I - T not isomorphism in non-Banach normed spaces

Question

So I know that if $T : X \to X$ is a bounded operator with $X$ a Banach space and that $\|T\| < 1$, then $I - T$ is an isomorphism. Is this true in general normed spaces?

Answer 1

For example, let $X$ be the polynomials with the norm $$ \|f\| = \sup_{x \in [0,1/2]} |f(x)|$$ and $$T(f)(x) = \int_0^x f(t)\; dt$$
It is easy to see that $\|T\| \le 1/2$. But $I-T$ is not an isomorphism, e.g. there is no $f$ such that $f - Tf = 1$ (as you see by differentiating the equation $f(x) - \int_0^x f(t)\; dt = 1$, leading to a differential equation initial value problem whose solution is not a polynomial).