How could I search the inverse operator $(Af)^{-1}(x)$

Question

I am try to search $A^{-1}$ when I define $A:L^2[0,2] \rightarrow L^2 [0,2] $ when $$(Af)(x)=x^{-1/4}f (\sqrt {2x}) $$

---

What I do: I consider that $ (Af)^{-1}((Af)(x))=Ix=x \Longleftrightarrow (Af)^{-1}(x^{-1/4}f (\sqrt {2x}))=x$ then.. $$(Af)^{-1}(x)=x^{9/4}\frac{1}{f (\sqrt {2x})}$$

But it is not so.. help?

Answer 1

Let $B(f)(x)={x^{1/2}\over 2^{1/4}}f({x^2\over 2})$, we have $B(A(f))=B(x^{-1/4}f(\sqrt{2x}))$ $={x^{1/2}\over 2^{1/4}}(({x^2\over 2})^{-1/4})f(\sqrt{2{x^2\over 2}})=f(x)$

On the other hand, $A(B(f))=A({x^{1/2}\over 2^{1/4}}f({x^2\over 2}))=x^{-1/4}({{\sqrt{2x}}^{1/2}\over 2^{1/4}})f({\sqrt{2x}^2\over 2})=f(x).$