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I saw this statements in a proof, but I would like to see the details.

Let $U \subset \mathbb{R}^{n}$ and $u$ sperharmonic function in the sense that \begin{equation} \int_{D} \langle \nabla u(x) ,\nabla \varphi(x) \rangle dx \quad \mbox{for any nonnegative} \quad \varphi \in C^{\infty}_{0}(U). \end{equation}

  1. Is $u$ lower semicontinuous?

  2. if $u$ is lower semicontinuous and $\varphi \in C^{\infty}_{0}(U)$. Is $\{u > \varphi\}$ an open set?

Thank you.

  • 1
    2 Is easy: $v=u-\varphi $ is lsc , hence v>0 is open. To prove 1, consider convolution of $u$ with a radially symmetric mollifier and argue that it (a) is superharmonic, and (b) converges to $u$ in an increasing manner, as the radius of mollification tends to zero.2012-07-17

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