Is it true if $\|T\|>1$, where $T \in B(X)$ for some Banach space $X$, then $T^{-1}$ exists?
I suppose that for $\|T\|=1$ this isn't true? Because, if we suppose that inverse exists for such operators, we can look at some operator $A$ which doesn't have inverse and then neither $\frac{A}{\|A\|}$ has inverse, contradiction.
I know that if $\|T\|<1$ then $(I-T)^{-1}$ exists and it is equal to $\sum_{n=0}^{\infty} T^n$, but I don't think we can use that fact.
On
It is not true, and it is easy to construct counterexamples. For instance, take $X = \Bbb R^2$ with the usual norm $\Vert y \Vert = \langle y, y \rangle^{1/2}$, where $\langle \cdot, \cdot \rangle$ denotes the standard real inner product, viz. $\langle (w_1, w_2), (v_1, v_2) \rangle = w_1v_1 + w_2 v_2$; let $T$ be the matrix
$T = \begin{bmatrix} a & 0 \\ 0 & 0 \end{bmatrix}, \tag{1}$
where $\vert a \vert > 1$. Then
$Ty = T(y_1, y_2)^T = (ay_1, 0)^T, \tag{2}$
whence
$\Vert Ty \Vert^2 = a^2 y_1^2; \tag{3}$
taking $y = (y_1, 0)^T$ yields
$\Vert T y \Vert^2 = a^2y_1^2 = a^2 \Vert y \Vert^2= \vert a \vert^2 \Vert y \Vert^2, \tag{4}$
whence
$\Vert Ty \Vert = \vert a \vert \Vert y \Vert > \Vert y \Vert, \tag{5}$
which implies we must have
$\Vert T \Vert \ge \vert a \vert > 1, \tag{6}$
but inspecting (1), we see that
$\det(T) = 0, \tag{7}$
so $T$ is not invertible.
This example can obviously be generalized.
Invertible and nonivertible operators may have arbitrary norms: just note that $\|cT\|=|c|\|T\|$ and $T$ is ivertible if and only if $cT$ is invertible (provided $c \neq 0$)