Does my operator $C([0,1]) → C([0,1])$ have an inverse? Consider the Banach space $X := C_\mathbb{C}([0,1])$ with the supremum norm. Consider the bounded linear operator $T : X → X$ given by
$$ (Tf)(t) = \int_0^t 2sf(s)ds + f(1).$$
I want to know whether $4I - T$ (with $I$ the identity operator) has an inverse in $B(X)$. I know this is the case whenever $\lVert(4I-T) - I\rVert = \lVert3I-T\rVert < 1$. Indeed, the Neumann series $\sum_{n=0}^∞ (T - 3I)^n$ would then be a candidate, and by uniqueness, the inverse of $4I-T$. Now if I perform the calculation, I end up with, if I have interpreted all the norms in the right ways,
$$\lVert 3I - T\rVert = \sup_{\lVert f \rVert ≤ 1} \sup_{t ∈ [0,1]} \left\{\left|3f(t) -2\int_0^t 2sf(s)ds + f(1) \right|\right\} ≤ \left|3 - \int_0^1 2sds + 1\right|≤ 5.
$$
Now, although this is not less than $1$, why could I not simply multiply by, say, ⅙, obtain an operator with norm ⅚ $< 1$, and later scale up by $6$ again, which should not matter much in the vector space setting we are in.
Perhaps I'm approaching this the complete wrong way, though.
EDIT: For a further question (whether $T$ is compact), I know that, since $[0, 1]$ is compact, $\overline{\mathcal{B}_{00}(C_{\mathbb{C}}([0,1]))} = \mathcal{B}_0(C_{\mathbb{C}}([0,1]))$. That is, if $T$ is compact, we can approximate $T$ by a sequence of finite-rank operators. One easy way to define a sequence that converges to $T$ is to define, for every $n ∈ ℕ$, the operators $T_n$ given by
$$ (T_nf)(t) = \int_0^{t(1-\frac{1}{n})}2sf(s)ds + f(1),$$
but I don't know whether these are of finite rank, and I generally have trouble with such arguments in infinite-dimensional vector spaces..
 A: Notice that
$$|Tf(t)|\leq\int^t_0|f(s)|2s\,ds+|f(1)|\leq 2\|f\|_\infty$$
and taking $f(t)\equiv1$, we conclude that $\|T\|=2$. Then for all $\lambda\in\mathbb{C}$ with $|\lambda|>2$ is $\lambda I-T=\lambda\big(I-\tfrac1\lambda T\big)$ invertible. This answers the OP completely.

One can further question for which other values of $\lambda\in\mathbb{C}$ is $(\lambda I -T)$ invertible. Notice that
$T$ is a compact operator as $f\mapsto \int^t_02s f(s)\,ds$ and $f\mapsto f(1)$ are compact operator on $C([0,1])$. Thus, $0I-T=-T$ is not invertible (it can be seen that $T$ is injective, however).
Suppose $\lambda\neq0$. Fix any polynomial $p(t)=a_0+\ldots +a_nt^n$. If $f$ solves equation
$$\begin{align}
\lambda f(t)-Tf(t)=\lambda f(t)-\int^t_02sf(s)\,ds- f(1)=p(t)\tag{0}\label{zero}
\end{align}$$
then $f$ is differentiable, and
$$\begin{align}
\lambda f(0)-f(1)=a_0\tag{1}\label{one}
\end{align}$$ Consequently, $f$ solves
$$\begin{align}
\lambda f'(t)-2tf(t)= p'(t)\tag{2}\label{two}
\end{align}$$
Solving equation \eqref{two} we obtain that
$$\begin{align}
\frac{d}{dt}e^{-t^2/\lambda}f(t)=\frac1\lambda e^{-t^2/\lambda}p'(t)\\
\end{align}$$
$$\begin{align}
f(1)e^{-1/\lambda}&-e^{-t^2/\lambda}f(t)=\frac1\lambda\int^1_te^{-s^2/\lambda}p'(s)\,ds
\end{align}$$
and so, after integration by parts we obtain
$$\begin{align}
f(t)&=e^{t^2/\lambda}\Big(f(1)e^{-1/\lambda}-\frac1\lambda\Big(e^{-\tfrac1\lambda}p(1)-e^{-\tfrac{t^2}{\lambda}}p(t)+\\
&\qquad\qquad\frac1\lambda\int^1_t2se^{-\tfrac{s^2}{\lambda}}p(s)\,ds\Big)\Big)\tag{3}\label{three}
\end{align}$$
and
$$\begin{align}
f(1)e^{-1/\lambda}- f(0)&=\frac1\lambda\int^1_0e^{-\tfrac{s^2}{\lambda}}p'(s)\,ds\tag{1'}\label{onep}
\end{align}
$$
As long as $\lambda\neq0$ and
$$\begin{align}
\operatorname{det}\begin{pmatrix}\lambda & -1\\-1 & e^{-1/\lambda}\end{pmatrix}=\lambda e^{-\tfrac1\lambda}-1\neq0\tag{4}\label{four}
\end{align}
$$
conditions \eqref{one} and \eqref{onep} are compatible.
Putting things together, we have that
\begin{align}
\lambda f(t)-\lambda f(0)-\int^t_02s f(s)\,ds=p(t)-a_0
\end{align}
and so,
$$(\lambda I-T)f(t) + f(1)-\lambda f(0)=p(t)-a_0$$
i.e.,  $(\lambda I-T)f=p$.
When $p\equiv0$, once can see that $f(1)=0=f(0)$ and so $f(t)\equiv0$. This means that $\lambda I-T$ is injective.
We have shown that if $\lambda\neq0$ satisfies \eqref{four}, then $(\lambda I-T)$ is injective and has dense range. As $T$ is compact, range of $(\lambda I-T)$ is closed; hence, by the Open Mapping Theorem, $\lambda I-T$ has a bounded inverse. Using \eqref{three} and solving the linear system \eqref{one} and \eqref{onep}, we can in fact derive an explicit expression form $(\lambda I-T)^{-1}$.
A: First off, the question in the title -if any operator has an inverse - of course not every operator has inverse. For example, consider operators which have finite dimensional image (or, more general compact operators).
Now, regarding $T$. One easily checks that $||T|| \leq 2$. Hence, $||\frac{1}{4}T|| \leq \frac{1}{2}$. It is a general fact that if $||A|| < 1$ then $I-A$ has inverse.
One just writes formula for it (which is analogous to number like formula) that is
$$(I-A)^{-1} = \sum A^{k}$$. The series one the right side converges because of assumption on $||A||$.
