Existence of adjoint of the inverse

Question

Let $H$ be a Hilbert space over $\mathbb{F}$ and $V$ be an inner product space over $\mathbb{F}$.

Let $T:H\rightarrow V$ be a bounded linear bijection.

If $V$ is a Hilbert space, then the open mapping theorem implies that $T^{-1}$ is a bounded operator hence $T^{-1}$ has the adjoint. However, I'm curious about the case when $V$ is just an inner product space.

Question 1

Is $T^{-1}$ necessarily a bounded operator?

Question 2

Assuming $T^{-1}$ is a bounded operator, does $T^{-1}$ have the adjoint?

Answer 1

1. No. Let $H=\ell^2$, and let $V$ be the subspace of $\ell^2$ consisting of the sequences $(x_n)_{n=1}^\infty$ such that $\sum n^2 |x_n|^2<\infty$. The subspace $V$ inherits the inner product from $H$. The operator sending $(x_n)$ to $(x_n/n)$ is bounded on $H$, and maps $H$ bijectively onto $V$. But its inverse is not bounded, since $T^{-1}(e_n)=ne_n$.

2. Yes. If $T^{-1}$ is bounded, then $V$ is isomorphic to $H$ and therefore is itself a Hilbert space. The usual construction of adjoint operator applies.