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A barrier differential inequality implies a distributional differential inequality

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I encountered the following statement while reading an article about Ricci flow:

Thus $\Delta d_{K, p}^2\leq4$ in the barrier sense, hence in the viscosity sense and in the distributional sense.

I couldn't find the definition for viscosity sense while searching for references. The definitions of barrier sense and distributional sense are as follows:

Given $f\in C(M)$ and a function $g$ on $M$, one says that $\Delta f \leq g$in the sense of barriers if for all $q\in M$ and $\epsilon > 0$, there is a neighborhood $V$ of $q$ and some $f_\epsilon \in C^2(V)$ so that $f_\epsilon(q) = f(q)$, $f_\epsilon \geq f$ on $V$ and $\Delta f_\epsilon \leq g+\epsilon$ on $V$.

The other sense of a differential inequality is the distributional sense, i.e. $\Delta f ≤ g$ if for every nonnegative compactly-supported smooth function $\phi$ on $M$,$$\int_M(\Delta \phi)f~dV \leq \int_M \phi g~dV.$$

I tried to directly use the definition of barrier sense to obtain distributional sense. But I found I need a uniform convergence:$$\Delta f\leq g {\rm ~in~the~barrier~sense}\implies\Delta f_\epsilon\leq g+\epsilon$$$$\implies\int_M(\Delta \phi)f_\epsilon~dV \leq \int_M \phi (g+\epsilon)~dV\overset{\rm uniform~convergence}\implies\int_M(\Delta \phi)f~dV \leq \int_M \phi g~dV$$$$\implies\Delta f\leq g {\rm ~in~the~distributional~sense}.$$Then I don't know what to do else. Any help would be appreciated!


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