 Solution

The lower bound in (ii) makes the image of $I+T$ closed: if $(I+T)x_n$ converges, the bound makes $(x_n)$ Cauchy; completeness gives its limit and hence a image preimage. The same bound applies to $I+T^{\ast }$ by (iii), so its kernel is zero. The orthogonal complement identityshows that the image is dense, using the orthogonal decomposition by a closed subspace for its closure. Being both closed and dense, it equals $X$. Thus $I+T$ is onto and bounded below, and by part 3(a)(iii) it has a bounded inverse. Moreover