is only mentioned at one point: where we make use of the Bolzano-Weierstraß theorem. We used it to show that any sequence from the domain of definition contains a convergent subsequence. Hence, the proof arguments hold true, as long as the domain of definition allows for the usage of the Bolzano-Weierstraß theorem.

So we can generalize the above theorem. It will hold not only on closed intervals $[a,b]$, but on all sets satisfying the Bolzano-Weierstraß theorem. We will call these sets satisfying the Bolzano-Weierstraß theorem sequentially closed sets:

If the domain of definition $D$ of a continuous function $f:D\to ℝ$ is sequentially compact, then the function $f$ must fulfill the extreme value theorem. The generalization of sequential compactness from real numbers to other sets of mathematical objects is one of the topics dealt with in topology.

Exercise: Image of polynomials of even degree

File:Aufgabe – Bild einer Polynomfunktion.webm Math for Non-Geeks: Template:Aufgabe

Exercise: Continuous functions on [0,1]

File:Aufgabe – Stetige Funktionen auf kompakte Intervalle.webm Math for Non-Geeks: Template:Aufgabe

In addition, the statement of the above exercise can be generalized in multiple ways:

1. We have shown that on a compact interval $[a,b]$ , there is no continuous function attaining each value twice.
2. Similarly, one may show that there is no continuous function $f:ℝ\to ℝ$attaining each of its values twice.
3. And for each given number $n\in \mathbb{N}$, $n>1$ , one can show that there is no continuous function $f:[a,b]\to ℝ$ attaining each of its values exactly $n$ times.

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