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

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

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