TITLE: $\| U+\|V\|_r\|_m =\| U+V \|_m$ if and only if $m=r=2$.
QUESTION [5 upvotes]: Let $\|U \|_p = \left( E[|U|^p] \right)^{\frac{1}{p}}$. 
Is the following result statement true? 

Let $U$ and $V$ bet two independent, symmetric, non-degenerate random
  variables.  Then, \begin{align*} \| U+\|V\|_r\|_m =\| U+V \|_m
 \end{align*}  if and only if $r=m=2$.

Note that the "if"  directions is trivial since
\begin{align}
E[(U + \|V\|_2)^2]= E[U^2]+E[V^2]=E[(U+V)^2].
\end{align}
The question is how to show the "only if" direction. 
I also feel that this should have come up somewhere. For example, is it related somehow to the fact that $L^2$ norm is the only $L_p$ norm induced by the inner product?  Anyway, this is just a thought and the question is not about inner products.

REPLY [2 votes]: This fact seems to be false. Let $U$ and $V$ be random variables with support over $\{-1,1\}$. 

Since $|V|=1$, conclude that $||V||_{r}=1$. Therefore, $|U+||V||_{r}|=|U+1|$. Since $U$ has a symmetric distribution, $P(|U+1|=0)=P(|U+1|=2)=0.5$. Conclude that
$P(|U+||V||_{r}|=0)=P(|U+||V||_{r}|=2)=0.5$
Since $U$ and $V$ are symmetric and independent $P(|U+V|=0)=P(|U+V|=2)=0.5$. 

Since $|U+||V||_{r}|$ and $|U+V|$ have the same distribution,
conlude that $||U+||V||_{r}||_{m}=||U+V||_{m}$. 
Note that $m$ and $r$ were arbitrary.