I was wondering if my hypothesis is correct: Let $X=L^2[0,1]$, $f\in X$, $g:[0,1]\to[0,1]$ definited as $g(x)=\sqrt x$
Is $f\circ g\in L^2[0,1]$ necessarily?
Thanks a lot
2026-04-02 03:58:47.1775102327
composition of measureable function with $\sqrt x$ in $L^2[0,1]$
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Yes. Using the change of variable $x=\sqrt t$, $$ \int_0^1|f(\sqrt t)|^2\,dt=2\int_0^1x\,|f(x)|^2\,dx\leq2\int_0^1|f(x)|^2\,dx<\infty $$ The change of variable is obvious for continuous $f$, and can be deduced for arbitrary $f$ by approximation with continuous functions and using dominated convergence.