# 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?

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

• @LSpice Thanks for the fix. Aug 5, 2022 at 19:25
• Maybe i misunderstood but on a finite dimensional vector space ($\mathbb{R}^d)$ all norms are equivalent... Aug 6, 2022 at 17:41
• @Toni Yes, but there is a subtlety here: the constants in the equivalence should be independent of the dimension $d$ ... Aug 6, 2022 at 18:00
• @dohmatob. This is rather confusing. Perhaps, I’m also misunderstanding you but, for instance, your constant $a$ is very dependent on $d$ by your own definition. Secondly, if you desire them to be independent of $d$, then you’re essentially working in the sequence space $(c_{00},\eta)$ of sequences with finite support (i.e. with only finitely many nonzero terms and not it’s canonical finite-dimensional subspaces $\mathbb{R}^d$ for all $d$. Aug 6, 2022 at 18:22
• @ToniMhax Please read the question carefully, then I can clarify any confusions which remain. I'm fine with $a$ being dependent on $d$, and I actually wrote $a=a(d,p,r)$ explicitly to stress this... Aug 6, 2022 at 18:48

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 $$p implies $$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