Why Existence Matters
When studying absolute extrema, it is natural to ask which inputs give the largest and smallest function values. But before searching for those inputs, there is a more basic question: are such values guaranteed to exist at all?
For instance, \(f(x)=x\) has no largest value on the open interval \((0,1)\), even though its values get arbitrarily close to \(1\). By contrast, a continuous function on a closed, bounded interval cannot behave this way: it must attain both a largest and a smallest value. This guarantee is the Extreme Value Theorem.
The theorem is about existence, not about the method for finding the points where the extrema occur. Its hypotheses are sufficient to guarantee extrema; when a hypothesis fails, the guarantee may fail too, but extrema can still exist in particular cases.
Absolute Extrema and the Theorem
The point \(c\) or \(d\) must belong to the domain. A number that is merely approached by function values is not an absolute maximum or minimum unless the function actually takes that value.
Here, “closed” means that the interval includes its endpoints, and “bounded” means that it has finite endpoints. Continuity is required at every point of the interval, including its endpoints (using the appropriate one-sided behavior there). The conclusion is that both extreme values are attained at points in the interval.
Why the Hypotheses Give Attained Values
A continuous function cannot become unbounded on a closed, bounded interval. Also, the interval’s closedness ensures that limits of sequences of its points remain in the interval. Together, these facts prevent the largest or smallest function value from “escaping” without being attained.
The proof uses the Bolzano–Weierstrass theorem: every bounded sequence of real numbers has a convergent subsequence. For a sequence in \([a,b]\), the limit of that subsequence is still in \([a,b]\), because the interval is closed.
Because the set of values \(f([a,b])\) is nonempty and bounded, it has a supremum \(M\), its least upper bound. For each \(n\in\mathbb{N}\), the definition of supremum lets us choose \(x_n\in[a,b]\) such that
The sequence \((x_n)\) has a convergent subsequence \(x_{n_k}\to c\in[a,b]\). By continuity, \(f(x_{n_k})\to f(c)\). The displayed bounds show that \(f(x_{n_k})\to M\), so \(f(c)=M\). The supremum is therefore attained and is an absolute maximum. Applying the same argument to \(-f\) shows that \(-f\) attains a maximum, which means \(f\) attains a minimum. This proves the theorem.
The two hypotheses play distinct roles in this reasoning. Boundedness of the interval supplies convergent subsequences, while closedness keeps their limits in the domain. Continuity transfers convergence of inputs to convergence of function values. If any of these protections is absent, the proof’s guarantee can break down.
Worked Examples: A Continuous Function on a Closed Interval
Worked Example: Verifying the extrema directly
Consider \(f(x)=(x-1)^2+2\) on \([0,3]\). The function is a polynomial, so it is continuous, and the interval is closed and bounded. The Extreme Value Theorem guarantees that an absolute minimum and maximum exist.
For every \(x\), a square is nonnegative, so \((x-1)^2\geq 0\). Hence \(f(x)\geq 2\), with equality at \(x=1\), which belongs to \([0,3]\). Thus the absolute minimum is \(2\), attained at \(x=1\).
On \([0,3]\), we have \(-1\leq x-1\leq 2\), so \((x-1)^2\leq 4\). Equality occurs at \(x=3\). Therefore \(f(x)\leq 6\), and \(f(3)=6\). The absolute maximum is \(6\), attained at \(x=3\).
Check: \(f(1)=(1-1)^2+2=2\), and \(f(3)=(3-1)^2+2=4+2=6\). These values satisfy \(2\leq f(x)\leq 6\) throughout the interval.
When an Endpoint Is Missing
On an open interval, a function may approach an extreme value near an excluded endpoint without taking that value. The interval is bounded, but it is not closed, so the Extreme Value Theorem does not apply.
Worked Example: A bounded interval that is not closed
Let \(f(x)=x\) on \((0,1)\). This function is continuous on its domain, and the interval is bounded, but neither endpoint belongs to the interval.
For every \(x\in(0,1)\), \(x<1\), so no value of \(f\) equals \(1\). Yet values of the function can be made arbitrarily close to \(1\): for \(n\geq 2\), let \(x_n=1-\frac{1}{n}\). Then \(x_n\in(0,1)\) and \(f(x_n)=1-\frac{1}{n}\to1\). Therefore there is no absolute maximum.
Likewise, every value is greater than \(0\), but \(f(x_n)=\frac{1}{n}\to0\) for \(x_n=\frac{1}{n}\), with \(n\geq2\). Thus there is no absolute minimum either. The function’s values approach both bounds without attaining them.
Check: \(0<f(x)<1\) for every \(x\in(0,1)\), and neither \(0\) nor \(1\) is a function value on this domain. This confirms that neither bound is attained.
When the Interval Is Unbounded
A closed interval with no finite endpoints is not bounded. Even when a function is continuous on an unbounded interval, its values can approach an unattained bound or grow without limit.
Worked Example: A continuous function with no absolute maximum
Consider \(f(x)=\frac{1}{x+1}\) for \(x\in[0,\infty)\). The function is continuous at every point in its domain, but the interval is unbounded.
For every \(x\geq0\), \(x+1\geq1\), so \(0<f(x)\leq1\). At \(x=0\), \(f(0)=1\), so the absolute maximum is \(1\).
There is no absolute minimum: \(f(x)\) is always positive. For positive integers \(n\), \(f(n)=\frac{1}{n+1}\to0\), but no domain point gives a function value of \(0\). Hence the infimum is \(0\), but it is not attained.
Check: \(f(0)=\frac{1}{0+1}=1\), and \(f(n)=\frac{1}{n+1}>0\) for every \(n\geq0\). This confirms the attained maximum and the unattained lower bound.
When Continuity Fails
Continuity is also essential to the theorem’s guarantee. A function can be defined at every point of a closed, bounded interval and still fail to attain an absolute maximum if it is discontinuous.
Worked Example: A discontinuity prevents a maximum
Define \(f:[0,1]\to\mathbb{R}\) by \(f(x)=x\) when \(0\leq x<1\), and \(f(1)=0\). The domain is closed and bounded, but \(f\) is discontinuous at \(x=1\): as \(x\) approaches \(1\) from the left, \(f(x)\) approaches \(1\), whereas \(f(1)=0\).
For every \(x<1\), choose \(y=\frac{x+1}{2}\). Then \(x<y<1\), and \(f(y)=y>x=f(x)\). At the remaining point \(x=1\), \(f(1)=0\), while \(f(\frac{1}{2})=\frac{1}{2}>0\). Thus every point in the domain has another point with a larger function value; no point gives an absolute maximum.
The absolute minimum is \(0\), attained at \(x=0\) and \(x=1\), since \(f(x)=x\geq0\) for \(x<1\) and \(f(1)=0\). Check: \(f(0)=0\) and \(f(1)=0\), while values for \(0\leq x<1\) never exceed \(1\) and approach \(1\) without reaching it. So the minimum is attained, but the maximum is not.
What the Hypotheses Do—and Do Not—Say
The theorem’s conditions form a reliable guarantee, not a list of necessary conditions for every individual function. For example, \(f(x)=x\) on \([0,1)\) has an absolute minimum at \(x=0\), despite the interval not being closed. A function that fails one hypothesis may still have one or both absolute extrema.
The correct conclusion is more limited: if \(f\) is continuous on a closed, bounded interval, then both extrema must exist. If continuity, closedness, or boundedness is missing, the theorem cannot be used to guarantee both. The examples show that a missing condition can matter; they do not imply that extrema are impossible in every such case.
Why the Theorem Matters
The Extreme Value Theorem gives a firm starting point for optimization. On a closed, bounded interval, once continuity is established, we know that there is a greatest and a least function value to locate. Without that guarantee, a search for “the” maximum or minimum may have no answer: values might approach a bound that is never reached.
Before relying on the theorem, check three things: the function is defined throughout the interval, the interval is closed and bounded, and the function is continuous there. If a condition fails, analyze the specific function and domain instead of assuming either that extrema exist or that they do not.
Check Your Understanding
For each question, identify which hypothesis or conclusion is relevant and explain your reasoning.
- State the Extreme Value Theorem, including its hypotheses and conclusion.
- Why does closedness matter when a sequence of points in an interval converges?
- For \(f(x)=x\) on \((0,1)\), explain why the supremum is not an absolute maximum.
- Give the missing hypothesis in the example \(f(x)=\frac{1}{x+1}\) on \([0,\infty)\).
- Does failure of continuity prove that a function has no absolute extrema? Explain.
