Fix positive $t$. Construct $a_n \in \mathbb R^n$ such that $(\inf_x \|x-a_n\|_2 + t\|x\|_1 )/\min(\|a_n\|_2,t\|a_n\|_1) \to 0$

Question

For any positive number $t$, nonnegative integer $n$, and nonzero vector $a \in \mathbb R^n$, define

$$ \begin{split} K_n(a,t) &:= \inf_{x \in \mathbb R^n} \|x-a\|_2 + t\|x\|_1,\\ M_n(a,t) &:= \min(\|a\|_2,t\|a\|_1),\\ R_n(a,t) &:= K_n(a,t)/M_n(a,t). \end{split} $$

Note. $K_n$ defines the Peetre's K-functional between $(\mathbb R^n,\ell_2)$ and $(\mathbb R^n,\ell_1)$, where the $\ell_p$ norm of a vector $x=(x^1,\ldots,x^n)$ is defined by $\|x\|_p := (\sum_i |x^i|^p)^{1/p}$.

Question. Is it possible to construct $a_n \in \mathbb R^n$ for each $n$, such that $\lim_{n \to \infty} R_n(a_n,t) = 0$ ?

Motivation

Clearly, one always has $K_n(a,t) \le M_n(a,t)$ with equality when $n=1$. What is not clear is whether one can construct $a$ with growing dimension $n$ such that $K_n(a,t) \ll M_n(a,t)$ eventually.

Answer 1

This is to extend Christian Remling's comment to all real $t>0$, with an explicit lower bound on $K/M$, where $K:=K_n(a,t)$ and $M:=M_n(a,t)$.

$\newcommand\norm[1]{\lVert#1\rVert}$The key here, as in Christian Remling's comment, is the observation that $\norm x_1\ge\norm x_2$ for all $x\in\mathbb R^n$. Indeed, this observation implies $$K \ge\inf_{x\in\mathbb R^n}\bigl(\lvert\norm x_2-\norm a_2\rvert+t\norm x_2\bigr) =\inf_{u\ge0}\bigl(\lvert u-\norm a_2|+tu\bigr) =\min(1,t)\norm a_2$$ and $$M\le\norm a_2,$$ whence $$\frac KM\ge\min(1,t).$$ (In particular, $K\ge M$ if $t\ge1$.)