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Continuity on a closed bounded interval gives a largest and smallest value, both attained; and what each hypothesis is doing.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to prove that a continuous function on a closed bounded interval is bounded and attains its extreme values, using Bolzano-Weierstrass twice, and to say which hypothesis each standard counterexample removes. You will also be able to distinguish a supremum of the values, which boundedness alone supplies, from a maximum, which is what the theorem adds.
Bolzano-Weierstrass, the approximation property of a supremum, and the sequential form of continuity. This lesson uses all three and adds nothing new, which is a good sign that unit 2 was the right place to spend the effort.
A value is attained when some point of the domain takes it. A maximum is an attained upper bound of the values, as distinct from their supremum, which exists whenever they are bounded. A subset of the line is compact when it is closed and bounded, which is the pair of words this theorem needs.
The extreme value theorem. A continuous function on a closed bounded interval $[a,b]$ is bounded and attains both a largest and a smallest value.
The two claims are separate and are proved in that order.
Bounded. Suppose not. Then for each $n$ there is $x_n \in [a,b]$ with $|f(x_n)| > n$. The $x_n$ are bounded, so by Bolzano-Weierstrass a subsequence converges to some $p$, and $p \in [a,b]$ because the interval is closed. Continuity makes $f$ along that subsequence converge, hence bounded — contradicting values exceeding every $n$.
Attained. The set of values is non-empty and bounded above, so it has a supremum $M$. By the approximation property choose $x_n$ with $f(x_n) > M - 1/n$. Extract a convergent subsequence with limit $p \in [a,b]$. Continuity gives $f(p) = \lim f(x_{n_k}) = M$.
What each hypothesis does. Bounded is what makes Bolzano-Weierstrass available. Closed is what keeps the limit point inside the domain. Continuous is what transports the limit from the points to the values. Every hypothesis in this subject is doing work somewhere in the proof. The way to know a theorem is to know which example breaks when a hypothesis is dropped.
Another way: picture
A curve over a closed interval has a highest point on it, and the two solid dots at the ends are part of the graph. Delete one end dot and let the curve climb towards it: the ceiling is still there and nothing on the graph reaches it. That single missing dot is the whole content of the word closed.
Another way: steps
Before claiming a maximum exists:
All three yes means a maximum and a minimum exist. Any no, and they may or may not; the theorem is then silent and an argument is needed.
| Function | Domain | What fails | Symptom |
|---|---|---|---|
| $1/x$ | $(0, 1]$ | not closed | unbounded |
| $x$ | $\mathbb{R}$ | not bounded | unbounded |
| $x$ | $[0, 1)$ | not closed | bounded, supremum not attained |
| $x$ below $1$, else $0$ | $[0, 2]$ | not continuous | bounded, supremum not attained |
The third and fourth rows are the instructive ones. Nothing is unbounded and nothing is infinite; the supremum exists perfectly well and is simply not a value. The theorem is about attainment, and only the first two rows are about boundedness at all.
The general statement is that a continuous function maps a compact set to a compact set. Closed and bounded is what compact means for a subset of $\mathbb{R}$, and this theorem is the reason that notion is worth naming.
The intermediate value theorem says a continuous function on an interval omits no value between the ones it takes. The extreme value theorem says that on a closed bounded interval it takes a largest and a smallest.
Put together: a continuous function maps a closed bounded interval onto a closed bounded interval — namely $[\min f, \max f]$. The extreme value theorem supplies the two ends, and the intermediate value theorem fills in everything between them.
That single sentence is the reason optimisation is possible in one variable. It says the search for a maximum is a search for a point rather than a hunt for a supremum that may not be there, and it is why the next unit can differentiate at an interior maximum and expect the derivative to vanish.
Treating a supremum as a maximum. The supremum of the values always exists when they are bounded. That it is a value is exactly the extra claim, and on an open interval it usually fails.
Expecting the extremum at an end. Ends are candidates, not answers. The maximum of $x(c - x)$ is strictly inside and the ends give the minimum.
Applying it on an open interval. $\tan$ on $(-\pi/2, \pi/2)$ is continuous and unbounded. Nothing about the theorem survives the missing ends.
Expecting uniqueness. A constant function attains its maximum everywhere. The theorem produces at least one point and never claims one only.
Does $f(x) = x^3 - 3x$ have a largest value on $[-2, 2]$? It is a polynomial, so continuous; the interval is closed and bounded.
The three hypotheses, checked first.
So a maximum exists and is taken somewhere. Only now is it worth looking for it.
Existence before search.
The candidates are the ends and the interior points where the derivative vanishes: $f(-2) = -2$, $f(2) = 2$, $f(-1) = 2$, $f(1) = -2$. The largest value is $2$, taken at two points.
Attained, and not uniquely.
Let $f(x) = x$ on $[0, 1)$. It is continuous and bounded above by $1$.
Two of the three hypotheses hold.
The supremum of the values is $1$, and the values climb towards it.
The supremum exists, as always.
No point of the domain has value $1$. One missing end point, and the conclusion is gone.
Closedness is not a formality.
The theorem does not apply: the domain is not bounded, so nothing is promised either way.
Check the hypotheses before guessing.
But the value at $0$ is $1$, and every other value is strictly below $1$ since $1 + x^2 > 1$.
Argue directly instead.
So the maximum is attained at $0$ after all — and the minimum is not attained, since the values approach $0$ without reaching it. A theorem that does not apply gives no verdict in either direction, which is exactly why this example is worth doing.
Build the proof that a continuous function on a closed bounded interval attains its largest value.
This task has no paper form; do it on a device.
$f(x) = x(9 - x)$ on the closed interval from $0$ to $9$. What is its largest value?
Answer:
Each function below fails to attain a largest value. Match it to the hypothesis it removes.
| the domain is not closed, and the function is unbounded | the domain is not bounded | the domain is not closed, though the function is bounded | the function is not continuous | |
|---|---|---|---|---|
| $1/x$ on the interval from $0$ to $1$, left end excluded | ||||
| $x$ on the whole real line | ||||
| $x$ on the interval from $0$ to $2$, right end excluded | ||||
| the function equal to $x$ below $1$ and to $0$ from $1$ on, on the closed interval from $0$ to $2$ |
Put the steps of the proof that a continuous function on the closed interval from $0$ to $8$ is bounded into order.
Number the steps in order (write the number in the box):
Let $h(x) = x$ on the interval from $0$ to $6$ with the right end excluded. Here is an argument that $h$ attains a largest value. Mark the one step that does not follow.
This task has no paper form; do it on a device.
A continuous $f$ on the closed interval from $0$ to $8$ is already known to be bounded. What does the extreme value theorem add?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Build the proof that a continuous function on a closed bounded interval attains its largest value.
This task has no paper form; do it on a device.
You can prove boundedness and attainment on a closed bounded interval, name the hypothesis each counterexample removes, and tell a supremum apart from a maximum. Say in your own words what the closedness of the interval is doing in the proof. Next: a strengthening of continuity that the same hypotheses buy.
10. Your turn: does $f(x) = \dfrac{1}{1 + x^2}$ attain a maximum on the whole real line?, step 3