Possible Frechet-Kolmogorov compactness theorem equivalent in $L^\infty$?

Question

The Frechet-Kolmogorov(-Sudakov) compactness theorem states that in $L^p(\mathbb{R}^n)$ for $1\leq p<+\infty$, a set of functions $\mathcal{F}$ is totally bounded if and only if $\mathcal{F}$ is

1. equitight : $\forall \varepsilon>0, \exists R>0, \forall f\in\mathcal{F},\quad \lVert f\cdot 1_{\{|x|>R\}}\rVert_p \leq \varepsilon\,;$
2. equicontinuous : $\forall \varepsilon>0,\exists\delta>0,\forall|y|<\delta, \forall f\in\mathcal{F},\quad \lVert T_y f - f\rVert_p\leq\varepsilon\,.$

where $T_y f(x)=f(x-y)$.

I am studying the (im)possibility of such a theorem in $L^\infty$. I have found conterexamples for :

• totally bounded $\Longrightarrow$ 1, with $f=1$;
• totally bounded $\Longrightarrow$ 2, with $f=1_{[0,1]}$;

Now, i am wondering if the converse might be true. Here's my heuristic reasoning towards why it could be true :

1. Condition 2 provides that each function $f_i\in\mathcal{F}$ has a continuous representative $\hat{f_i}$.
2. Condition 1 provides that each $\hat{f_i}$ is smaller that $\varepsilon$ outside $B=\{\lVert x\rVert\leq R\}$.
3. Then, for each $\hat{f_i}$, create a continous function $g_i$ with support in $B$ that $\varepsilon$-approximates $\hat{f_i}$ with respect to the supremum norm.
4. Now we apply the Arzelà-Ascoli theorem to the set of all function $g_i$, which provides that $G=\{g_i\mid i\in I\}$ is totally bounded in $(\mathcal{C}(B),\lVert\cdot\rVert_\infty)$, thus is also totally bounded in $(L^\infty(\mathbb{R}),\lVert\cdot\rVert_\infty)$.
5. Take a finite cover of $\overline{G}$ with balls of constant radius : this is a finite cover of $\mathcal{F}$ as well, which completes the proof that $\mathcal{F}$ is totally bounded.

Is this reasoning valid ? Thanks for any insight.