I am trying to show counter examples to the Open Mapping Theorem. In this particular case, I am trying to show that both spaces need to be Banach.
First the OMT:
Let $X, Y$ be Banach spaces. Let $T : X \rightarrow Y$ be a surjective
bounded linear operator. $T$ is an open mapping.
Now, for the counterexample, let $X = l^1$ equipped with $||\cdot||_1$. Let $Y = l^1$ equipped with $||\cdot||_\infty$. Claim: Y is not complete. Let $T: X \rightarrow Y$ be the identity operator defined by $Tx = x$. Clearly T is bijection and is bounded (thus continuous). Claim: T is not an open mapping.
Basically, the main question is that $X$ is Banach, $Y$ is not and I want to show that the operator is not an open mapping.
I am also having a hard time proving that $Y$ is not complete (I know its not, I just cant prove it).
Best Answer
For a counterexample with incomplete $X$ you can use e.g. a Hamel basis of $\ell^1$ to produce a discontinuous linear functional $f:\ell^1\to\mathbb R$. Then define a norm on $X=\ell^1$ by $\|x\|_f=\|x\|_1+|f(x)|$. The identity $(\ell^1,\|\cdot\|_f)\to (\ell^1,\|\cdot\|_1)$ is continuous and surjective but not open since otherwise $f$ would be continuous with respect to $\|\cdot\|_1$.