Let $X,Y$ be Banach spaces and $T \in \mathcal{L}^*(X,Y)$ bijective show that $$R(T)=N(T^{*})^{\perp}\Leftrightarrow \inf_{||x||=1} ||T(x)||>0$$
I try to use these facts:
$T^{*}$ is injective iff $R(T)$ is dense
And
$R(T)=N(T^{*})^{\perp}\Leftrightarrow$ $R(T)$ is closed
But I don't sure to these, any suggestion or help I will very grateful
If $R(T)$ is closed then Open Mapping Theorem shows that $T^{-1}$ is a bounded operator. Hence, $1=\|x\|=\|T^{-1}(Tx)\| \leq \|T^{-1}\| \|Tx\|$ whenever $\|x\|=1$. This proves that $\inf_{||x||=1} ||T(x)||>0$.
Conversely, if $c\equiv\inf_{||x||=1} ||T(x)||>0$ then $R(T)$ is closed: Suppose $(Tx_n)$ is a sequence in $R(T)$ converging to some $y$. Note that $\|Tx\| \geq c$ whenver $\|x\|=1$. This implies (by replacing $x$ by $\frac x {\|x\|}$ ) that $\|Tx \| \geq c \|x\|$ for all $x$. Hence, $\|T(x_n-x_m)\| \geq c \|x_n-x_m\|$. This proves that $(x_n)$ is Cauchy, hence convergent. If $x_n \to x$ the $y=\lim T(x)n)=Tx$.