I have the linear operator $A_\lambda :g(t)\mapsto \sqrt \lambda g(\lambda t)$ on $C[0,1]$ with $\lambda\in (0,1)$ (which extends to $L^2[0,1]$). I need to find its adjoint and then prove that while $A_\lambda A_\lambda^\ast=1$, $A_\lambda ^\ast A_\lambda $ is a projection. I also need to find what subspace it projects to.
After calculating I found $A_\lambda^\ast:g(t)\mapsto \frac1{\sqrt\lambda}g(\frac t\lambda)\cdot \chi_{[0,1]}(\frac t\lambda)$, since $$ \left\langle A_\lambda f,g \right\rangle =\sqrt \lambda\int _{[0,1]}f(\lambda t)g(t)dt=\frac1{\sqrt \lambda}\int _{[0,1]}f(t^\prime)g(\frac {t^\prime}\lambda)\chi_{[0,\lambda]}(t^\prime)dt^\prime$$ and $\chi_{[0,\lambda]}(t^\prime)=\chi_{[0,1]}(\frac {t^\prime}\lambda)$.
But then $A_\lambda A_\lambda^\ast=1=A_\lambda ^\ast A_\lambda $ and I don't see my mistake...
By the way, I calculated $\|A_\lambda\|=\lambda$. Is this correct?