$\newcommand\norm[1]{\lVert#1\rVert}$For any $p \in [1,2]$, $r \ge 0$, and integer $d \ge 1$, define a mixed-norm $\eta:\mathbb R^d \to \mathbb R$ by $\eta(x) := \norm x\_2 + r\norm x\_p$, for any $x \in \mathbb R^d$.


Do there exist scalars $a = a(d,p,r) \ge 0$ and $q=q(p,r) \in [1,\infty]$ such that $c\_1 \le \dfrac{\eta(x)}{a\norm x\_{q}} \le c\_2$ for every nonzero $x \in \mathbb R^d$, where $c\_1$ and $c\_2$ are absolute constants?


I'm particularly interested in the case $p=1$.



Is a mixture of $\ell_p$-norms $\eta(x):=\lVert x\rVert_2 + r\lVert x\rVert_p$ always dimensionlessly equivalent to some $\ell_q$-norm?
This is impossible unless the following holds:



> 
> $q=2$ and either $p=2$ or $r=0$.
> 
> 
> 


The above exception is due to the fact that it implies $\frac{1}{1+r}\eta(\cdot)=\|\cdot\|\_q=\|\cdot\|\_2$, for which the desired inequality trivially exists.


We argue ad absurdum; hence, suppose the desired inequality exists. If $r=0$, then $p$ is redundant in your definition of $\eta$; hence, in that case, we appropriate $p$ for the $2$ in the $2$-norm $\|\cdot\|\_2$. A *fortiori*, assume in the proof below that $p=2$ when $r=0$. (Note that this assumption implies that the exception above simplifies to $p=q=2$).


First, observe that the constant $a(d,p,r)$ and the absolute constants $c\_1,c\_2$ are essentially positive constants; hence, necessarily we must have that $a\_1:=\liminf\_{d\to\infty}\frac{1}{a(d,p,r)}>0$ and $a\_2:=\limsup\_{d\to\infty}\frac{1}{a(d,p,r)}<\infty$ (otherwise, either $c\_1$ will be forced to be $0$ or $c\_2$ will be forced to be $\infty$ in your desired inequality because $x\in\mathbb{R}^d$ implies $x\in\mathbb{R}^{d’}$ for all $d’>d$ via the canonical imbedding of $\mathbb{R}^d$ in $\mathbb{R}^{d’}$).


It follows from the above that for all non-zero $x\in\mathbb{R}^d$ and all $d\ge1$, we must have
\begin{array}
\label{(1)}
&a\_1c\_1\le\frac{\eta(x)}{\|x\|\_q}\le a\_2c\_2\,.
\end{array}
In other words, the norms $\eta$ and $\|\cdot\|\_q$ are equivalent on the real vector space $c\_{00}$ of finitely supported sequences (i.e. real sequences with only finitely many nonzero terms). However, this is impossible because of the following well-known (and in any case easily proven) result:



> 
> $c\_{00}$ is dense in the Banach space $(X\_q,\|\cdot\|\_q)$, where $X\_q=\ell^q$ for any $q\in[1,\infty)$ and $X\_q=c\_0$ for $q=\infty$.
> 
> 
> 


Recall that whenever $1\le q\le q’\le\infty$, then $X\_q\subseteq X\_{q’}$; thus, because $p\le2$ and $\eta(\cdot)=\|\cdot\|\_2+r\|\cdot\|\_p$, it follows that $(\ell^p,\eta)$ is a well-defined Banach space as the completion of $(c\_{00},\eta)$—note that this is true when $r=0$ since we assume $p=2$ in that case.


Now, let $x= (k^{-1/p})\_{k\ge1}$ and $y=(k^{-1/q})\_{k\ge1}$, and consider their truncated sequences in $c\_{00}$,that is
\begin{align}
&x\_n=(1,2^{-1/p},\ldots,n^{-1/p},0,0,0,\ldots)\,,\\
&y\_n= (1,2^{-1/q},\ldots,n^{-1/q},0,0,0,\ldots)\,.
\end{align}
Thanks to the divergence of the harmonic series, we know that $\|x\_n\|\_p\to\infty$ and $\|y\_n\|\_q\to\infty$; however, observe that $p<q$ implies $x\_n\to x$ in $(X\_q,\|\cdot\|)$ and $p>q$ implies $y\_n\to y$ in $(\ell^p,\eta)$, either of which contradicts Inequality (1). We must therefore necessarily have that $p=q$. However, this implies that $q=2$, because otherwise $p=q<2$ and this time considering $z=(k^{-\frac{2}{2+q}})\_{k\ge1}$ and its truncated sequences in $c\_{00}$,
$$z\_n=(1,2^{-\frac{2}{2+q}},\ldots,n^ {-\frac{2}{2+q}},0,0,0,\ldots)\,,$$
then Inequality (1) implies that
$$(a\_1c\_1-r)\|z\_n\|\_q\le\|z\_n\|\_2\le(a\_2c\_2-r)\|z\_n\|\_q\,,$$
which leads to a contradiction as $n\to\infty$ (because $\|z\_n\|\_2\to\|z\|\_2\ne0$ whereas $\|z\_n\|\_q$ is unbounded).


Q.E.D


