Let us define $\|A\|_1$ the element wise norm 1 of a matrix $A \in \mathbb{R}^{n \times m}$ as $$ \|A\|_1= \sum_{i,j} |A_{i,j}|. $$ Obviously, this function is convex over $\mathbb{R}^{n \times m}$. Is it true that the function $f:S^n_{++} \longrightarrow \mathbb{R}$, defined as $f(A) = \|A^{-1}\|_1$, is convex?
Some observation:
Unfortunately matrix inversion is $S_+$ - convex while $\|.\|_1$ is non decreasing with respect to the cone $R^{n \times m}_+$ and not with respect to $S_+$ (it is easy to find counterexamples). Hence theorems on combination of convex functions are not applicable.
I have tried to formulate function $f$ as $max\{ \ trace(M_i A^{-1}) \ \}_{i\in\mathcal{I}}$ where the $M_i\in S^n$ live in a family of matrices with elements equal to 1 or -1 so as to cover all the possible combinations of signed sums of elements of $A^{-1}$ but unfortunately not all such $M_i$ are PSD and therefore not all $trace(M_i A^{-1})$ are convex in $A$. Therefore I was not able to define $f$ as the pointwise max of convex functions.
I have also tried to consider $$ \|A^{-1}\|_1= \sum_{i,j}=\frac{|\det A_{\hat{\imath}\hat{\jmath}}|}{detA} $$ where $\det A_{\hat{\imath}\hat{\jmath}}$ is the minor associated to the $n-1 \times n-1$ sub-matrix obtained by eliminating row $i$ and column $j$ from $A$. But I have no intuition on how to go further...
Any thought?