I would like to find the adjoint operator in the Hilbertspace $L^2(0,\infty)$ of the operator $$ (Ax)(t)=x(at), x\in L^2(0,\infty), a>0. $$

My calculation is the following; I use the substitution $u=at$.

$$ \langle Ax,y\rangle_{L^2}=\int\limits_0^{\infty}x(at)\overline{y(t)}\, dt=\int\limits_{0}^{\infty}x(u)\overline{\frac{1}{a}y\left(\frac{u}{a}\right)}\, du $$

So the adjoint operator $A^*\colon L^2(0,\infty)\to L^2(0,\infty)$ should be given by

$$ (A^*x)(u)=\frac{1}{a}x\left(\frac{u}{a}\right). $$

  • 1
    $\begingroup$ Looks good to me. $\endgroup$ – Christopher A. Wong Feb 6 '13 at 21:34
  • $\begingroup$ And your comment looks good to me. Thank you! $\endgroup$ – math12 Feb 6 '13 at 21:44
  • $\begingroup$ One thing, in your definition of $A^*x$, you accidentally have a $y$ on the right-hand side. $\endgroup$ – Christopher A. Wong Feb 6 '13 at 21:46
  • $\begingroup$ Oh, I replace it by x. $\endgroup$ – math12 Feb 6 '13 at 21:49

Your Answer

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

Browse other questions tagged or ask your own question.