Lipschitz continuity of the Moreau Envelope of a closed, convex, proper function with respect to the smoothing parameter

775 Views Asked by At

Let $g$ be a function in $\Gamma_0$, the cone of closed, convex, proper functions from some Hilbert space $\mathcal{H}$ to $\mathbb{R}\cup+\infty$. We define the Moreau Envelope of such a function to be,

$$g^{\beta}(x) = \inf\limits_{y\in\mathcal{H}} \left\{g(y) + \frac{\beta}{2}\|y-x\|^2\right\}.$$

My question is the following: what can we say about the Lipschitz continuity of $g^\beta(x)$ as a function of $\beta$? Do we have something like,

$$|g^{\beta}(x)-g^{\alpha}(x)|\leq C|\beta-\alpha|$$?

More precisely, does anyone know of a reference or text where I can learn about existence/nonexistence of such a result?

1

There are 1 best solutions below

6
On BEST ANSWER

In fact, $g^{\beta}(x)$ is convex, and thus locally Lipschitz, in the parameter $\beta$. To see this we write $g^\beta(x)$ in the following way:

$$g^{\beta}(x) = (g^*+\frac{\beta}{2}\|\cdot\|^2)^*(x)$$

where $*$ denotes the fenchel conjugate. This gives,

$$g^{\beta}(x) = \sup\limits_{y}\langle x, y\rangle -g^*(y)-\frac{\beta}{2}\|y\|^2$$

which is the supremum of affine functions of $\beta$ and thus convex.

With this convexity we can have the following estimate:

$$g^{\beta}(x)\geq g^{\beta'}(x)+(\beta-\beta')\frac{\partial}{\partial \beta}g^{\beta'}(x)\\ \implies g^{\beta'}(x)-g^{\beta}(x)\leq (\beta'-\beta)\frac{\partial}{\partial\beta}g^{\beta'}(x)$$

It is also possible to say something about $\frac{\partial}{\partial \beta}g^{\beta'}(x)$ but it comes from Hamiltonian mechanics/PDEs. We can see the following,

$$ \frac{\partial}{\partial\beta}g(x,\beta) = -\frac{1}{2}\|\nabla_x g (x,\beta)\|^2$$

as a Hamilton Jacobi equation with solution $g^{\beta}(x)$. So $g^\beta (x)$ is differentiable with respect to $\beta$ as well.