I am looking for a straightforward way to upper bound the covering number of a $d$-dimensional euclidean ball by $\ell_\infty$-balls of radius $\varepsilon$, which I will call cubes of sidelength $2\varepsilon$ for clarity. Let us denote this number by $\mathcal N(\varepsilon)$.

An elementary upper bound is to say $\mathcal N(\varepsilon) \leqslant 1 / \varepsilon^d$, as we can cover the whole cube of sidelength 2 with this many small cubes. However, I know that the precise asymptotic behavior, up to multiplicative and additive constants, is the following $$ \log \mathcal N(\varepsilon) \approx \left\{\begin{split} \frac{1}{\varepsilon^2} \log d \varepsilon^2\quad \text{if} \quad \varepsilon \geqslant 1/\sqrt{d}\\ d \log \frac{1}{d\varepsilon^2} \quad \text{if} \quad \varepsilon \leqslant 1/\sqrt{d} \, . \end{split}\right. \ $$ Therefore the easy upper bound of $d \log 1/ \varepsilon$ is loose when $\varepsilon$ is large. The change of regime happens when the diagonal of the smaller cubes become comparable to the radius of the larger ball.

The only proof I found, in online teaching notes [1], is convoluted. It consists in first bounding the covering number of $\ell_1$ balls by $\ell_2$ balls, thanks to Maurey's empirical method, and then to appeal to a duality result of [2].

While the proof is elegant and sophisticated, the inconvenient is that it yields non-explicit constants. Also, it does not seem to take advantage of the ``easiness'' of the problem: covering with hypercubes should not be too hard as the cubes fit well together. I tried counting the number of cubes needed to cover the ball in the natural covering (cutting the cube of sidelength $2$ into small cubes of length $2\varepsilon$ ) but I have no idea how one would do so. For all I know, it might even be that the optimal covering has some overlap between the cubes.

EDIT: A nice improvement on the trivial bound by Rémi Peyre. Considering the natural covering, it suffices to count the cubes that lie inside the ball of radius $1 + 2 \varepsilon \sqrt{d}$. Therefore \begin{equation} \mathcal N(\varepsilon) \leq \frac{1}{2 \epsilon^d} \mathrm{Vol}\big(B(1 + 2\varepsilon\sqrt{d}) \big) \leq \frac{1}{\sqrt{d\pi}} (2 \pi e)^{d/2} \Big( 1 + \frac{1}{2\varepsilon \sqrt{d}} \Big)^d \, . \end{equation} using a non-asymptotic version of Stirling. Taking logs we obtain something of order $d ( c + \log(1 + 1/(\varepsilon \sqrt{d}) )$. This is an improvement but is still far from the refined bounds above, when $\varepsilon \sqrt{d}$ is large.

[1] http://www.stat.yale.edu/~yw562/teaching/598/lec15.pdf

[2] Duality of metric entropy, S. Artstein, V. Milman, and S. J. Szarek https://annals.math.princeton.edu/wp-content/uploads/annals-v159-n3-p07.pdf

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