1
$\begingroup$

Suppose I have a linear operator $L$ s.t. $Ly(x-a)=\exp(-iax)Ly(x)$ and $L^2y(x)=ky(-x)$ where $a, k\in \mathbb R.$ Then what is $L^2y(x-a)$? May be it is really straightforward, but I don't know how to combine them!  Thank you.

  • 0
    Where is $L$ supposed to be defined? (it seems to be a space of function, but I want to be sure)2011-11-16
  • 0
    @DavideGiraudo: Yes, it is the space of functions.2011-11-16

1 Answers 1

1

$(L^2y)(x-a)=z(x-a)$ with $z=L^2y$ hence $z(x)=(L^2y)(x)=ky(-x)$ for every $x$ and $z(x-a)=ky(-(x-a))$, that is $(L^2y)(x-a)=ky(a-x)$.