Proof of Theorem 1 — the basic inequality $\|\hat A^\lambda-A_0\|^2_{L_2(\Pi)}\le\|A-A_0\|^2_{L_2(\Pi)}+2\Delta\|\hat A^\lambda-A\|_1+\lambda(\|A\|_1-\|\hat A^\lambda\|_1)$

Work in the trace regression model $\mathbb E(Y_i\mid X_i)=\langle A_0,X_i\rangle$, $i=1,\dots,n$, with square-integrable design entries and integrable responses. Fix a realization of the sample, a set $\mathbb A$ of $m_1\times m_2$ matrices, $\lambda>0$, and an estimator $\hat A^\lambda\in\operatorname{argmin}_{A\in\mathbb A}L_n(A)$ for the penalized empirical risk

Let $\mathbf M=\frac1n\sum_{i=1}^n(Y_iX_i-\mathbb E(Y_iX_i))$ and $\Delta=\|\mathbf M\|_\infty$ (operator norm). Then for every $A\in\mathbb A$,

No condition relating $\lambda$ and $\Delta$ and no structure of $\mathbb A$ is needed. This inequality is the starting point of the proof of Theorem 1; under $\lambda\ge2\Delta$ it gives (2.3) immediately.

Formalization Note The model hypothesis is the conditional-expectation form of (1.1); the sample regularity (measurability, square-integrable design entries, integrable $Y_i$ and $Y_i(X_i)_{jk}$) is added so that the expectations are meaningful. Independence of the pairs is not assumed (the statement holds realization by realization).