For all $f\in C[0,1]$ and $x\in[0,1]$ we define bounded linear operator $$(Tf)(x)=f(0)+\int_0^x f(t)dt,$$ with norm $||T||=2$ (right?). I have already determined its image $imT:=\{g\in C^1[0,1]| g'(0)=g(0)\} $. Question: Is $imT$ Banach space with respect to the norm $||.||_\infty$? Prove $imT$ is infinite dimensional. Find all eigenvalues of $T$.
If $T$ is bounded from below (i.e. $\exists M.||Tx||\geq M||x|| \quad \forall x$), then $imT$ is closed. (When is the image of a linear operator closed?)
Is $T$ any different than Volterra operator: Volterra Operator is compact but has no eigenvalue
For eigenvalues I suspect it has none.
