$\| U+\|V\|_r\|_m =\| U+V \|_m$ if and only if $m=r=2$.

Question

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.

Answer 1

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.