Let $H$ be a Hilbert space and $T:H\rightarrow H$ be a injective, bounded linear operator. Suppose the range of $T$ is contained inside a closed subspace of $H$, is that closed subspace $H$ itself?
It seems to me that this must be true. Since $T$ is injective, it maps every vector to a unique vector, and the range is not dense if the closed subspace is proper. But it seems so good to be true.
Best Answer
On $\ell^{2}$ define $T(x_n)=(0,x_1,x_2,....)$. Then $T$ is bounded, injective and its range is $\{(x_n): x_1=0\}$ which is a proper closed subsapce.