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

Consider a differential equation

\begin{eqnarray*} \frac{d}{d\tau}q^{\tau}\left(x\right)=Aq^{\tau}\left(x\right)+h\star q^{\tau}=Aq^{\tau}\left(x\right)+\int h\left(x-u\right)q^{\tau}\left(u\right)du, \end{eqnarray*}

where $q^{\tau}:R\rightarrow R^{n}$ is an bounded, and integrable vector-valued function, $A\in M_{n}$ is a square $n$-matrix, and $h:R\rightarrow M_{n}$ is a continuous, bounded, and integrable matrix valued function. Suppose that the real parts of the eigenvalues of matrices $A$, and for each $t$, \begin{eqnarray*} A+\hat{h}\left(t\right)=A+\int e^{itx}h\left(x\right)dx \end{eqnarray*}

are negative and uniformly bounded away from 0.

Because of the assumption on the eigenvalues of the Fourier coefficients, the Plancherel Theorem implies that $q^{\tau}$ converges to 0 in $L^{2}$-norm when $\tau\rightarrow\infty$. Is it possible that $q^{\tau}$ does not converge in $L^{1}$? The problem is that I do not know how to bound the $L^{1}$-norm of the inverse Fourier transform.

The question can be phrased as a convergence of the convolution exponential \begin{eqnarray*} \exp_{\star}\left(\left(A\delta+h\right)\tau\right) & = & \sum_{n}\frac{\tau^{n}}{n!}\left[\left(A\delta+h\right)\star...\star\left(A\delta+h\right)\right]_{n\mbox{ times}}\\ & = & \exp\left(A\tau\right)\delta+h^{\tau}, \end{eqnarray*} where $h^{\tau}$ is continuous, bounded, and integrable function, and $\delta$ is the Dirac's delta. Again, the assumptions imply that $\left\Vert h^{\tau}\right\Vert _{L^{2}}\rightarrow0$. Is it possible that $\left\Vert h^{\tau}\right\Vert _{L^{1}}\nrightarrow0$?

share|cite|improve this question
Yes, you have convergence of $q^\tau$ to $0$ in $L^1$ under your assumptions. I'll post a full solution later (unless someone beats me to it). The trick goes back to Wiener and is the same as in the proof that a non-vanishing on the circle function from the Wiener class (absolutely summable Fourier series) is invertible in the Wiener class. – fedja Jul 12 '12 at 2:24

Your Answer


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

Browse other questions tagged or ask your own question.