Closed complement to kernel implies closed range? Is there any reason this would not work in $l_\infty$? 
If $T:l_\infty \to l_\infty$ has complemented kernel and its complement is closed, then $Im T$ is closed. 
My idea here: 
If $y^{(n)}\in Im T$ and it converges to $y$, then there exist $x^{(n)}$ such that $T(x^{(n)}) = y^{(n)}$ for all $\mathbb{N}$. If $y^{(n)} = 0$, the proof is trivial, otherwise $x^{(n)} \in Y$ (the complement to $kerT$). If $x^{(n)}$ converges to $x$, then $x \in Y$ and so by uniqueness of limit $y = T(x)$. However, does $x^{(n)}$ necessarily converge? 
Thank you for any help.
 A: If the claim were true then any injective operator $T$ would have closed range, since 0 is certainly a complemented subspace with closed complement.  But that's false.
Consider for instance the operator $T : l^\infty \to l^\infty$ defined by $(Tx)_k = \frac{1}{k^2} x_k$, which is injective.  Clearly $c_{00}$, the space of sequences with only finitely many nonzero terms, is contained in the range of $T$.  The closure of $c_{00}$ is $c_0$, the space of sequences that converge to 0.  In particular, the sequence $y$ defined by $y_k = 1/k$ is in $c_0$.  But $y$ is not in the range of $T$, since if $Tx=y$, we would have to have $x_k = k$ and that is not an element of $l^\infty$.
Note that the problem with your proposed argument is exactly what you identified: if you let $y^{(n)}_k = 1/k$ for $k \le n$ and $0$ otherwise, then $y^{(n)} \to y$ in $l^\infty$, and we have $Tx^{(n)} = y^{(n)}$ where $x^{(n)}_k = k$ for $k \le n$ and $0$ otherwise.  But $\{x^{(n)}\}$ does not converge in $l^\infty$.
