The differentiable loss has gradient $X^T(X\beta-Y)/n$. The <subdifferential sum rule> and part (b) give
$$
\partial Q(\beta)=\frac1nX^T(X\beta-Y)+\lambda\sqrt m\,u,
$$
where blockwise
$$
u^{(k)}=\frac{\beta^{(k)}}{\lVert\beta^{(k)}\rVert_2}
\quad\text{if }\beta^{(k)}\ne0,
\qquad
\lVert u^{(k)}\rVert_2\leq1
\quad\text{if }\beta^{(k)}=0.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 205 / 1 / c / Solution

For proper convex functions under a standard relative-interior qualification,
$$
\partial(f+g)(x)=\partial f(x)+\partial g(x).
$$
In particular, if $f$ is differentiable at $x$, then $\partial(f+g)(x)=\nabla f(x)+\partial g(x)$.


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