Suppose $A\in R^{n\times n}$ and $B\in R^{m\times m}$ are symmetric positive definite matrices. Is the following set convex and compact?
$\mathcal{S}^{+}=\left\{C\in R^{n\times m}\Big|\begin{bmatrix}A&C\\C^T&B\end{bmatrix}>0\right\}$
How about the following set:
$\mathcal{S}=\left\{C\in R^{n\times m}\Big|\begin{bmatrix}A&C\\C^T&B\end{bmatrix}\geq 0\right\}$
Is $\mathcal{S}$ homeomorphic to a closed ball in $\mathbb{R}^{p}$ for an arbitrary norm, where $p=n\times m$.
On
Both of them are bounded. Suppose the opposite is true. Because of the equivalence of norms, we look at the Frobenius norm. As $m$ and $n$ are finite, by the pigeonhole principle, there exists at least one entry, say $(i,j)$, of $C$ the set of the values of which is unbounded. The $2\times 2$ principle minor $\begin{bmatrix}A_{i,i} & C_{i,j} \\ C_{i,j} & B_{j,j} \end{bmatrix}$ of the original large matrix should be positive (semi-)definite, which stipulates that $C_{i,j}^2\le A_{i,i}B_{j,j}$. We have a contradiction.
$\mathcal S$ is closed, so it is compact. $\mathcal S^+$ is not closed as $C$ can reduce the rank of the original large matrix by continuously changing it value, so it is not compact.
As mentioned by Eric Brown, there is another way to prove the boundedness of $\mathcal S$ and $\mathcal S^+$.
The original large matrix is symmetric positive (semi-)definite iff $$\begin{bmatrix}A^{-\frac12} & 0 \\ 0 & B^{-\frac12}\end{bmatrix}\begin{bmatrix}A & C \\ C^T & B\end{bmatrix}\begin{bmatrix}A^{-\frac12} & 0 \\ 0 & B^{-\frac12}\end{bmatrix}=\begin{bmatrix}I & P:=A^{-\frac12}CB^{-\frac12} \\ P^T=B^{-\frac12}CA^{-\frac12} & I\end{bmatrix}$$ is also iff its Schur complement $F:=I-P^TP\succ(\succeq)0$. Diagonalization leads to the spectral norm of $\|P\|<(\le)1$. This is actually a generalization and stronger version of the enry-wise inequality derived in my previous answer $A_{i,i}^{-1}C_{i,j}^2B_{j,j}^{-1}<(\le)1$. As a corollary, the diagonal entries of $F$ should be positive (nonnegative), thus each column 2-norm of $P$ or $\sqrt{\sum_jP_{i,j}^2}<(\le)1$, so is each individual entry $P_{i,j}$ of $P$.
$\mathcal S$ is closed, so it is compact. $\mathcal S^+$ is not closed as $C$ can reduce the rank of the original large matrix by continuously changing it value, so it is not compact.