Commutativity and Translation Covariance of Convolution

Two Symmetries of Convolution

Convolution combines two functions by averaging products of their values over all possible decompositions of a point as a sum. Its definition singles out one input variable, but the resulting operation has symmetries that are not immediately visible from the formula. We prove that the inputs can be interchanged, and that translating either input translates the output by the same amount.

Throughout, the measure space is $(\mathbb{R}^n,\mathcal{L},m)$, with Lebesgue measure, which is complete and sigma-finite. The functions $f,g:\mathbb{R}^n\to\mathbb{C}$ are Lebesgue measurable and belong to $L^1(m)$, meaning $\int_{\mathbb{R}^n}|f|\,dm<\infty$ and $\int_{\mathbb{R}^n}|g|\,dm<\infty$. Real-valued functions are included as a special case. Equalities between $L^1$ functions, and the convolution identities below, are understood $m$-a.e. unless we explicitly discuss a particular point where the defining integrals exist.

Definition: For $f,g\in L^1(m)$, their convolution is given, wherever the integral is absolutely convergent, by $$ (f*g)(x)=\int_{\mathbb{R}^n} f(y)g(x-y)\,dm(y). $$ It defines an $L^1(m)$ function up to equality $m$-a.e. For $h\in\mathbb{R}^n$, define the translation $\tau_h f$ by $(\tau_h f)(x)=f(x-h)$.

Before using the formula, we check that it makes sense. The function $(x,y)\mapsto |f(y)||g(x-y)|$ is nonnegative and measurable on the product space. Tonelli’s theorem applies to this nonnegative function; translation invariance of Lebesgue measure then gives

\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|f(y)||g(x-y)|\,dm(y)\,dm(x) =\int_{\mathbb{R}^n}|f(y)|\left(\int_{\mathbb{R}^n}|g(x-y)|\,dm(x)\right)dm(y) =\|f\|_1\|g\|_1<\infty.

Consequently, the inner integral is finite for $m$-a.e. $x$, by the fact that a nonnegative function with finite integral is finite almost everywhere. The same calculation with Fubini’s theorem, now applicable because the absolute integral is finite, also yields $\|f*g\|_1\le\|f\|_1\|g\|_1$. We use these facts only to ensure that the expressions in the identities are defined almost everywhere.

Commutativity

Theorem (Commutativity): If $f,g\in L^1(m)$ on $\mathbb{R}^n$, then $f*g=g*f$ $m$-a.e. The functions may be real- or complex-valued.

Proof. By the absolute-integrability calculation above, for almost every $x$ the defining integral for $(f*g)(x)$ is absolutely convergent. Fix such an $x$. In that integral make the affine change of variables $z=x-y$. Its linear part is $-I$, whose determinant has absolute value $1$, so the change-of-variables formula preserves Lebesgue measure. Since $y=x-z$, we obtain

(f*g)(x)=\int_{\mathbb{R}^n}f(x-z)g(z)\,dm(z) =\int_{\mathbb{R}^n}g(z)f(x-z)\,dm(z) =(g*f)(x).

The middle equality uses commutativity of multiplication in $\mathbb{C}$. The change of variables also shows absolute convergence of the last integral. The equality therefore holds wherever the first integral is defined, and hence $m$-a.e. $\square$

Worked Example: Two Interval Indicators

On $\mathbb{R}$ with Lebesgue measure $m$, let $f=\chi_{[0,1]}$ and $g=\chi_{[0,2]}$. Both functions are in $L^1(m)$. Compute their convolution and verify the symmetry directly.

The integrand $f(y)g(x-y)$ equals $1$ exactly when $0\le y\le1$ and $0\le x-y\le2$. The second condition is equivalent to $x-2\le y\le x$. Thus $(f*g)(x)$ is the length of $[0,1]\cap[x-2,x]$. For $x<0$ or $x>3$, this intersection has length zero. For $0\le x\le1$, its length is $x-0=x$. For $1\le x\le2$, it is all of $[0,1]$, with length $1$. For $2\le x\le3$, its length is $1-(x-2)=3-x$. Therefore

(f*g)(x)= \begin{cases} 0,&x<0,\\ x,&0\le x\le1,\\ 1,&1\le x\le2,\\ 3-x,&2\le x\le3,\\ 0,&x>3. \end{cases}

For $g*f$, the integrand is $1$ when $0\le y\le2$ and $0\le x-y\le1$, or equivalently when $y\in[0,2]\cap[x-1,x]$. Its intersection lengths are $x$ for $0\le x\le1$, $1$ for $1\le x\le2$, and $3-x$ for $2\le x\le3$, with zero length outside $[0,3]$. This gives the same formula. Check: at $x=\tfrac12,\tfrac32,\tfrac52$, the computed values are respectively $\tfrac12,1,\tfrac12$. The integral of the formula is $\tfrac12+1+\tfrac12=2$, agreeing with $\int f\,dm\int g\,dm=1\cdot2=2$.

Translation Covariance

Theorem (Translation Covariance): Let $f,g\in L^1(m)$ on $\mathbb{R}^n$, and let $h\in\mathbb{R}^n$. Then $$ \tau_h(f*g)=(\tau_h f)*g=f*(\tau_h g) $$ $m$-a.e.

Proof. Translation preserves Lebesgue measure: for an integrable function $q$, the change of variables $z=y-h$ gives $\int q(y-h)\,dm(y)=\int q(z)\,dm(z)$. In particular, $\tau_h f$ and $\tau_h g$ are in $L^1(m)$. The convolution existence calculation applies to each pair in the statement. Also, $f*g$ is defined at $x-h$ for almost every $x$, since translating a null exceptional set leaves it null.

For almost every $x$, the defining integral for $(\tau_h f)*g$ is absolutely convergent. Using $(\tau_h f)(y)=f(y-h)$, then substituting $z=y-h$, we have

\begin{aligned} ((\tau_h f)*g)(x) &=\int_{\mathbb{R}^n}f(y-h)g(x-y)\,dm(y)\\ &=\int_{\mathbb{R}^n}f(z)g(x-h-z)\,dm(z)\\ &=(f*g)(x-h)=(\tau_h(f*g))(x). \end{aligned}

The substitution is a translation, so its Jacobian has absolute value $1$; it also carries the absolutely integrable integrand to an absolutely integrable one. For the other input, the defining formula directly gives, for almost every $x$,

(f*(\tau_h g))(x) =\int_{\mathbb{R}^n}f(y)g(x-y-h)\,dm(y) =(f*g)(x-h) =(\tau_h(f*g))(x).

Each equality has been shown on a full-measure set on which its integrals exist. The intersection of the finitely many full-measure sets is still a full-measure set. Thus all three functions agree $m$-a.e. $\square$

Worked Example: Translating an Interval Convolution

On $\mathbb{R}$ with Lebesgue measure, take $f=g=\chi_{[0,1]}$ and $h=2$. The overlap calculation gives $$ (f*g)(x)=T(x)= \begin{cases} x,&0\le x\le1,\\ 2-x,&1\le x\le2,\\ 0,&x<0\text{ or }x>2. \end{cases} $$ The values agree at the shared endpoint $x=1$, and the formula is zero outside $[0,2]$.

Translating the output gives $T(x-2)$, supported on $[2,4]$, with value $x-2$ on $[2,3]$ and $4-x$ on $[3,4]$. Meanwhile $\tau_2 f=\chi_{[2,3]}$. In $((\tau_2 f)*g)(x)$, the integrand is $1$ when $y\in[2,3]$ and $x-y\in[0,1]$, that is, when $y\in[2,3]\cap[x-1,x]$. The intersection has length $x-2$ for $2\le x\le3$, length $4-x$ for $3\le x\le4$, and zero outside $[2,4]$. For $f*(\tau_2g)$, the conditions are $y\in[0,1]$ and $x-y\in[2,3]$, or $y\in[0,1]\cap[x-3,x-2]$. These intersection lengths are again $x-2$ on $[2,3]$, $4-x$ on $[3,4]$, and zero elsewhere. Thus both translated-input convolutions equal $T(x-2)$. Check: at $x=\tfrac52, \tfrac72$, all three formulas give $\tfrac12$; at $x=1$, all give $0$.

Worked Example: A Signed Input and Commutativity

Let $f=\chi_{[0,1]}$ and $g=\chi_{[0,1]}-\chi_{[1,2]}$ on $\mathbb{R}$. These are real-valued $L^1(m)$ functions; the negative part of $g$ is allowed. Define $T=\chi_{[0,1]}*\chi_{[0,1]}$, so the same overlap calculation gives $T(u)=u$ for $0\le u\le1$, $T(u)=2-u$ for $1\le u\le2$, and $T(u)=0$ otherwise.

The second interval indicator is the translate $\tau_1\chi_{[0,1]}=\chi_{[1,2]}$ up to endpoint values, which do not affect integrals. By linearity of the integral and the interval formula, $f*g=T(x)-T(x-1)$. Evaluating on successive intervals gives

(f*g)(x)= \begin{cases} x,&0\le x\le1,\\ 3-2x,&1\le x\le2,\\ x-3,&2\le x\le3,\\ 0,&x<0\text{ or }x>3. \end{cases}

For example, on $1\le x\le2$, $T(x)=2-x$ and $T(x-1)=x-1$, so their difference is $3-2x$. On $2\le x\le3$, $T(x)=0$ and $T(x-1)=3-x$, so the difference is $x-3$. Directly computing $g*f$ gives the overlap for $[0,1]$ minus the overlap for $[1,2]$; these lengths are exactly $T(x)$ and $T(x-1)$, respectively. Check: the integral of the displayed formula is $\tfrac12+0-\tfrac12=0$. Also $\int f\,dm=1$ and $\int g\,dm=1-1=0$, so the integral-of-convolution identity gives the same result.

Tracking the Translation and Avoiding a Sign Error

The convention $\tau_h f(x)=f(x-h)$ means that the graph, and the support of an indicator, moves in the positive direction by $h$. For example, $\tau_2\chi_{[0,1]}=\chi_{[2,3]}$ up to endpoint values. The covariance formula matches this convention: translating either input to the right by $h$ translates the convolution to the right by $h$.

A common mistake is to replace $x-h$ with $x+h$ in the output. With $f=g=\chi_{[0,1]}$ and $h=1$, the convolution $T=f*g$ is supported on $[0,2]$. The correct translated output is $T(x-1)$, supported on $[1,3]$. At $x=2$, it has value $T(1)=1$. The incorrect expression $T(x+1)$ has value $T(3)=0$ at $x=2$, and is supported on $[-1,1]$, the wrong direction.

The formulas are $m$-a.e. identities because $L^1(m)$ functions are defined up to changes on null sets, and convolution integrals are guaranteed to exist only almost everywhere. Translation preserves null sets, so those choices do not affect the identities as $L^1$ functions. These symmetries are useful when moving or exchanging factors in explicit calculations; they do not assert pointwise equality for arbitrary representatives at every $x$.

Check Your Understanding

For each question, keep track of the translation convention and state whether the claimed identity is pointwise or almost everywhere.

For $f=\chi_{[0,1]}$ and $g=\chi_{[0,2]}$ on $\mathbb{R}$, compute $(f*g)(x)$ by finding the interval of $y$-values satisfying both indicator conditions.
In the commutativity proof, what affine substitution changes $\int f(y)g(x-y)\,dm(y)$ into the defining integral for $g*f$, and why does it preserve Lebesgue measure?
For $f=g=\chi_{[0,1]}$, identify the support of $\tau_{-1}(f*g)$. Is it a shift to the left or to the right?
Explain why Tonelli’s theorem applies to the absolute-value integrand used to establish that $f*g$ exists almost everywhere.
For the signed example, verify from its piecewise formula that $\int(f*g)\,dm=0$, and compare this with $(\int f\,dm)(\int g\,dm)$.

Commutativity and Translation Covariance of Convolution

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