$T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?

Let $T:H_1\rightarrow H_2$ be a bounded linear operator where $H_1$ and $H_2$ are Hilbert spaces. The Hilbert-adjoint is defined to the the operator $T^*:H_2\rightarrow H_1$ such that $\langle Tx,y\rangle_{H_2}=\langle x,T^*y\rangle_{H_1}$ for all $x\in H_1$ and $y\in H_2$. It can be shown that $\ker T^*=\left(\operatorname{im}\, T\right)^\perp$ and $\left(\ker T^*\right)^\perp=\overline{\operatorname{im}\, T}$ (see, e.g., here). It follows that $$T \text{ surjective} \Rightarrow \operatorname{im}\, T = H_2 \Rightarrow \ker T^* = H_2^\perp = \{0\}\Rightarrow T^* \text{ injective}$$ In the finite-dimensional case, we have that $$T \text{ surjective} \iff T^* \text{ injective}$$ In the general infinite-dimensional case, does the same statement hold, or is there a counterexample where $T^*$ is injective but $T$ is not surjective?

Solution 1:

The main problem, in those things, is that linear operators in infinite dimensional spaces need not have closed image. So even if $T^\star$ is injective, meaning that $\ker T^\star=\{0\}$, we can only infer that $\overline{\text{im}\ T}=H_2$, which is not necessarily the same as $\text{im}\ T=H_2$.

For example, consider the operator $$T\colon \ell^2\to \ell^2,\qquad T\mathbf{x}=\left(x_1, \frac{x_2}{2}, \frac{x_3}{3}\ldots \right).$$ This operator is self-adjoint, so $T^\star=T$, and it is not surjective even if it is injective.