Fredholmness and invertibility in a C* algebra generated convolution-type operators - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:13:48Z http://mathoverflow.net/feeds/question/52065 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52065/fredholmness-and-invertibility-in-a-c-algebra-generated-convolution-type-operato Fredholmness and invertibility in a C* algebra generated convolution-type operators Matt Heath 2011-01-14T12:09:10Z 2011-01-14T12:35:07Z <p>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 $a,b\in [PC,SO]$.</p> <p>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 $x\ge v$.</p> <p>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 <code>$T_v:=\sum_{i=1}^N \prod_{j=1}^M (a_{i,j})_v\mathcal F^{-1}(b_{i,j})_v\mathcal F$</code> is Fredholm.</p> <p>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.</p> <p>My motivation is to extend a Fredholm-index formula from $\mathcal A_{PC}$ to $\mathcal A_{[PC,SO]}$.</p>