Converse of Bounded Inverse Theorem

Question

Are there examples of a bijective bounded linear operator $T\colon X\longrightarrow Y$ where $X$ is a Banach space but not $Y$, yet $T$ has a bounded inverse?

Answer 1

Since the condition of boundedness of $T$ was added to the question, this makes my old answer irrelevant. It can still be found below.

It is not possible to find such a $T$ because it would be an isomorphism of normed spaces which preserves properties like completeness:

If $(y_n)_n$ is a Cauchy sequence in $Y$, then $(T^{-1}y_n)_n$ is a Cauchy sequence in $X$ because $T^{-1}$ is bounded, thus the sequence $(T^{-1}y_n)_n$ converges in $X$. But then, by boundedness of $T$, the sequence $(y_n)_n = (TT^{-1}y_n)_n$ converges in $Y$, hence $Y$ is complete.

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Let $(X, \|\cdot\|_1)$ be an arbitrary infinite dimensional Banach space. Then choose an arbitrary discontinuous functional $\varphi$ on $X$.¹ Define $\|x\|_2 := \|x\|_1 + |\varphi(x)|$. Then $\|\cdot\|_2$ is a norm on $X$ with the property $\|\cdot\|_1\leq \|\cdot\|_2$, hence for $T:(X,\|\cdot\|_1)\to (X,\|\cdot\|_2),x\mapsto x$, the inverse $T^{-1}$ is bounded. $(X,\|\cdot\|_2)$ is not a Banach space because this would imply $|\varphi(x)|+\|x\|\leq C\|x\|$ for some $C>0$ and all $x\in X$ by the bounded inverse theorem. This implies $|\varphi(x)|\leq (C-1)\|x\|$ however, making $\varphi$ bounded which contradicts the choice of $\varphi$.

¹If $(b_i)_{i\in I}$ is a base of $X$ with $\|b_i\|_1 = 1$ for all $i$ and w.l.o.g. $\mathbb N\subset I$. Then set $\varphi(b_i) = i$ if $i\in\mathbb N$ and $\varphi(b_i)=0$ otherwise.