Titchmarsh convolution theorem

Remarks

Equivalently, write $\lambda(\varphi)$ for the largest $\lambda$ such that $\varphi$ vanishes almost everywhere on $(0, \lambda)$. Then for integrable $\varphi, \psi$ on $[0, T]$ one has $\lambda(\varphi * \psi) = \lambda(\varphi) + \lambda(\psi)$ whenever the right side is less than $T$. A standard form concerns functions or distributions with compact support on $\R$. It says that the convex hull of the support of $\varphi * \psi$ is the sum of the convex hulls of the supports of $\varphi$ and $\psi$. The classical proof uses complex analysis: the Laplace or Fourier transforms of such functions are entire functions of exponential type, and the theorem reduces to growth estimates for them. Real-variable proofs are also known.

Example: with $T = 2$ and $\varphi = \psi = \mathbf{1}_{(1, 2)}$, the convolution vanishes on $(0, 2)$, and the conclusion holds with $\lambda = \mu = 1$, so $\lambda + \mu = T$ exactly.

The easy direction is the converse: if $\varphi = 0$ on $(0, \lambda)$ and $\psi = 0$ on $(0, \mu)$, then each product $\varphi(t) \psi(x - t)$ with $x < \lambda + \mu$ vanishes, so $\varphi * \psi = 0$ on $(0, \lambda + \mu)$.

Going deeper

For a student

Letting $T \to \infty$ gives the corollary that the convolution ring of continuous functions on the half-line has no zero divisors. This is the algebraic foundation of the Mikusiński operational calculus, which builds the field of fractions of that ring. The support form of the theorem is used in the theory of distributions and in the uniqueness theory of partial differential equations.