I am working on this problem:
Consider the space $H=\{f:\mathbb{R}\to\mathbb{C} : \int_{-\infty}^{+\infty} e^{-|x|} |f(x)|^2 dx <\infty\}$, this is a Hilbert space with the inner product $<f,g >=\int_{-\infty}^{+\infty} e^{-|x|} f(x) \overline{g(x)} dx$.
Consider the operator $T:H\to H$ defined as $Tf(x)=f(x+1)$. Find the adjoint of T.
I have calculated $<Tf,g>$ by the substitution $x+1=y$. I split the integrals but I don't know how to reassemble them. Could anyone help me? Thanks a lot.
We start with $$ \langle Tf,g\rangle = \int_{\mathbb R} e^{-|x|}f(x+1)g(x)dx. $$ In order to get something that looks like $\langle f, h\rangle$, we have to transform the integral to contain $e^{-|x|}f(x)$. Now do a change of coordinates $x\mapsto x-1$, then $$\begin{split} \langle Tf,g\rangle &= \int_{\mathbb R} e^{-|x-1|}f(x)g(x-1)dx\\ &=\int_{\mathbb R} e^{-|x|} e^{|x|-|x-1|}f(x)g(x-1)dx\\ &=\int_{\mathbb R} e^{-|x|} \cdot f(x) \cdot (e^{|x|-|x-1|} g(x-1))dx\\ &=\langle f, T^*g\rangle \end{split} $$ so $$ (T^*g)(x)=e^{|x|-|x-1|} g(x-1) $$