MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Signal Processing books, a fundamental theorem is that linear time invariant systems can be represented as a convolution with a distribution. Could you give a mathematically rigorous statement of this theorem, or refer a book that includes it?

Edit: For example, would the following be a correct statement?

"Let S' be the space of tempered distributions. If L is a linear operator on S' that commutes with translations, then there exists a distribution h in S' such that Lf = f*h for all f in S'"

share|cite|improve this question
up vote 3 down vote accepted

I think the result you are looking for is the following: Let T be a linear continuous and translation invariant operator mapping S into S' (rather than S' into S'). Then there exists a distribution K s.t. Tf = f*K, for every f in S.

The continuity of T is referred to the usual Frechet topology on S and the weak dual topology on S' (you want f -> to be a continuous linear functional on S for every g in S). You can find a proof (by Sobolev embedding) on Introduction to Fourier analysis on Euclidean spaces (E. Stein).

From this you can prove analogous results for L^p spaces by embedding (translation invariantly) S in L^p and L^q in S'.

To rephrase everything in the language of multipliers it suffices to remember that the Fourier transform F is a topological isomorphism of S' and that F(f*K) = F(f)F(K) whenever K is a tempered distribution and f is a Schwartz function. Than the operator T is a multiplier operator F(Tf) = m F(f) for m = F(K).

share|cite|improve this answer

How about: let T be a linear transformation on some appropriate space of functions such that if g(x) = f(x+r) for some fixed r, then (Tg)(x) = (Tf)(x+r). Then there is a kernel k such that Tf = f*k for all f.

Note that the translation map τr which sends f to (x->f(x+r)) is a linear transformation, and the above hypothesis on T can be stated as Tτr = τrT for all r, i. e., that T commutes with all translations. Waving hands, linear operators that all commute with one another tend to have other things in common, like being diagonal in the same basis. In this case, that translates (no pun intended) to all being representable by a convolution (i. e., diagonal after a Fourier transform).

share|cite|improve this answer

An abstract result in Banach algebra theory, known as Wendel's Theorem, tells us that the multiplier algebra of L^1(G) is M(G), the measure algebra, for any locally compact group G.

So, if G=R the reals, this says that if T:L^1(R) \rightarrow L^1(R) is a bounded linear map which commutes with translations, then there is some measure \mu on R such that T(f) = \mu * f for all f\in L^1(R). (And maybe this special case was known before Wendel?)

I don't know much about distributions, but this general area falls into the theory of "Multipliers" I believe.

share|cite|improve this answer

The Schwartz kernel theorem seams relevant here. You might recall from your signal processing books that in the linear but non-time-invariant case we still get the output as a integral $\int K(x,y) f(y)dy$ where $f$ is the input. The kernel theorem makes this rigorous as I recall where, $K$ then can be a distribution.

Once you have that theorem it is probably easy to get the statement you want.

share|cite|improve this answer

As a more "down to earth" answer, I would say that linear systems have linear solutions, and convolution is a linear operator (or possibly bi-linear, based on the type of convolution) and as such the solutions of these equations can be represented as convolutions. The time-invariant property probably imposes some additional restrictions on the properties of the convolution.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.