Derivative of Moreau envelope in Hilbert space with respect to regularization parameter $\lambda$ using Hopf-Lax formula?

Question

Let $\mathcal H$ be a Hilbert space, $f \colon \mathcal H \to (- \infty, \infty]$ a proper, convex, lower semi-continuous function and $\lambda > 0$. The $\lambda$-Moreau envelope of $f$ is $$ f_{\lambda} \colon \mathcal H \to \mathbb R, \qquad x \mapsto \min_{y \in H} f(y) + \frac{1}{2\lambda} \| x - y \|_H^2. $$ We have $\frac{\partial}{\partial x} f_{\lambda}(x) = \frac{1}{\lambda} \left( x - \text{prox}_{\lambda f}(x)\right)$, (where $\text{prox}$ is the proximal operator, which maps $x$ to the minimzer $y$ in the above formula) but I am interested in $\frac{\partial}{\partial \lambda} f_{\lambda}$.

For $\mathcal H = \mathbb R^n$ the Hopf-Lax formula (see e.g. Evans: PDE, thm. 4 in section 3.3.2b) with convex, smooth Hamiltonian $H \colon \mathcal H \to \mathbb R$, $x \mapsto \frac{1}{2} \| x \|_2^2$ (which has superlinear growth at infinity) states that if $f$ is Lipschitz continuous, then $(t, x) \mapsto f_t(x)$ is the unique viscosity solution of the Hamilton-Jacobi-Bellmann equation $$ \begin{cases} \partial_t u_t(x, t) + \frac{1}{2} \| \nabla_x u(t, x) \|_2^2 = 0, \qquad x \in \mathbb R^n, t > 0, \\ u(0, x) = f(x), \qquad x \in \mathbb R^n \end{cases}, $$ which shows that $\frac{\partial}{\partial \lambda} f_{\lambda}(x) = - \frac{1}{2 \lambda^2} \| x - \text{prox}_{\lambda f}(x) \|_2^2$ for $\lambda > 0$ and $x \in \mathbb R^n$.

Is there a corresponding result in Hilbert spaces and can the Lipschitz continuity assumption on $f$ be relaxed?

Edit. Can we perhaps use an envelope theorem, by defining $$ g \colon \text{dom}(f) \times \mathcal H \times \mathbb R, \qquad (x, y, \lambda) \mapsto - f(x) - \frac{1}{2 \lambda} \| x - y \|_{\mathcal H}^2 $$ and the value function $$ V \colon \mathcal H \times \mathbb R \to \mathbb R, \qquad (y, \lambda) \mapsto \sup_{x \in \text{dom}(f)} g(x, y, \lambda) = - f_{\lambda}(y). $$ Since $\partial_y g$ and $\partial_{\lambda} g$ exist, the multidimensional version of the envelope theorem should somehow state that if $V$ is totally differentiable, then $\nabla V(y, \lambda) = (\partial_y g, \partial_{\lambda} g)$, yielding the desired result.

Answer 1

The algebra in equation (155) of [1] (stated for convex functions of real variable) is elementary and applies to Hilbert spaces as well.

Let $\lambda>0$ and $y$ be such that $\lambda+y>0$. Let $p_0=prox[\lambda f](x)$ and $p_y=prox[(\lambda+y)f](x)$. By definition, the objective function at $y+\lambda$ of the minimizer $p_y$ is smaller than the same objective at $p_0$, \begin{align} h(y) &= f_{\lambda+y}(x)-f_\lambda(x) + \frac{y}{2\lambda^2}\|x-p_0\|^2 \\&\le \frac{\|x-p_0\|^2}{2(\lambda+y)} + f(p_0) - f(p_0) - \frac{\|x-p_0\|^2}{2\lambda} + \frac{y}{2\lambda^2}\|x-p_0\|^2 \\&= \|x-p_0\|^2 \frac{y^2}{2\lambda^2(y+\lambda)}. \end{align} Because $\lambda\mapsto f_\lambda(x)$ is convex in $\lambda$, $y\mapsto h(y)$ is also convex and $h(0)=0$ gives $$h(y)\ge - h(-y)\ge - \|x-p_0\|^2 \frac{y^2}{2\lambda^2(-y+\lambda)}.$$ This gives $h(y)=O(y^2)$ hence the desired derivative.

Reference

[1] Chistos Thrampoulidis, Ehsan Abbasi, Babak Hassibi, "Precise error analysis of regularized M-estimators in high dimensions", IEEE Transactions on Information Theory 64, No. 8, 5592-5628 (2018), arxiv:1601.06233v1[cs.IT], DOI:10.1109/TIT.2018.2840720, MR3832326, Zbl 1401.94051.