Positivity and Support of Convolutions

Positivity and the Geometry of Support

Convolution combines values of two functions at pairs of points whose sum is fixed. This makes positivity natural: if both factors are nonnegative, every product in the defining integral is nonnegative. It also makes support geometry natural: a contribution at \(x\) can occur only when \(x=y+(x-y)\), with \(y\) in the support of the first function and \(x-y\) in the support of the second. The relevant set is therefore a sum of supports, not their intersection.

Throughout, the underlying space is \((\mathbb{R}^n,\mathcal{L},m)\), where \(m\) is Lebesgue measure. Functions are real-valued measurable functions in \(L^1(m)\), considered up to equality \(m\)-a.e. For representatives \(f,g\), the convolution is given by the Lebesgue integral \( (f*g)(x)=\int_{\mathbb{R}^n} f(y)g(x-y)\,dm(y)\) where this integral is defined; the resulting \(L^1\) function is understood as an a.e. equivalence class. The convolution results below concern this Lebesgue integral, not a pointwise product.

The definition records where a function is nonzero in a measure-sensitive way: changing its values on a null set does not change its essential support. In particular, \(f=0\) a.e. on the complement of its essential support. To see why, take the countable basis of open balls with rational centers and radii. Every point outside the essential support lies in one of these balls on which \(\int |f|\,dm=0\). A nonnegative function with integral zero vanishes a.e.: indeed, each set \(\{|f|>1/k\}\) has measure at most \(k\int |f|\,dm=0\), and \(\{|f|>0\}\) is the union of these sets over positive integers \(k\). The complement of the essential support is covered by countably many such null-integral balls, so \(f=0\) a.e. there. This reasoning also covers the zero function, whose essential support is empty.

Existence and the Main Support Result

For \(f,g\in L^1(m)\), Tonelli's theorem applies to the nonnegative measurable function \((x,y)\mapsto |f(y)||g(x-y)|\). Translation invariance of Lebesgue measure gives, for each fixed \(y\), \(\int_{\mathbb{R}^n}|g(x-y)|\,dm(x)=\|g\|_1\). Consequently,

This is finite, so the inner integral of the absolute value is finite for a.e. \(x\). Thus \(f*g\) is defined by an absolutely convergent integral for a.e. \(x\). Fubini's theorem, with this absolute integrability established, also gives \(\|f*g\|_1\le \|f\|_1\|g\|_1\). The finiteness and the estimate are statements about a.e. \(x\); an integral at an exceptional point can be infinite.

Proof. For positivity, choose representatives that are nonnegative everywhere by replacing \(f\) and \(g\) by their positive parts. This does not alter their \(L^1\) classes or convolution integrals: for any fixed \(x\), changing \(f\) on a null set changes the integrand only on that null set of \(y\), and changing \(g\) on a null set \(N\) changes it only for \(y\in x-N\), also a null set by translation invariance. For every \(x\), the integrand \(f(y)g(x-y)\) is now nonnegative, so its integral is nonnegative as an extended value. The absolute-integrability calculation above shows this integral is finite for a.e. \(x\). Therefore the convolution, as an \(L^1\) function, is nonnegative a.e.

For the support claim, the definition of essential support and the countable-basis argument above give \(f=0\) a.e. on \(A^c\) and \(g=0\) a.e. on \(B^c\). Choose representatives with these vanishing properties outside null sets \(N_f\subseteq A^c\) and \(N_g\subseteq B^c\). Fix \(x\notin A+B\). For any \(y\notin N_f\cup(x-N_g)\), if \(y\in A\), then \(x-y\notin B\); otherwise \(x=y+(x-y)\) would belong to \(A+B\). Thus either \(f(y)=0\) or \(g(x-y)=0\). The exceptional set \(N_f\cup(x-N_g)\) is null, so the integrand vanishes a.e. in \(y\), and \((f*g)(x)=0\). This holds for every \(x\notin A+B\) for these representatives. Since \(A+B\) is compact (it is the image of the compact set \(A\times B\) under the continuous addition map), its complement is open. The convolution vanishes there, so its essential support is contained in \(A+B\). If either support is empty, the corresponding function is zero a.e. and the same argument gives zero convolution. \(\square\)

Positivity is an a.e. conclusion for the \(L^1\) convolution class; no assertion of strict positivity at every point is implied. Support containment is also an a.e. statement: values assigned to a representative on a null set do not enlarge its essential support.

Worked Interval Convolutions

A Common Support Mistake

A tempting but incorrect rule is to use the intersection of the supports. For example, take \(f=\chi_{[0,1]}\) and \(g=\chi_{[2,3]}\). Their support intersection is empty, but the convolution is not zero: at \(x=3\), the conditions \(y\in[0,1]\) and \(3-y\in[2,3]\) hold for every \(y\in[0,1]\). Therefore \((f*g)(3)=\int_0^1 1\,dm(y)=1\). The supports combine by addition: \([0,1]+[2,3]=[2,4]\), which contains \(3\). The mistake comes from comparing the two input locations directly instead of enforcing the convolution relation \(x=y+(x-y)\).

Another distinction is between nonnegative and strictly positive. The first example has a nonnegative convolution that vanishes at \(x=0\) and \(x=2\), even though those points belong to the Minkowski sum. The theorem guarantees nonnegativity and support containment; it does not promise a positive value at every point of that sum. In applications, this distinction matters when interpreting convolution as spreading or combining mass: the geometric sum identifies where nonzero values can occur, while the integral determines their actual size.