# Example: operator injective, then the adjoint is NOT surjective

Let $T: V \rightarrow W$ be a bounded operator on normed spaces $V,W$. Now, there is a unique adjoint operator $T': W' \rightarrow V'$ defined by $T'(\alpha) = \alpha \circ T$. In finite dimensional spaces, we have that if $T$ is injective, then $T'$ is surjective. From basic Hilbert space theory, I suspect that the Banach space adjoint must not be surjective in general. Probably, $T'$ has only dense range. Since we don't know that the range of $T'$ is closed( which is trivial in the finite-dimensional case), $T'$ does not have to be onto. Does anybody know an example of this situation, where $T$ is injective, but $T'$ not surjective?

If anything is unclear, please let me know.

• Every compact injective operator will do. Compact operators cannot have closed range. – daw Jan 8 '15 at 13:33

Take the self-adjoint diagonal operator $Te_k=\frac1k e_k$ on a Hilbert space with orthonormal base $\{e_k, k\in\mathbb N\}.$ The kernel of $T$ is $\{0\}$ but $T$ is not surjective, since e.g. $\sum_k \frac1k e_k$ is not in the range.