Convergence of roots in $L^{p}$

Question

Assume that we have a collection of sequences, all converging with respect to the $L^{p}-$norm. Taking a product of roots of those sequences, such that the exponents add up to $1$, can we conclude, that this product will also converge to in the $L^{p}-$norm?

Answer 1

Assuming that $u_n,v_n\ge0$, then $$ u_n^{1/2}v_n^{1/2}-u^{1/2}v^{1/2}=u_n^{1/2}v_n^{1/2}-u_n^{1/2}v^{1/2}+u_n^{1/2}v^{1/2}-u^{1/2}v^{1/2}\\=u_n^{1/2}(v_n^{1/2}-v^{1/2})+ v^{1/2}(u_n^{1/2}-u^{1/2}) $$ and hence $$ \|u_n^{1/2}v_n^{1/2}-u^{1/2}v^{1/2}\|_p\le \|u_n^{1/2}(v_n^{1/2}-v^{1/2})\|_p+ \|v^{1/2}(u_n^{1/2}-u^{1/2})\|_p. $$ Cauchy-Schwarz inequality implies that $$ \|U^{1/2}V^{1/2}\|_p^2\le\|U\|_p\|V\|_p $$ Also, $|a^{1/b}-b^{1/2}|^2\le |a-b|$, for all $a,b\ge0$. Thus $$ \|u_n^{1/2}v_n^{1/2}-u^{1/2}v^{1/2}\|_p\le \|u_n^{1/2}(v_n^{1/2}-v^{1/2})\|_p+ \|v^{1/2}(u_n^{1/2}-u^{1/2})\|_p\\ \le \|u_n\|_p\|(v_n^{1/2}-v^{1/2})^2\|_p+ \|v\|_p\|(u_n^{1/2}-u^{1/2})^2\|_p \\ \le \|u_n\|_p\|v_n-v\|_p+ \|v\|_p\|u_n-u\|_p\to 0, $$ as $n\to\infty$, since $\|u_n\|$ remains bounded (Uniform Boundedness Principle).

In general, if $w_{k,n}\to w_k$ in the $p-$norm (non-negative functions) and $\sum_{k=1}^m\frac{1}{q_k}=1$, then $$ \prod_{k=1}^mw_{k,n}^{1/q_k}\to \prod_{k=1}^mw_{k}^{1/q_k}, $$ in the $p-$norm.