When the Feasible Boundary Is Excluded
Many optimization problems have feasible values in an interval that does not include one or both of its boundary points. For example, a variable may be required to satisfy \(0<x<1\), or \(x>0\). In these settings, a function can approach a best possible value as \(x\) approaches an excluded boundary without ever taking that value.
This distinction matters: a limiting value is not automatically an attained maximum or minimum. The supremum and infimum describe the best upper and lower bounds of the function's values; a maximum or minimum must itself be a value of the function at a feasible input.
Maximum, Minimum, Supremum, and Infimum
Let \(f:D\to\mathbb{R}\) be an objective function on a feasible domain \(D\). A number \(M\) is a maximum of \(f\) on \(D\) if \(f(x)\leq M\) for every \(x\in D\), and \(f(c)=M\) for at least one \(c\in D\). The input \(c\) is a maximizer. A minimum is defined by reversing the inequalities: \(m\leq f(x)\) for every \(x\in D\), and \(f(c)=m\) for some feasible \(c\).
The supremum is the least upper bound of the function's values: it is an upper bound, and no smaller number is an upper bound. The infimum is the greatest lower bound. A supremum need not be a value of the function, and neither does an infimum. When a supremum is attained, it is a maximum; when an infimum is attained, it is a minimum.
| Question | What it asks |
|---|---|
| Is \(M\) an upper bound? | Does \(f(x)\leq M\) hold for every feasible \(x\)? |
| Is \(M\) the supremum? | Is it the least upper bound? |
| Is \(M\) a maximum? | Is it the supremum and does some feasible input satisfy \(f(x)=M\)? |
These definitions also clarify optimization language. Saying that a function has a supremum does not say that there is a feasible input that maximizes it. Always check attainment separately.
What an Excluded Boundary Limit Tells Us
For a feasible interval \((a,b)\), the endpoint \(a\) is not feasible, even if \(f(x)\) has a finite right-hand limit there. If \(\lim_{x\to a^+}f(x)=L\), values of \(f\) approach \(L\) as feasible inputs approach \(a\). This makes \(L\) relevant to the function's upper or lower bounds, but does not by itself prove that \(L\) is either bound: interior values may be larger or smaller.
The same reasoning applies at \(b\), using the left-hand limit. For an interval such as \((a,\infty)\), also examine behavior as \(x\to\infty\). A finite limit at infinity is approached along feasible inputs, but is not attained merely because it is a limit. If \(f(x)\to+\infty\), the function is unbounded above and has no finite maximum. If \(f(x)\to-\infty\), it is unbounded below and has no finite minimum.
When an optimum is attained at an interior point where \(f\) is differentiable, the derivative must be zero. This is a necessary condition, not a guarantee that the point is an optimum. In an open interval there are no feasible endpoints to substitute into the objective, so derivative-based interior candidates must be considered alongside boundary limits. For an unbounded interval, the behavior at infinity must be considered as well.
Worked Example: A Function on an Open Finite Interval
Worked Example: A Function on an Open Finite Interval
Find the supremum and infimum of \(f(x)=x\) for \(0<x<1\), and determine whether either is attained.
Every feasible input satisfies \(0<x<1\). Since \(f(x)=x\), this gives \(0<f(x)<1\). Thus \(0\) is a lower bound and \(1\) is an upper bound.
To show that \(1\) is the least upper bound, take \(x_n=1-\frac{1}{n}\) for integers \(n\geq2\). Each \(x_n\) is feasible, and \(f(x_n)=1-\frac{1}{n}\to1\). No number smaller than \(1\) can be an upper bound: given \(U<1\), choose \(n\) large enough that \(\frac{1}{n}<1-U\). Then \(f(x_n)=1-\frac{1}{n}>U\). Hence \(\sup f=1\). But \(f(x)<1\) for every feasible \(x\), so the supremum is not attained and there is no maximum.
Similarly, \(x_n=\frac{1}{n}\) is feasible and \(f(x_n)=\frac{1}{n}\to0\). Since every feasible value is positive, \(0\) is the greatest lower bound, but no feasible input gives \(f(x)=0\). Therefore \(\inf f=0\), and there is no minimum.
Check: The endpoint limits agree with these bounds: \(\lim_{x\to0^+}f(x)=0\) and \(\lim_{x\to1^-}f(x)=1\). Both endpoints are excluded, so neither limiting value is attained.
A Precise Result for Finite Open Intervals
There is a useful way to make the boundary-limit comparison precise when both endpoints are finite. Suppose \(f\) is continuous on \((a,b)\), where \(a<b\), and both finite one-sided limits exist. Define \(F\) on the closed interval \([a,b]\) by using \(f\) in the interior and setting \(F(a)=\lim_{x\to a^+}f(x)\) and \(F(b)=\lim_{x\to b^-}f(x)\). Then \(F\) is continuous on \([a,b]\).
The extreme value theorem applies to the continuous function \(F\) on the closed, bounded interval \([a,b]\). It gives a point \(c\in[a,b]\) where \(F\) has its largest value, say \(M\). For every \(x\in(a,b)\), \(f(x)=F(x)\leq M\), so \(M\) is an upper bound for the values of \(f\).
It remains to show that no smaller number is an upper bound. If \(c\in(a,b)\), then \(f(c)=F(c)=M\), so the supremum is \(M\) directly. If \(c=a\), choose a sequence of interior inputs approaching \(a\). By the definition of the right-hand limit, their function values approach \(F(a)=M\). An upper bound smaller than \(M\) could not bound values approaching \(M\). If \(c=b\), the same argument uses interior inputs approaching \(b\) and the left-hand limit. Thus \(\sup f=M\) in every case.
Applying the same argument to the minimum of \(F\) shows that \(\inf f=\min_{[a,b]}F\). An endpoint value can determine the supremum or infimum, but an endpoint is not in the original domain \((a,b)\). Therefore, attainment on the open interval occurs exactly when an interior point has the relevant bound as its function value. This proves the theorem and explains why a closed-interval boundary value cannot simply be reported as an attained optimum on an open interval.
Worked Example: An Unbounded Domain with an Attained Minimum
Worked Example: An Unbounded Domain with an Attained Minimum
Find the minimum of \(f(x)=x+\frac{1}{x}\) on \(x>0\).
First check the limits at the two ends of the feasible interval. As \(x\to0^+\), \(\frac{1}{x}\to+\infty\), so \(f(x)\to+\infty\). As \(x\to\infty\), the term \(x\to+\infty\) and \(\frac{1}{x}\to0\), so again \(f(x)\to+\infty\). These limits show that neither end provides a smaller limiting value, but they do not by themselves locate a minimum.
For \(x>0\), the derivative is \(f'(x)=1-\frac{1}{x^2}\). Setting it equal to zero gives \(1-\frac{1}{x^2}=0\), so \(x^2=1\). Of the solutions \(x=\pm1\), only \(x=1\) is feasible. The derivative is negative for \(0<x<1\), since \(x^2<1\), and positive for \(x>1\), since \(x^2>1\). Thus \(f\) decreases to \(x=1\) and increases after it.
To verify the value globally, for every \(x>0\) calculate \(f(x)-2=x+\frac{1}{x}-2=\frac{x^2-2x+1}{x}=\frac{(x-1)^2}{x}\geq0\). Equality holds at \(x=1\). Therefore \(f(x)\geq2\) for every feasible input, and \(f(1)=1+1=2\). The minimum is \(2\), attained at \(x=1\).
Check: Substituting the candidate gives \(f(1)=2\), and the nonnegative-square calculation proves that no feasible input gives a smaller value. Although the domain is unbounded and both end limits are infinite, an interior minimum is attained.
Worked Example: An Unbounded Domain with an Attained Maximum
Worked Example: An Unbounded Domain with an Attained Maximum
For \(f(x)=\frac{x}{1+x^2}\) on \(x>0\), find its supremum and infimum, and determine which are attained.
The function is positive for every \(x>0\). Its endpoint limits are \(\lim_{x\to0^+}\frac{x}{1+x^2}=0\) and \(\lim_{x\to\infty}\frac{x}{1+x^2}=0\). The second limit follows, for example, by dividing numerator and denominator by \(x^2\), giving \(\frac{1/x}{1+1/x^2}\to0\).
For the maximum, compare the denominator with \(2x\). Since \((x-1)^2\geq0\), we have \(1+x^2\geq2x\). All quantities being positive, division gives \(\frac{x}{1+x^2}\leq\frac12\). Equality holds when \((x-1)^2=0\), that is, at the feasible input \(x=1\). Substitution confirms \(f(1)=\frac{1}{2}\). Thus the supremum is \(\frac12\), and it is an attained maximum.
For the lower bound, \(f(x)>0\) at every feasible input, so \(0\) is not attained. The limit \(f(x)\to0\) as \(x\to\infty\) shows that no positive number can be a lower bound: eventually the function values are smaller than any chosen positive number. Hence the infimum is \(0\), but there is no minimum.
Check: The inequality proves every value is at most \(\frac12\), with equality at \(x=1\); positivity and the limit at infinity verify that \(0\) is an unattained infimum.
A Reliable Optimization Check
Boundary limits are evidence about the possible bounds, not replacements for comparing all relevant values. A function may approach a limit at an excluded boundary while attaining a better value in the interior. Likewise, a finite limit as \(x\to\infty\) does not automatically give a supremum or infimum: interior values may lie on either side of that limit.
Mark which finite endpoints are excluded and whether the domain extends to infinity.
At an interior local optimum where the derivative exists, the derivative must be zero. Compare any resulting candidates using the objective values.
Compute one-sided limits at excluded finite endpoints and limits at infinity when relevant.
After identifying the best upper or lower bound, check whether any feasible input actually gives that value.
A common mistake is to treat an excluded endpoint as a feasible input. For \(f(x)=x\) on \((0,1)\), one might incorrectly write, “The maximum is \(f(1)=1\).” The calculation \(f(1)=1\) is correct for the formula, but \(1\notin(0,1)\), so that input is not feasible. The correction is to say that the supremum is \(1\), approached as \(x\to1^-\), and that no maximum exists on the stated domain.
It is also a mistake to decide attainment from a limit alone. A limit describes values near a boundary, not necessarily what happens throughout the interval. Compare it with interior values and then test the defining condition for a maximum or minimum: the bound must be reached at a feasible input.
Check Your Understanding
For each question, distinguish the bound of the function's values from whether that bound is attained.
- For \(f(x)=2-x\) on \(0<x<2\), find the supremum and infimum. Are either attained?
- What is the difference between saying that \(M\) is an upper bound and saying that \(M\) is a maximum?
- For \(f(x)=x+\frac{1}{x}\) on \(x>0\), what do the limits at \(0^+\) and infinity tell you, and what additional check verifies the minimum?
- Can a finite limit as \(x\to\infty\) be called a maximum just because the function approaches it? Explain.
- For \(f(x)=\frac{x}{1+x^2}\) on \(x>0\), which of the supremum and infimum is attained, and how can you verify your answer?
