injective bounded operator with a dense image and a bounded inverse

Question

I am wondering whether such an operator can exist.

It is a bounded linear operator defined on a Hilbert space. It is injective. It is not surjective (its image is not the whole of the Hilbert space). Its inverse defined on the image is bounded.

I am asking because in Lax's book 'functional analysis', chapter 20, he states that 'by definition, a bounded linear operator $M$ is invertible iff it maps H onto H in a one-to-one fashion', and that $M$ can fail to be invertible only in two ways: (i) $M$ is not 1-1; (ii) $M$ is not onto.

The problem is, while for $M$ to be invertible, of course it should be one-to-one, but why should it also be onto?

Answer 1

If $T$ has a bounded inverse on the range of $T$ then there exist $C \in (0,\infty)$ such that $\|x\|\leq C\|Tx\|$ for all $T$. If $Tx_n \to y$ then $(Tx_n)$ is Cauchy which implies $\|x_n-x_m\|\leq C\|T(x_n-x_m)\| \to 0$. So $x_n $ converges to some $x$ and $y=\lim Tx_n=Tx$. Hence, the range of $T$ is necessarily closed in $H$. Since it is also dense $T$ must be surjective.

Answer 2

Let $\{e_n\}_{n\in\mathbb N}$ be an orthonormal basis of the Hilbert space $H$.

Define an operator $T:H\to H$, as $Te_n=e_{n+1}$, and thus $$ T\Big(\sum_{k=1}^\infty a_ke_k\Big)=\sum_{k=1}^\infty a_ke_{k+1}. $$ Clearly $\|Tx\|=\|x\|$, for all $x\in H$, and $$ \mathrm{Ran}(T)=\{e_1\}^\perp=\{x\in H: \langle x,e_1\rangle=0\}. $$ Now $\{e_1\}^\perp$ is a closed subspace of $H$, and hence a Hilbert space, and $T: H\to \{e_1\}^\perp$ is not just a bounded operator, but an isometric isomorphism, and $T^{-1}$ is hence bounded with $\|T^{-1}\|=1$.