$\mathbf {The \ Problem \ is}:$ If $f$ & $h$ are $L^1(\mathbb T)$ functions with $f\star h=0$ identically on $\mathbb T.$ Then can we say either $f\equiv 0$ or $h\equiv 0 ?$

$\mathbf {My \ approach}:$ By convolution theorem, $\widehat f(n)$.$\widehat h(n)=0$ for all $n\in \mathbb N.$

But, can we find $f,h \in L^1(\mathbb T)$ with $\widehat f$ is supported on odd integers & $\widehat h$ on even integers ?

I couldn't find anything . Thanks in adv. for a hint .

  • 2
    $\begingroup$ You may consider $f(x)=e^{2\pi ix}$ and $g(x)=1$. In general, choose $(a_n)_{n\in\mathbb{Z}}$ such that $\sum_n |a_n|<\infty$ and then set $$f(x)=\sum_{n\text{ odd}}a_n e^{2\pi i n x}\qquad\text{and}\qquad g(x)=\sum_{n\text{ even}}a_n e^{2\pi i n x}.$$ $\endgroup$ Jan 11 at 11:57
  • $\begingroup$ Easy example (special case of the above) - if we set $\mathbb T=[0,1]$, $f(x)=\sin(2\pi x)$, $h(x)=\sin (4\pi x)$, then $\int f(s)h(s-x)\ dx=0$. quick verification on Desmos: desmos.com/calculator/rqmudg7azm $\endgroup$ Jan 11 at 12:00
  • 2
    $\begingroup$ The similar result that I know, due to Titchmarsh: $C([0,\infty))$ under convolution is a division algebra. $\endgroup$
    – GEdgar
    Jan 11 at 12:08

Note that a complex division Banach algebra is isomorphic to the complex numbers. Even if you consider the real scalars, the complexification gives the answer immediately.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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