If $f^3_n\to f^3$ in $L^2(0,\pi)$ then $f_n\to f$ in $L^6(0,\pi)$?

Question

This question is brief, does this statement hold:

If $f^3_n\to f^3$ in $L^2(0,\pi)$ then $f_n\to f$ in $L^6(0,\pi)$?

and how about this

If $g^3_n\to g^3$ in $L^2(0,\pi;L^2(0,\pi))$ then $g_n\to g$ in $L^2(0,\pi;L^6(0,\pi))$?

Answer 1

The operation $f \mapsto f^{1/3}$ maps $L^2$ into $L^6$, hence it is continuous: After extracting subsequence, $f_n\to f$ pointwise a.e. We have the upper bounds $$ |f_n-f|^6 \le 2^5(|f_n|^6+|f|^6). $$ The right-hand side converges pointwise. Moreover its integral converges too: $$ \int |f_n|^6 = \int |f_n^3|^2 \to \int |f|^6. $$ Then by a variant of dominated convergence: Variant of dominated convergence theorem, does it follow that $\int f_n \to \int f$? we get the convergence $$ \int |f_n-f|^6 \to 0. $$ Therefore, the first claim is true.

This is a nice property of such nonlinear superposition operators on $L^p$: as soon as they map from $L^p$ to $L^q$ they are also continuous. See the book by Appell, Zabrejko 'Nonlinear superposition operators'.

For the second: we get $g_n\to g$ in $L^6(L^6)$, which implies convergence in $L^2(L^6)$.