$\newcommand{\R}{\mathbb R}$For natural $n$, $a\in\R^n$, and real $t>0$, let  

\begin{equation\*}
 K:=K\_{n,t}(a):=\inf\_{x\in\R^n}(\|a-x\|\_2+t\|x\|\_1),
\end{equation\*}
\begin{equation\*}
 M:=M\_{n,t}(a):=\min(\|a\|\_2,t\|a\|\_1), 
\end{equation\*}
and (for nonzero $a$)
\begin{equation\*}
 R:=R\_{n,t}(a):=\frac KM, 
\end{equation\*}
where $\|x\|\_p:=(\sum\_1^n|x\_i|^p)^{1/p}$ for $x=(x\_1,\dots,x\_n)\in\R^n$.


So, the function $K\_{n,t}$ is a norm on $\R^n$, which is the [infimal convolution](https://en.wikipedia.org/wiki/Convex_conjugate#Infimal_convolution) of the norms $\|\cdot\|\_2$ and $t\|\cdot\|\_1$. The function $M\_{n,t}$ is a norm only for $t\ge1$ (and then $M\_{n,t}=\|\cdot\|\_2$) and for $t\le1/\sqrt n$ (and then $M\_{n,t}=t\|\cdot\|\_1$).


Clearly, $K\le M$.
It was [previously asked](https://mathoverflow.net/q/432168/36721) whether, for each $t>0$,  

\begin{equation\*}
 \inf\_{a\in\R^n\setminus\{0\}}R\_{n,t}(a)\to0
\end{equation\*}
as $n\to\infty$.


It was then shown that this is [not true for $t=1$](https://mathoverflow.net/questions/432168/fix-positive-t-construct-a-n-in-mathbb-rn-such-that-inf-x-x-a-n-2#comment1112378_432168) and also [not true for any real $t>0$](https://mathoverflow.net/a/432184/36721), because
$$\frac KM\ge\min(1,t).$$


It was [further asked](https://mathoverflow.net/questions/432168/fix-positive-t-construct-a-n-in-mathbb-rn-such-that-inf-x-x-a-n-2#comment1112883_432184) if
\begin{equation\*}
 \inf\_{a\in\R^n\setminus\{0\}}R\_{n,t\_n}(a)\to0
\end{equation\*}
as $n\to\infty$ assuming that $t\_n\to0$.


A somewhat surprising answer to this question will be given below.



On the infimal convolution of two norms on $\mathbb R^n$
$\newcommand\ka\kappa\renewcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand\ep\varepsilon$Take any nonzero $a=(a\_1,\dots,a\_n)\in\R^n$. We have  

\begin{equation\*}
 K=\inf\_{x\in\R^n}\ka(x),\quad \ka(x):=\ka\_a(x):=(\|a-x\|\_2+t\|x\|\_1). \tag{1}\label{1}
\end{equation\*}


Since the norms $\|\cdot\|\_p$ are orthant-symmetric, without loss of generality (wlog) $a\_i\ge0$ for all $i\in[n]:=|[1,\dots,n\}$. Since the function $\ka$ is continuous and $\ka(x)\to\infty$ as $\|x\|\_2\to\infty$, the infimum in \eqref{1} is attained at some $x=(x\_1,\dots,x\_n)\in\R^n$.


In what follows, (unless specified otherwise) let $x$ be such a minimizer.


If now $x\_j<0$ for some $j\in[n]$ then, recalling that $a\_i\ge0$ for all $i\in[n]$ and replacing the $j$th coordinate of $x$ by $-x\_j$, we get another minimizer of $\ka$. So, wlog $x\_j\ge0$ for all $j\in[n]$. Let
\begin{equation\*}
 J:=\{j\in[n]\colon x\_j>0\}. 
\end{equation\*}
Then, differentiating the function $\ka$ at the minimizer $x$, we get  

\begin{equation\*}
 a\_j-x\_j=ct\quad\text{for all}\quad j\in J, 
\end{equation\*}
with $x\_j=0$ for $j\notin J$, where
\begin{equation\*}
 c:=\|a-x\|\_2=\sqrt{kc^2t^2+\sum\_{j\notin J}a\_j^2}, 
\end{equation\*}
whence
\begin{equation\*}
 c=\sqrt{\frac{\sum\_{j\notin J}a\_j^2}{1-kt^2}}, 
\end{equation\*}
where
\begin{equation\*}
 k:=|J|,
\end{equation\*}
the cardinality of $J$; here it is assumed that $k<1/t^2$.


Thus, letting
\begin{equation\*}
 A\_1:=\sum\_{j\in J}a\_j,\quad A\_2:=\sum\_{j\notin J}a\_j,\quad 
 B\_1:=\sqrt{\sum\_{j\in J}a\_j^2},\quad B\_2:=\sqrt{\sum\_{j\notin J}a\_j^2}, 
\end{equation\*}
after some algebra we get
\begin{equation\*}
 K=\sqrt{1-kt^2}B\_2+tA\_1,\quad M=\min(\sqrt{B\_1^2+B\_2^2},tA\_1+tA\_2). 
\end{equation\*}


Take now any real $\ep\in(0,1)$ and consider the case when
\begin{equation\*}
 t\le\frac{1-\ep}{\sqrt n}; \tag{2}\label{2}
\end{equation\*}
note that then $k\le n\le(1-\ep)^2/t^2$, so that the condition $k<1/t^2$ is satisfied.
Suppose now that $K<<M$; we write $A<<B$ or, equivalently, $B>>A$ if $A=o(B)$, and we write $A\ll B$ or, equivalently, $B\gg A$ if $A=O(B)$. Then
\begin{equation\*}
 tA\_1\le K<< M\le tA\_1+tA\_2,
\end{equation\*}
so that $tA\_1<<tA\_2$ and $tA\_1+tA\_2\ll tA\_2$. Also, $A\_2\le B\_2\sqrt{n-k}\le B\_2\sqrt n$ and hence
$tA\_2\le\frac{1-\ep}{\sqrt n}B\_2\sqrt n\le B\_2$. So,
\begin{equation\*}
 M\le tA\_1+tA\_2\ll tA\_2\ll B\_2. 
\end{equation\*}
On the other hand, \eqref{2} implies that $kt^2\le nt^2\le(1-\ep)^2$ and hence $1-kt^2\ge1-(1-\ep)^2>0$, so that
\begin{equation\*}
 K\gg B\_2+tA\_1\ge B\_2.
\end{equation\*}
We conclude that, in the case \eqref{2}, the assumption $K<<M$ leads to $K\gg M$. Thus,
\begin{equation\*}
 t\le\frac{1-\ep}{\sqrt n}\implies K\gg M. \tag{3}\label{3}
\end{equation\*}


Consider finally the case when, for some real $\ep>0$,  

\begin{equation\*}
 1>>t\ge\frac{1+\ep}{\sqrt n}. \tag{4}\label{4}
\end{equation\*}
For all $j\in[n]$, let then
\begin{equation\*}
 a\_j:=1(j\le m)+b\,1(j>m),\quad x\_j:=1(j\le m)(1-tC),
\end{equation\*}
where
\begin{equation\*}
 m:=\Big\lceil\frac1{t^2}\Big\rceil-1,\quad C:=b\sqrt{\frac{n-m}{1-mt^2}}, 
\end{equation\*}
and a real $b$ varies with $n$ and $t$ so that
\begin{equation\*}
 \frac{tm}{\sqrt n}<<b<<\frac m{\sqrt n}. \tag{5}\label{5}
\end{equation\*}
Note that $m>>1$,
\begin{equation\*}
 n-m\gg n \tag{6}\label{6}
\end{equation\*}
by \eqref{4},
\begin{equation\*}
 1-mt^2\le t^2, \tag{7}\label{7}
\end{equation\*}
and
\begin{equation\*}
 b>>\frac{tm}{\sqrt n}\ge(1+\ep)\frac mn\ge\frac mn \tag{8}\label{8}
\end{equation\*}
by \eqref{5} and \eqref{4}.
Next,  

\begin{equation\*}
 K\le\ka\_a(x)=K\_1+K\_2,\quad M=\min(M\_1,M\_2), \tag{9}\label{9}
\end{equation\*}
where
\begin{equation\*}
 K\_1:=\sqrt{1-mt^2}\, b\sqrt{n-m},\quad K\_2:=tm,
\end{equation\*}
\begin{equation\*}
 M\_1:=\sqrt{m+(n-m)b^2},\quad M\_2:=tm+t(n-m)b. 
\end{equation\*}
Further,
\begin{equation\*}
 K\_1\le tb\sqrt{n-m}<<b\sqrt{n-m}\le M\_1 \tag{10}\label{10}
\end{equation\*}
by \eqref{7} and \eqref{4};
\begin{equation\*}
 K\_2<< b\sqrt n\ll \sqrt{(n-m)b^2}\le M\_1 \tag{11}\label{11}
\end{equation\*}
by \eqref{5} and \eqref{6};
\begin{equation\*}
 K\_1\le tb\sqrt{n-m}<<t(n-m)b\le M\_2 \tag{12}\label{12}
\end{equation\*}
by \eqref{7} and \eqref{6};
\begin{equation\*}
 K\_2=tm<< tbn\ll tb(n-m)\le M\_2 \tag{13}\label{13}
\end{equation\*}
by \eqref{8} and \eqref{6}.


It follows from \eqref{9}--\eqref{13} that $K<<M$ in the case \eqref{4}.


Summarizing, for all $t=t\_n>0$ we have
\begin{equation}
 \inf\_{a\in\R^n\setminus\{0\}}R\_{n,t\_n}(a)\to0\quad\text{if}\quad 1>>t\_n\ge\frac{1+\ep}{\sqrt n}
\end{equation}
and
\begin{equation}
 \inf\_{a\in\R^n\setminus\{0\}}R\_{n,t\_n}(a)\asymp1 \quad\text{if}\quad 0<t\_n\le\frac{1-\ep}{\sqrt n}
 \quad\text{or}\quad t\_n\gg1. 
\end{equation}


