If $T:X\to Y$ is a bijective bounded linear operator and $X$ is Banach then $T$ is invertible Let $X$ and $Y$ be normed linear spaces and $T:X\to Y$ a bounded linear operator.
Notation:

*

*$X'$ denote the dual space of $X$, and $T'$ the dual map of $T$.

*If $W\subset X$, then ${}^\circ W=\{f\in X'\mid f(x)=0,\forall x\in W\}$ is the annhilator of $W$

*If $Z\subset X'$, then $Z^\circ=\{x\in X\mid f(x)=0,\forall f\in Z\}$ is the annhilator of $Z$.

Then I need to prove that

*

*$\text{im}\,T$ is closed iff $\text{im}\,T={}^\circ\ker T'$

*$\ker T'=\{0\}$ iff $\text{im}\,T$ is dense in $Y$

*Let $X$ be Banach. Show that if $\inf_{x\neq 0}\frac{||Tx||}{||x||}>0$ and $\ker T'=\{0\}$ then $T$ is bijective and boundedly invertible.

I proved the first two parts, and I managed to show that $T$ is bijective, but I'm not sure how to prove that it has a bounded inverse.
 A: If $X,Y$ are Banach then the inverse of a bijective bounded operator is automatically bounded, it is a consequence of the open mapping theorem. If $X$ or $Y$ is not Banach there are counter-examples so in your case, $a$ $priori$, you cannot apply this result.
You can show by hand that $T^{-1}$ is continuous, let
$$
c = \inf_ {x \in X \setminus \{ 0 \}} \frac{||Tx||}{||x||} > 0,
$$
as $T^{-1}$ is linear it is continuous if and only if there is a constant $c' > 0$ such that
$$
\forall y \in Y,\quad ||T^{-1}y|| \leq c'||y||.
$$
But forall $y \in Y$ not null there is $x \in X$ not null such that $y = Tx$ and
$$
||T^{-1}y|| = ||x|| \leq \frac{1}{c}||Tx|| = \frac{1}{c}||y||.
$$
Actually there is another way to answer your question that is more "functionnal analysis", we can show that $Y$ is Banach. Indeed if $(y_n) = (Tx_n)$ is a Cauchy sequence of $Y$ by definition of $c$
$$
||y_{n+p} - y_n|| = ||Tx_{n+p} - Tx_n|| = ||T(x_{n+p}-x_n)|| \geq c ||x_{n+p} - x_n||
$$
so $(x_n)$ is Cauchy and converges because $X$ is Banach.
