# Can I smooth a map to the circle, while preserving the Lipschitz constant?

Let $$M$$ be a closed Riemannian manifold (perhaps even a surface, at first), and let $$u: M \to \mathbf S^1$$ be a Lipschitz map to the circle, with Lipschitz constant $$L$$. We can find an approximation $$(u_n) \subset C^\infty$$ to $$u$$ uniformly, such that Lipschitz constants $$L_n$$ converge to $$L$$. However, I am interested in an approximation which preserves or contracts the Lipschitz constant, thus $$u_n \to u$$ uniformly and $$L_n \leq L$$.

If $$M$$ is a flat torus, then this holds. Let $$u(t)$$ be the solution of the heat equation with Cauchy data $$u$$ and $$u_n := u(1/n)$$. Since $$M$$ is a flat torus, the heat kernel $$K(t, x, y)$$ is actually of the form $$K(t, x, y) = K(t, x - y)$$ so $$e^{t\Delta}$$ is a convolution operator and hence commutes with $$\partial_i$$. In particular, $$\partial_i u(t) = K(t) * \partial_i u$$ and the claim follows from Young's inequality. However, on a general closed Riemannian manifold, the heat kernel takes the form $$K(t, x, y) = \sum_{n=1}^\infty e^{-\lambda_n t} \varphi_n(x) \varphi_n(y)$$ where $$\lambda_n$$ is the $$n$$th eigenvalue of $$M$$, with eigenfunction $$\varphi_n$$. This is not a convolution operator! In fact, I'm pretty sure we can think of the Riemann curvature tensor as "the obstruction to the existence of an orthonormal frame of vector fields which commute with the heat flow".

So is there another way to obtain a smooth approximation to a Lipschitz map, which contracts the Lipschitz constant?

• Iirc this is proven in $C^\infty$ approximations of convex, subharmonic, and plurisubharmonic functions by Greene & Wu. (full text here) Commented May 29, 2023 at 18:15
• @Kajelad Thanks! (Sorry for the very late response, I thought I already responded to this.) I guess that some modification of the methods of Section 2 of that paper can be used to get a convolution that preserves maps to the circle. Commented Jun 8, 2023 at 18:34