Convolution of viscosity solutions and subharmonic functions

Question

Suppose that $u : \mathbb{C}^n \rightarrow \mathbb{R}$ is continuous. We say that $u$ is a viscosity subsolution (resp. viscosity supersolution) for the Laplace's equation if for all $\varphi \in C^2$ such that $u-\varphi$ has a local maximum (resp. local minimum) at $z_0$, we have $\Delta \varphi(z_0) \geq 0$ (resp. $\Delta \varphi(z_0) \leq 0$).

Note that when $u \in C^2$ is a viscosity subsolution (resp. viscosity supersolution), $\Delta u(z) \geq 0$ (resp. $\Delta u(z) \leq 0$) for all $z \in \mathbb{C}^n$.

In addition, denote the standard convolution of $u$ as follows $$[u]_r(z) := \int_{\mathbb{C}^n} u(z+r\tilde{z}) \eta(\tilde{z}) \Omega(\tilde{z}) = \int_{\mathbb{C}^n} u(z') \eta\left(\frac{z'-z}{r}\right)\frac{1}{r^{2n}} \Omega(z'),$$ where $\Omega$ is the standard volume form on $\mathbb{C}^n$, and $\eta$ is a standard mollifier with $\eta \geq 0$, $\text{supp}(\eta) \in B_1(0)$, and $\int_{\mathbb{C}^n} \eta(\tilde{z}) d\tilde{z} = 1$.

The following are my questions:

(i) Let $u$ be a continuous viscosity subsolution. Is $[u]_r$ still a viscosity subsolution?

(ii) Let $u$ is a continuous bounded subharmonic function on $\mathbb{C}^n$ with $\sup_{\mathbb{C}^{n}} u =\alpha$. Does $\lim_{r \rightarrow \infty} [u]_r(z) = \alpha$?

Answer 1

The answer to both questions is yes. The key is that being a viscosity subsolution is equivalent to satisfying the sub-mean value property $$u(x) \leq \frac{1}{|\partial B_r|}\int_{\partial B_r(x)} u \,dA$$ for all $r$ and $x$. Indeed, if $u$ is a viscosity subsolution then one can compare with the harmonic function with the same values on $\partial B_r(x)$. The other direction comes directly from the definition of viscosity subsolution.

The statement (i) follows quickly by Fubini. For (ii), note that by integrating in $r$ one also has the sub-mean value property in solid balls. Let $(f)_r(x)$ denote the average of $f$ in $B_r(x)$. It follows that $(\alpha - u)_r(x) \leq 2^{2n}\inf_{B_r(x)}(\alpha - u)_{2r} \leq 2^{2n}\inf_{B_r(x)}(\alpha - u) \rightarrow 0$ as $r \rightarrow \infty$. As a consequence, $[\alpha - u]_r \rightarrow 0$ as $r \rightarrow \infty$.