Let, $1 ≤ p ≤ ∞$. Show that, for any $x∈ \Bbb R$ and any $f∈L^p(\lambda)$ ,one has $T_x(f) (:= f(x+·) ∈ L^p$ and $||T_x(f)||_p = ||f||_p$
I started with indicator functions.
Let, $f=1_B$ for some Borel set, then $\int|T_x(f)|^p d\lambda = \int(1_{B+x})^p d\lambda= \int 1_{B+x} d\lambda = \lambda(B+x)=\lambda(B)=\int|f|^p d\lambda $
Similarly, for simple functions.
Taking increasing limits, same is true for non-negative measurable functions.
Let, $f$ be measurable and thus, $f=f^+ - f^-$. Similarly, $T_x(f)={T_x(f)}^+ -{T_x(f)}^-$
$\int|T_x(f)|^p d\lambda = \int|T_x(f)^+ - T_x(f)^- |^p d\lambda =$
The terminal terms fit in nicely for the equality with terms of same form for $f$ , but how to deal with the mixed terms, like there will be terms of the form, $\int |T_x(f)^+|^l |T_x(f)^-|^{p-l} d\lambda$ , then how to manipulate these terms?
You should observe that at least one of $T_x(f)^+$ and $T_x(f)^-$ equals zero. Thus $$|T_x(f)|^p = |T_x(f)^+ - T_x(f)^-|^p = |T_x(f)^+|^p + |T_x(f)^-|^p.$$