Convergence of roots in $L^{p}$ 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?
 A: 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.
