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?

Question

$\newcommand\norm[1]{\lVert#1\rVert}$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 $p=1$.

Answer 1

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