Consider the operator norm of the following $n \times n$ matrix:
$$ \|(I + 11^* + X^*X)^{-1} 11^* (I + 11^* + X^*X)^{-1}\|, $$ where X is a $n \times n$ matrix.
Is it bounded by $\frac{1}{n}$?
I know it is true without the matrix $X^*X$, but how to deal with the matrix $X^*X$?
Here is a reformulation of the problem. Let $P=I+XX^\ast$ and $y=P^{-1}e$. By Sherman-Morrison formula, $$ x:=(I+XX^\ast+ee^T)^{-1}e =(I+ye^T)^{-1}y =\left(I-\frac{ye^T}{1+e^Ty}\right)y =\frac{y}{1+e^Ty}=\frac{P^{-1}e}{1+e^TP^{-1}e}. $$ By a change of orthonormal basis, the expression on the RHS can be rewritten as $$ \frac{Dv}{1+v^\ast Dv} $$ where $\|v\|=\|e\|=\sqrt{n}$ and $D=\operatorname{diag}(d_1,\ldots,d_n)$ with $0<d_i\le1$ for each $i$ (the $d_i$s are the eigenvalues of $P^{-1}$). The problem thus reduces to showing that $$ \|xx^\ast\|=\|x\|^2=\left\|\frac{Dv}{1+v^\ast Dv}\right\|^2 =\frac{\sum_id_i^2|v_i|^2}{\left(1+\sum_id_i|v_i|^2\right)^2} $$ is bounded above by $\frac1n$ when $\|v\|=\sqrt{n}$ and $0<d_i\le1$ for each $i$.