Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $PC$ be the algebra of complex-valued, piecewise-continuous functions from $[-\infty,+\infty]$, $SO$ be the algebra of bounded, continuous, complex-valued functions on $\mathbb R$ which are slowly oscillating at infinity and $[PC,SO]$ be the closed subalgebra of $L^\infty (\mathbb R)$ generated by $PC$ and $SO$. Let $\mathcal A_{[PC,SO]}$ be the the closed subalgebra of $\mathcal B(L^2(\mathbb R))$ generated by operators $a\mathcal F^{-1}b\mathcal F$ where $\mathcal F$ is the Fourier transform and $a,b\in [PC,SO]$.

For $v\in(0,\infty)$, $a\in [PC,SO]$ let $a_v\in PC$ be given by: $f_v(x)=f_v(-v+0)$ if $x\le -v$; $f_v(x)=f(x)$ if $x\in (-v,v)$ ; $f_v(x)=f_v(v-0)$ if $x\ge v$.

The question I need to answer is if $T=\sum_{i=1}^N \prod_{j=1}^M a_{i,j}\mathcal F^{-1}b_{i,j}\mathcal F$ ($a_{i,j}, b_{i,j}\in [PC,SO]$) is invertible, does it follow that $T_v:=\sum_{i=1}^N \prod_{j=1}^M (a_{i,j})_v\mathcal F^{-1}(b_{i,j})_v\mathcal F$ is Fredholm.

The answer to the above must be "yes" if $T\mapsto T_v$ extends continuously to $\mathcal A_{[PC,SO]}$. In that case we would have that $T_v$ would have to be invertible. My instinct is that this is true, probably allowing much more general functions. I also guess the answer must be fairly obvious to anyone with a better understanding of harmonic analysis.

My motivation is to extend a Fredholm-index formula from $\mathcal A_{PC}$ to $\mathcal A_{[PC,SO]}$.

share|improve this question
    
what do you mean exactly by "slowly oscillating at infinity"? thanks. –  Pietro Majer Jan 14 '11 at 21:42
    
I forgot that had different defintions; thanks for catching that. For a subset $I\subset \mathbb R$ and a function $f:\mathbb R\rightarrow \mathbb C$ we let $\textrm{osc}(f,I)=\sup_{t,s\in I}|f(t)-f(s)|$. We say $f$ is slowly oscillating at infinty if $\lim_{x\rightarrow\infty}\textrm{osc}(f,[-2x,-x]\cup[x,2x])=0$ –  Matt Heath Jan 17 '11 at 11:06

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.