On the convexity of the element-wise norm 1 of a pseudoinverse

Question

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 $\mathbb{R}^{n \times m}$. Let us now consider $S^n_+ - \{0\}$ the set of non-zero positive semidefinite $n \times n$ matrices. Let us denote $A^{\dagger}$ the pseudoinverse of a matrix $A \in S^n_+$. Is it true that the function $f:S^n_+ - \{0\} \longrightarrow \mathbb{R}$, defined as $f(A) = \|A^{\dagger}\|_1$, is convex?

Answer 1

Let $A(t)=\begin{pmatrix}1&0 \\ 0& t\end{pmatrix}$ where $t\ge 0$. Then $A(t)^\dagger=\begin{pmatrix}1&0 \\ 0& t^{-1}\end{pmatrix}$ if $t>0$ and $A(0)^\dagger=\begin{pmatrix}1&0 \\ 0& 0\end{pmatrix}$. So the function $\|A(t)\|_1$ is not even continuous, let alone convex.

Generally, you should not expect anything good from pseudoinverse as a function in the regions where the matrix changes its rank.