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A Gaussian net test turns a finite covering of an alternative into a test with an explicit covering-number cost. In the Gaussian sequence model $Y=\theta +n^{-1/2}g$, compare a null ${\theta }_0$ with centers ${\theta }_1,\dots ,{\theta }_N$ in L2 space satisfying $\Vert {\theta }_j-{\theta }_0{\Vert }_2\ge r>0$. Put ${\delta }_j={\theta }_j-{\theta }_0$ and reject if anyEach pairing is well-defined as an isonormal Gaussian process pairing even though the noise sequence itself is not in L2 space. Under the null, it is centered normal with variance $\Vert {\delta }_j{\Vert }_2^2/n$ before subtracting the threshold. The Gaussian tail bound and union bound giveIf $\Vert \theta -{\theta }_j{\Vert }_2\le r/4$ for some center, the mean of that center's displayed statistic is at least $\Vert {\delta }_j{\Vert }_2^2/4$, by the Cauchy-Schwarz inequality. ThusA radius-$r/4$ net with centers in the separated alternative supplies such a test for its entire covered set. To make the first error small, the squared separation $n{r}^2$ must dominate $\log N$. This is why metric complexity, rather than just individual Gaussian tails, enters composite testing.

Write $K=K_n$, $ϵ={ϵ}_n$, and chooseTake a $\delta$-net $({\theta }^1,\dots ,{\theta }^N)$ of $H_1$ in ${\ell }^2$, with centers in $H_1$. Since every alternative lies in the $K$-dimensional Euclidean ball of radius $n$, the volumetric covering bound givesfor all large $n$. Put $v_j={\theta }^j-{\theta }_0$ and define the Gaussian net testThe inner products are well-defined Gaussian linear functionals because $v_j\in {\ell }^2$. Under $H_0$,and $\Vert v_j{\Vert }_2\ge Mϵ$. A Gaussian tail bound and a union bound give

For any $\theta \in H_1$, choose $j$ with $\Vert \theta -{\theta }^j{\Vert }_2\le \delta$. ThenAnother Gaussian tail bound givesBecause $n{ϵ}_n^2=K_n\log n$, the entropy term in the type-I bound is dominated by the signal exponent when $M$ is sufficiently large. Also ${ϵ}_n^2\to 0$ implies $K_n<n$ eventually, so $\log n\ge \log K_n$. Given $c_1>0$, choose $M$ large enough to obtainEnlarging $M$ if necessary handles the finitely many initial $n$.