TITLE: On the convexity of element-wise norm 1 of the inverse
QUESTION [8 upvotes]: Question first asked on math.stackexchange here: https://math.stackexchange.com/questions/317209/on-the-convexity-of-element-wise-norm-1-of-the-inverse
On the convexity of element-wise norm 1 of the inverse
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? Here we denote with $S^n \supset S^n_+ \supset S^n_{++}$ respectively the set of Symmetric, Positive Semidefinite (PSD) and Positive Definite (PD) $n \times n$ matrices.
Some observations:
Unfortunately matrix inversion is convex with respect to $S_+$, 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?

REPLY [3 votes]: The function that you have is convex for unitarily invariant norms, but for the (basis dependent) elementwise absolute value, it can clearly break as a trivial counterexample below shows.
\begin{equation*}
 X = \begin{pmatrix}
 2 & 0 & 0\\\\
 0 & 1 & 0\\\\
 0 & 0 & 1
\end{pmatrix},\qquad Y = \begin{pmatrix}
 10  &    9 &    5\\\\
     9 &   10 &    5\\\\
     5 &    5 &    4
\end{pmatrix}
\end{equation*}
Now, simply define $Z = 0.5(X+Y)$, and consider $g(H)=\|H^{-1}\|_1$ as your function. 
Then, we see that $g(Z) = 3.1692$, while $0.5(g(X)+g(Y)) = 3$, clearly violating convexity.