Bounded inverse theorem when the domain is not complete The bounded inverse theorem states that a bijective bounded linear operator $T$ from a Banach space $X$ to a Banach space $Y$ has a bounded linear inverse $T^{-1}$.
I want to find a constructive counterexample to this when the domain $X$ is not complete but all other conditions are satisfied.
In particular, I want a non-complete normed linear space $X$, a Banach space $Y$, and a bijective bounded linear opeartor $T:X\to Y$ such that $T^{-1}$ is not bounded.
I found a counterexample when the codomain $Y$ is not complete or when the operator is not bijective,
but I want to maintain that $Y$ is complete and $T$ is bijective. I think there are hints in this post or this post, but in either, I could not come up with a specific constructive choice of $g$ or $\varphi$ that makes only $X$ non-complete but maintains all other conditions. (For the latter post, the constructive map $\varphi$ is not defined for non-basis elements of $X$.)
 A: Let $X$ be a VS with a Hamel basis with the cardinality of the continuum, let $\{ e_r\mid r\in\Bbb (0,1)\}$ be such a basis and give $X$ the norm
$$\left\|\sum_{r} x_r \ e_r\right\|_X:=\sum_r|x_r|$$
(remember that only finitely many terms in that sum are not zero). Now let $Y$ be an infinite dimensional Banach space of cardinality $|\Bbb R|$, like $\ell^1(\Bbb N)$. Then any Hamel basis of $Y$ has cardinality $|\Bbb R|$, suppose $\{ b_r \mid r\in(0,1)\}$ is such a basis and wlog assume $\|b_r\|_Y=1$, else replace $b_r$ by $b_r/\|b_r\|$.
Now the map
$$B:X\to Y, \qquad \sum_r x_r\ e_r\mapsto \sum_r r\,x_r \ b_r$$
is continuous because
$$\|B(\sum_r x_r \ e_r)\|_Y ≤\sum_r r |x_r| \ \|b_r\|≤ \sum_r |x_r| = \left\|\sum_r x_r \ e_r\right\|_X,$$
ie it is a contraction. Further it is a linear bijection by construction. However the inverse cannot be bounded simply because $\|B^{-1}( b_r/\|b_r\|) \| = \frac1r$ with $r$ ranging over $(0,1)$, hence the image of the unit ball under $B^{-1}$ is unbounded.
