Let $N\in\mathbb{N}$ and $P$ be a symmetric pattern; i.e. $P$ is a subset of $\{1,\dots,N\}\times\{1,\dots,N\}$ such that $(i,i)\in P$, for all $i\in\{1,\dots,N\}$, and $(i,j)\in P$ if and only if $(j,i)\in P$. Moreover, let $\mathcal{H}_P$ be the space of $N\times N$ Hermitian matrices with sparsity pattern $P$; i.e. if $H_N$ is the space of $N\times N$ Hermitian matrices, then $$\mathcal{H}_P=\{ A=[a_{ij}]_{1\leq i,j\leq N}\in H_N: \ a_{ij}=0, \ \forall(i,j)\notin P \}.$$ By endowing $H_N$ with the inner product $\left<A,B\right>=\mathrm{Tr}(AB)$, it can be verified that $$\mathcal{H}_P^\perp=\{ A=[a_{ij}]_{1\leq i,j\leq N}\in H_N: \ a_{ij}=0, \ \forall(i,j)\in P \}.$$ Now define the cone $$\mathcal{C}=\{A\in\mathcal{H}_P: \ \exists B\in\mathcal{H}_P^\perp \ \text{s.t. } A+B\geq0\},$$ where $A+B\geq0$ means that the matrix $A+B$ is positive semidefinite. I want to show that the cone $\mathcal{C}$ is closed. If $\{A_n\}_{n\in\mathbb{N}}\subseteq\mathcal{C}$ is a sequence that converges to $A$, then it can be checked indeed that $A$ is a Hermitian matrix of sparsity pattern $P$. But I am struggling a bit to find an appropriate $B\in\mathcal{H}_P^\perp$ such that $A+B\geq0$. Obviously, there also exists a sequence $\{B_n\}_{n\in\mathbb{N}}\subseteq\mathcal{H}_P^\perp$ such that $A_n+B_n\geq0$, for all $n\in\mathbb{N}$, but I don't see how to argue further. A couple of ideas:
- If $\{B_n\}_{n\in\mathbb{N}}$ is (uniformly) bounded, then it has a convergent subsequence so, we are done. But I don't know how to prove the boundedness.
- By contradiction. Assume that for every $B\in\mathcal{H}_P^\perp$, there exists $x_B\in\mathbb{C}^n$ such that $x_B^*(A+B)x_B<0$. But I don't really know how to relate this with the sequence $\{B_n\}_{n\in\mathbb{N}}$.
Any help is more than welcome.