Suppose that we have Banach spaces $X$ and $Y$ and a bounded linear operator $T\colon X\to Y$ that maps a dense subspace $D_{X}\subset X$ bijectively into a dense subspace $D_{Y}\subset Y$. Can we conclude that $T$ is an isomorphism?
Here is what I tried for proving surjectivity: Pick any $y\in Y$. Then there is a sequence $D_{Y}\ni y_{n}\to y$. By assumption there is a (unique) $x_{n}\in D_{X}$ such that $y_{n}=T(x_{n})$. Now I tried to prove that $x_{n}$ is Cauchy, so that it has a limit $x\in X$, and then I was hoping that $y=T(x)$. But I cannot find an estimation of the form $\|x_{i}-x_{j}\|\leq C\|T(x_{i})-T(x_{j})\|$.
Any help would be greatly appreciated!
Best Answer
Not even true when $X=Y$ is a Hilbert space. Any injective compact operator on an infinite-dimensional Hilbert space is an example.
For instance, take $X=Y=\ell^2(\mathbb N)$ and $$Tx=\big(x_1,\tfrac{x_2}2,\tfrac{x_3}3,\ldots\big),$$ on the dense subspace $$c_{00}=\{x\in X:\ \text{ finite support }\}.$$ The necessary and sufficient condition for $T$ to extend to an invertible operator on the full space is that $T^{-1}$ is bounded.