Convolution as an Overlap Length
For indicators of intervals, convolution has a direct geometric meaning: at a fixed location \(x\), it measures how much the two intervals overlap after one is reflected and shifted. As \(x\) varies, that overlap grows, may remain constant, and then shrinks. Equal interval lengths produce a triangle; unequal lengths produce a trapezoid with a flat top.
Work on \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure. This measure space is sigma-finite but not finite. For a measurable integrable function \(f:\mathbb{R}\to\mathbb{R}\), the convolution of \(f\) and \(g\), wherever the defining integral exists, is \[ (f*g)(x)=\int_{\mathbb{R}} f(x-y)g(y)\,dm(y). \] Here we use indicator functions of bounded intervals, which are measurable and integrable. For every fixed \(x\), the integrand is nonnegative and supported on a bounded interval, so its integral is finite. Thus the formulas below hold for every \(x\), not merely almost everywhere.
The Formula for Intervals Starting at Zero
Proof. First suppose \(a\ge b\). The integrand \(\chi_{[0,a]}(x-y)\chi_{[0,b]}(y)\) equals \(1\) exactly when \(y\in[0,b]\) and \(x-y\in[0,a]\). The second condition is equivalent to \(y\in[x-a,x]\). Therefore \[ (f*g)(x)=m\bigl([0,b]\cap[x-a,x]\bigr). \] If \(x\le0\), this intersection has measure zero. If \(0<x<b\), then \(x-a<0\), since \(x<b\le a\), and the intersection is \([0,x]\), of measure \(x\). If \(b\le x\le a\), then \(x-a\le0\) and \(x\ge b\), so the intersection is \([0,b]\), of measure \(b\). If \(a<x<a+b\), then \(x-a>0\) and \(x>b\), so the intersection is \([x-a,b]\), of measure \(b-(x-a)=a+b-x\). If \(x\ge a+b\), then \(x-a\ge b\), and the intersection has measure zero. These cases give the displayed formula when \(a\ge b\).
Now suppose \(b>a\). In the defining integral make the affine substitution \(z=x-y\), which preserves Lebesgue measure. Then \[ \int_{\mathbb{R}}\chi_{[0,a]}(x-y)\chi_{[0,b]}(y)\,dm(y) = \int_{\mathbb{R}}\chi_{[0,a]}(z)\chi_{[0,b]}(x-z)\,dm(z). \] The same overlap calculation now has the longer interval of length \(b\) and the shorter of length \(a\). It gives the increasing part \(x\), the plateau \(a\), and the decreasing part \(a+b-x\), with breakpoints \(a\) and \(b\). Since \(r=a\) and \(R=b\), this is exactly the stated formula. At each breakpoint, the adjacent expressions agree; at \(x=0\) and \(x=a+b\), the overlap is at most a singleton and has measure zero. Hence the formula holds at every \(x\). \(\square\)
Worked Example: Equal Lengths Give a Triangle
Take \(f=g=\chi_{[0,3]}\). Here \(a=b=3\), so \(r=R=3\) and \(a+b=6\). The theorem gives \[ (f*g)(x)= \begin{cases} 0, & x\le0,\\ x, & 0<x<3,\\ 3, & x=3,\\ 6-x, & 3<x<6,\\ 0, & x\ge6. \end{cases} \] Thus the function rises linearly from \(0\) to \(3\), then falls linearly back to \(0\).
For a pointwise check, take \(x=1\). The allowed values of \(y\) lie in \([0,3]\cap[x-3,x]=[0,3]\cap[-2,1]=[0,1]\), whose measure is \(1\); the formula also gives \(x=1\). At \(x=3\), the intersection is \([0,3]\cap[0,3]=[0,3]\), of measure \(3\), matching the peak. At \(x=5\), it is \([0,3]\cap[2,5]=[2,3]\), of measure \(1\), matching \(6-x=1\).
Check: At the support endpoints \(x=0\) and \(x=6\), the intersections are respectively \([0,3]\cap[-3,0]=\{0\}\) and \([0,3]\cap[3,6]=\{3\}\). Each singleton has Lebesgue measure \(0\), as the formula requires.
Worked Example: Unequal Lengths Give a Trapezoid
Take \(f=\chi_{[0,5]}\) and \(g=\chi_{[0,2]}\). Then \(r=2\), \(R=5\), and \(a+b=7\). The formula is \[ (f*g)(x)= \begin{cases} 0, & x\le0,\\ x, & 0<x<2,\\ 2, & 2\le x\le5,\\ 7-x, & 5<x<7,\\ 0, & x\ge7. \end{cases} \] The overlap increases to length \(2\), stays at that length while the shorter interval lies fully within the longer one, and then decreases.
At \(x=1\), the relevant intersection is \([0,2]\cap[x-5,x]=[0,2]\cap[-4,1]=[0,1]\), so its measure is \(1\), as the rising formula gives. At \(x=3\), it is \([0,2]\cap[-2,3]=[0,2]\), of measure \(2\), as the plateau gives. At \(x=6\), it is \([0,2]\cap[1,6]=[1,2]\), of measure \(1\), and the declining formula gives \(7-6=1\).
Check: At \(x=2\), the intersection is \([0,2]\cap[-3,2]=[0,2]\), of measure \(2\), agreeing with both adjacent formulas \(x=2\) and the plateau value \(2\). At \(x=5\), it is \([0,2]\cap[0,5]=[0,2]\), again of measure \(2\), agreeing with the plateau and the value \(7-5=2\).
Translated Intervals
Proof. In the convolution integral, the two indicator conditions are \(x-y\in[\alpha,\alpha+a]\) and \(y\in[\beta,\beta+b]\). Set \(z=y-\beta\), a translation preserving Lebesgue measure. Then \(z\in[0,b]\), and \(x-y\in[\alpha,\alpha+a]\) becomes \(x-\alpha-\beta-z\in[0,a]\). The resulting integral is precisely the convolution of \(\chi_{[0,a]}\) and \(\chi_{[0,b]}\) evaluated at \(x-\alpha-\beta\). Applying the theorem proves the identity for every \(x\). \(\square\)
Worked Example: A Shifted Trapezoid
Let \(f=\chi_{[-1,2]}\) and \(g=\chi_{[2,6]}\). Their lengths are \(a=3\) and \(b=4\), and \(\alpha+\beta=-1+2=1\). Thus \(r=3\), \(R=4\), and \(a+b=7\). The convolution equals \(H_{3,4}(x-1)\): it is zero for \(x\le1\), rises as \(x-1\) for \(1<x<4\), equals \(3\) for \(4\le x\le5\), decreases as \(8-x\) for \(5<x<8\), and is zero for \(x\ge8\).
Check \(x=2\). The formula gives \(x-1=1\). Directly, \(y\) must lie in \([2,6]\), while \(x-y\in[-1,2]\) requires \(y\in[x-2,x+1]=[0,3]\). The intersection \([2,6]\cap[0,3]=[2,3]\) has measure \(1\). At \(x=5\), the formula gives \(3\); directly, \(y\in[5-2,5+1]=[3,6]\), and \([2,6]\cap[3,6]=[3,6]\) has measure \(3\). At \(x=7\), the formula gives \(8-7=1\); directly, \(y\in[5,8]\), and \([2,6]\cap[5,8]=[5,6]\) has measure \(1\).
Check: The shift moves the support endpoints from \(0,7\) to \(1,8\), since \(x-1\) ranges from \(0\) to \(7\) exactly when \(x\) ranges from \(1\) to \(8\). The three direct overlap calculations agree with the formula.
Endpoint and Degenerate Cases
Whether an interval is written with closed or open endpoints does not change its indicator except at its endpoints. A finite set has Lebesgue measure zero, so changing these endpoint values does not change any convolution value: for fixed \(x\), the affected values of \(y\) form a finite set. The formula therefore applies to any choice of endpoint inclusion for finite intervals.
The theorem assumes positive lengths. If \(a=0\) or \(b=0\), the corresponding interval is a singleton and its indicator vanishes almost everywhere. The convolution is then identically zero: in the defining integral the integrand can be nonzero only on a singleton of \(y\)-values. This is the appropriate degenerate case, rather than a trapezoid of positive height.
A Common Mistake: Multiplying Instead of Overlapping
Convolution is not pointwise multiplication. For the intervals \([0,5]\) and \([0,2]\), the pointwise product of their indicators is \(\chi_{[0,2]}\), but that is a different operation. For example, at \(x=3\) the pointwise product is \(0\), since \(3\notin[0,2]\). The convolution at \(x=3\), however, is \[ \int_{\mathbb{R}}\chi_{[0,5]}(3-y)\chi_{[0,2]}(y)\,dm(y) =m([0,2]\cap[-2,3])=2. \] The correction is to keep the variable \(y\) in both conditions: \(y\) must belong to the second interval, and \(x-y\) must belong to the first. Measuring their intersection gives the convolution value.
The overlap description also explains the shape without relying on a sketch. Before the shorter interval begins to fit inside the longer one, the overlap grows at unit rate. Once it fits entirely, the overlap is constant at the shorter length. After it begins to leave, the overlap shrinks at unit rate. When the lengths agree, the constant part has zero width, leaving the triangular profile.
Check Your Understanding
In each question, use the interval-intersection description and include the endpoint cases in your reasoning.
- For \(\chi_{[0,4]}*\chi_{[0,4]}\), write the formula for every real \(x\), including the values at \(x=0,4,8\).
- Compute \((\chi_{[0,6]}*\chi_{[0,2]})(3)\) by finding the interval of allowed \(y\)-values.
- For intervals of lengths \(2\) and \(5\) starting at \(0\), state the interval on which the convolution is constant and give its value there.
- Find the shift in the convolution of \(\chi_{[3,7]}\) and \(\chi_{[-2,1]}\), and state its support endpoints.
- Explain why an interval of length zero gives the zero convolution, even though its indicator may equal \(1\) at one point.