When the Guarantee Is Unavailable
The Extreme Value Theorem gives a powerful guarantee: if a function is continuous on a closed and bounded interval, then it attains both an absolute maximum and an absolute minimum there. But many functions are considered on intervals that are open, unbounded, or contain a point of discontinuity. In those cases, the theorem’s guarantee is unavailable.
That does not automatically mean an extreme value fails to exist. It means only that this theorem cannot guarantee one. To understand the difference, examine what values a function actually takes on its domain.
Attained Values, Bounds, and Absolute Extrema
An absolute maximum value of \(f\) on a domain \(D\) is a value \(f(c)\), for some \(c\in D\), such that \(f(x)\leq f(c)\) for every \(x\in D\). An absolute minimum value is a value \(f(c)\), for some \(c\in D\), such that \(f(c)\leq f(x)\) for every \(x\in D\). The requirement that \(c\) belong to the domain matters: an extreme value must actually be attained.
A number \(U\) is an upper bound for the values of \(f\) if \(f(x)\leq U\) for every \(x\in D\). The least upper bound, or supremum, is the smallest such upper bound. A lower bound and greatest lower bound, or infimum, are defined in the corresponding way. A supremum need not be a value of the function, and an infimum need not be either.
Open Intervals Can Leave Out the Needed Endpoint
Worked Example: A Continuous Function with No Extrema on an Open Interval
Consider \(f(x)=x\) on \(D=(0,1)\). Since \(f\) is a polynomial, it is continuous at every point of its domain. The interval is not closed because it excludes 0 and 1, so the Extreme Value Theorem does not apply.
For any \(x\in(0,1)\), the number \(x/2\) also belongs to \((0,1)\), and
Since \(f(x/2)=x/2<f(x)\), no value \(f(x)\) can be an absolute minimum. Also, define \(y=(x+1)/2\). Because \(0<x<1\), we have \(x<y<1\), so \(y\) belongs to the domain and
Thus every function value has a larger value above it, so there is no absolute maximum either. The values of \(f\) approach 0 near the excluded left endpoint and 1 near the excluded right endpoint. Its infimum is 0 and its supremum is 1, but neither is attained: \(f(x)=x\) requires \(x=0\) to produce 0 and \(x=1\) to produce 1, and neither input is in \(D\).
The argument also shows why checking just a handful of inputs would not settle the question. For every proposed maximum or minimum, another point in the interval gives a value that is respectively larger or smaller.
An open interval does not always prevent extrema. It merely removes at least one endpoint that might otherwise supply an extreme value, so the outcome depends on the function. The example above proves that a continuous function on an open interval can fail to attain either extreme.
Unbounded Intervals Can Allow Values to Escape
Worked Example: A Continuous Function with a Missing Minimum
Consider \(g(x)=e^{-x}\) on \(D=[0,\infty)\). The function is continuous on this domain, and the interval includes 0 but is unbounded, so the Extreme Value Theorem does not apply.
For every \(x\geq 0\), \(e^{-x}>0\), and \(g(0)=e^0=1\). If \(x>0\), then \(-x<0\), so \(e^{-x}<1\). Therefore \(g(x)\leq 1=g(0)\) for every \(x\geq0\), and the absolute maximum is 1, attained at \(x=0\).
The function has no absolute minimum. For any \(x\geq0\), the point \(x+1\) is also in the domain, and
because \(0<e^{-1}<1\). Each function value therefore has a smaller value elsewhere in the domain. The values approach 0 as \(x\) increases without bound, but \(e^{-x}>0\) for every real \(x\), so 0 is the infimum and is not attained.
This example is a useful contrast with the open-interval example: the failure of the theorem’s conditions does not prevent an extreme value from existing. Here the maximum exists, while the minimum does not.
A Discontinuity Can Break the Guarantee on a Closed Interval
A closed and bounded interval alone is not enough for the Extreme Value Theorem. Continuity on the entire interval is also required. A discontinuity can create a gap at an extreme boundary: nearby function values can approach a value that the function never takes.
Worked Example: A Bounded Function with No Maximum
Define \(h\) on \([0,1]\) by
The interval \([0,1]\) is closed and bounded, but \(h\) is discontinuous at 1: as \(x\) approaches 1 from the left, \(h(x)=x\) approaches 1, while \(h(1)=0\).
For every \(x\in[0,1)\), \(h(x)=x<1\), and the remaining domain point gives \(h(1)=0<1\). So no point in the domain has function value 1. Yet for any \(x<1\) sufficiently close to 1, \(h(x)=x\) is correspondingly close to 1. In fact, the values \(h(1-1/n)=1-1/n\), for integers \(n\geq2\), approach 1 from below. Thus 1 is the supremum but is not attained, and \(h\) has no absolute maximum.
The absolute minimum is 0, since \(h(x)\geq0\) throughout the domain and \(h(0)=h(1)=0\). The function is bounded, but it fails to attain its upper bound.
This example shows precisely what changes when continuity is missing: approaching a boundary value does not require the function to take that value, even on a closed interval.
Why the Theorem’s Conditions Must Be Checked
The theorem has a one-way logical form: continuity on a closed and bounded interval guarantees attained extrema. If any condition is absent, the guarantee no longer follows from this theorem. That is not the same as proving that the function has no extrema. For example, \(g(x)=e^{-x}\) above does have an absolute maximum on its unbounded domain.
| Situation | What is missing? | What the examples show |
|---|---|---|
| Continuous function on \((0,1)\) | The interval is not closed | \(f(x)=x\) attains neither extreme |
| Continuous function on \([0,\infty)\) | The interval is not bounded | \(g(x)=e^{-x}\) has a maximum but no minimum |
| Function on \([0,1]\) with a discontinuity | The function is not continuous throughout the interval | \(h\) has no maximum, though the interval is closed and bounded |
Common Mistake: Confusing an Approached Value with an Attained One
A common incorrect argument is: “For \(f(x)=x\) on \((0,1)\), the largest value is 1 because the function approaches 1 as \(x\) approaches 1.” This confuses a supremum with an absolute maximum. The input 1 is not in the domain, so \(f(1)\) is not a function value on \((0,1)\).
The correction is to check whether the proposed extreme value occurs at an input in the domain. In this example, every \(x\in(0,1)\) satisfies \(f(x)=x<1\), so 1 is an upper bound but not an attained maximum. The same check applies at the lower end: 0 is the infimum, but \(f(x)>0\) throughout the interval, so 0 is not an attained minimum.
Check Your Understanding
For each question, distinguish carefully between a value being approached and a value being attained.
- For \(f(x)=x\) on \((0,1)\), what are the infimum and supremum, and does either one give an absolute extreme value?
- Why does \(g(x)=e^{-x}\) on \([0,\infty)\) attain an absolute maximum but not an absolute minimum?
- In the example \(h\) on \([0,1]\), at which point is the function discontinuous, and why does its supremum fail to be a maximum?
- Does failure of a hypothesis of the Extreme Value Theorem prove that a function has no absolute extrema? Explain.
- What is the difference between an absolute maximum value and a supremum that is not attained?
