From Partition Intervals to a Lower Sum
A bounded function can vary across an interval, so one value of the function does not describe its behavior everywhere. A partition breaks the interval into smaller pieces. On each piece, we can record the function’s lowest possible value and weight it by the width of that piece.
Combining these weighted values gives a lower Darboux sum. Constructing one requires care: the lowest value might occur at an endpoint, might occur inside the interval, or might not be attained at all. The infimum handles all three situations.
Partitions and Interval Infima
Let \(f:[a,b]\to\mathbb{R}\) be bounded, where \(a<b\). A partition \(P\) of \([a,b]\) is a finite increasing list of points
It divides \([a,b]\) into \(n\) subintervals. For each index \(i\) from \(1\) to \(n\), the \(i\)th subinterval is \([x_{i-1},x_i]\), and its width is \(x_i-x_{i-1}>0\).
On each subinterval, the infimum of \(f\) is its greatest lower bound: it is a number no larger than any function value there, and no larger number has that property. The infimum need not itself be a function value. Write the infimum on the \(i\)th subinterval as \(m_i\).
If \(f\) attains its least value on a subinterval, that value is its minimum there and equals the infimum. In that case one may find \(m_i\) by locating the minimum. If no least value is attained, one must still use the infimum. Because \(f\) is bounded, each of these infima is a finite real number.
The subintervals are written as closed intervals, so consecutive subintervals share a partition point. This does not cause a problem: each infimum is calculated separately on its own subinterval, and the formula weights it by that subinterval’s width.
How to Calculate a Lower Sum
For a given partition, calculate the sum one subinterval at a time. First find its width. Then determine the infimum of \(f\) over the entire subinterval, not just at a convenient sample point. Finally multiply the width by that infimum and add the contributions.
Read them from consecutive partition points, \( [x_{i-1},x_i] \).
Check the function’s behavior on the whole subinterval, including endpoint and interior minima.
For the \(i\)th subinterval, calculate \(m_i(x_i-x_{i-1})\).
Their sum is \(L(P,f)\).
This procedure is about the infimum on each complete subinterval. An endpoint value identifies the infimum only when the function’s behavior on that interval supports that conclusion.
Worked Example: Minima at Subinterval Endpoints
Worked Example: A Decreasing Function on Two Subintervals
Let \(f(x)=3-x\) on \([0,3]\), with partition \(P=\{0,1,3\}\). Find \(L(P,f)\).
The subintervals are \([0,1]\) and \([1,3]\). On \([0,1]\), \(3-x\) decreases from \(3\) to \(2\), so its infimum is \(m_1=2\), attained at \(x=1\). The width is \(1-0=1\), giving the contribution
On \([1,3]\), the function decreases from \(2\) to \(0\), so its infimum is \(m_2=0\), attained at \(x=3\). The width is \(3-1=2\), giving
Therefore,
Check: on \([0,1]\), \(x\leq 1\), so \(3-x\geq 2\); on \([1,3]\), \(x\leq 3\), so \(3-x\geq 0\). Thus the chosen values are lower bounds on their respective intervals, and the calculation gives \(2(1)+0(2)=2\).
Worked Example: A Minimum Inside a Subinterval
Worked Example: A Squared Distance with an Interior Minimum
Let \(f(x)=(x-1)^2\) on \([0,2]\), with partition \(P=\{0,\tfrac12,\tfrac32,2\}\). Find \(L(P,f)\).
On \([0,\tfrac12]\), \(x\) is at most \(\tfrac12\), so the distance from \(x\) to \(1\) is smallest at \(x=\tfrac12\). Hence
The first contribution is \(\frac14\cdot\frac12=\frac18\).
On \([\tfrac12,\tfrac32]\), the point \(x=1\) lies inside the interval. Since \(f(1)=(1-1)^2=0\) and a square is never negative, \(m_2=0\). The width is \(\tfrac32-\tfrac12=1\), so the second contribution is \(0\cdot1=0\).
On \([\tfrac32,2]\), the distance from \(x\) to \(1\) is smallest at \(x=\tfrac32\), giving
The third contribution is \(\frac14\cdot\frac12=\frac18\). Therefore,
Check: on the first and third intervals, \(|x-1|\geq\frac12\), so \(f(x)\geq\frac14\); on the middle interval, \(f(x)\geq0\) and equals \(0\) at \(1\). These confirm all three infima and the resulting sum.
Worked Example: An Infimum That Is Not Attained
Worked Example: A Function with No Minimum on the Whole Interval
Define \(f:[0,1]\to\mathbb{R}\) by \(f(0)=1\) and \(f(x)=x\) for \(0<x\leq1\). Use the single-interval partition \(P=\{0,1\}\) to find \(L(P,f)\).
Every value of \(f\) is positive, so \(0\) is a lower bound. For any \(\varepsilon>0\), choose \(x=\min(\varepsilon/2,1/2)\). Then \(0<x\leq1\) and \(f(x)=x<\varepsilon\). Function values can therefore be made arbitrarily close to \(0\), so no positive number is a lower bound. It follows that
The infimum is not attained: \(f(0)=1\), while \(f(x)=x>0\) whenever \(x>0\). The interval width is \(1-0=1\), so
Check: \(f(x)\geq0\) everywhere, which confirms that \(0\) is a lower bound; the arbitrarily small positive values confirm it is the greatest lower bound. The lower sum uses this infimum even though there is no point with function value \(0\).
Why the Construction Works
On each subinterval, the infimum is no greater than any value of \(f\) there. Thus the constant value \(m_i\) sits at or below the function throughout that subinterval. The lower sum records that constant level multiplied by the interval width. This explains why the infimum, rather than a typical value or the largest value, is used.
More precisely, for every \(x\in[x_{i-1},x_i]\), the definition of infimum gives \(m_i\leq f(x)\). Multiplying by the positive width preserves the relevant order, and the full sum is exactly the sum of these width-weighted infima:
The width is positive because the partition points are strictly increasing. The height \(m_i\), however, may be negative. The formula remains valid in that case: a subinterval’s contribution is its width multiplied by its infimum, with the sign retained.
A Common Mistake: Using Maxima Instead of Infima
A lower sum does not use the largest function value on each subinterval. For example, for \(f(x)=x^2\) on \([0,2]\) with partition \(P=\{0,1,2\}\), the maxima on the two subintervals are \(1\) and \(4\). Using those values gives the calculation
That is an upper-sum calculation, not a lower-sum calculation. The infimum on \([0,1]\) is \(0\), attained at \(x=0\); the infimum on \([1,2]\) is \(1\), attained at \(x=1\). Both widths are \(1\), so the correct lower sum is
Check: \(x^2\geq0\) on \([0,1]\) and \(x^2\geq1\) on \([1,2]\). These are the respective infima, so the contributions add to \(1\), not \(5\).
Why Lower Sums Are Useful
A lower Darboux sum condenses information about a function on an entire interval into one number for each partition subinterval. The construction is especially useful when the function has a clear minimum on each piece, but the infimum definition also works when minima are not attained. Its reliability depends on identifying the infimum on every piece before doing the arithmetic.
A quick final check is to ask whether each proposed \(m_i\) really lies below every value of \(f\) on its subinterval. If it does not, the resulting sum is not built from the required partition infima. Also check the widths: the contribution uses \(x_i-x_{i-1}\), not an interval endpoint by itself.
Check Your Understanding
For each question, identify the relevant subinterval infima before calculating a sum.
- For \(f(x)=x\) on \([0,2]\) with partition \(\{0,1,2\}\), what is the lower Darboux sum?
- For \(f(x)=(x-2)^2\) on \([1,3]\) with the single-interval partition, where does the minimum occur, and what is the lower sum?
- Can a lower Darboux sum use an infimum that is not attained by the function? Explain briefly.
- For a subinterval of width \(3\) with infimum \(-2\), what is its contribution to the lower sum?
- What must be checked before treating a subinterval endpoint value as its infimum?
