Norm of an operator on an infinite dimensional space Let $C[0,1]$ denote the space of all  real-valued continuous functions on $[0,1]$
equipped with the supremum norm $\left\lVert . \right\rVert _\infty$. Let $ T:C[0,1]\to C[0,1]$ be the linear operator defined by $T(f)(x)=\int\limits_0^x{e^{-y}f(y)}dy$.Then which of the following is/are true?
(A) $\, \left\lVert T \right\rVert =1$
(B) $\, I-T$ is invertible 
(C) $ \,T$ is surjective 
(D) $ \,\left\lVert I+T \right\rVert = 1+\left\lVert T \right\rVert$
Since f is a continuous function on a compact interval it is bounded. So I am able to conclude that the norm of T is bounded. More than that I am clueless. Please help me. 
 A: (a) No, the norm is not equal to $1$. Observe that for any $x\in[0,1]$ it is $|T(f)(x)|\leq\|f\|_\infty\cdot|\int_0^xe^{-y}dy|=(1-e^{-x})\|f\|_\infty$, so we have $\sup_{x\in[0,1]}|T(f)(x)|\leq(1-e^{-1})\|f\|_\infty$, i.e. $\|T(f)\|_\infty\leq(1-e^{-1})\|f\|_\infty$, thus $\|T\|\leq(1-e^{-1})<1$.
(b) Yes, $I-T$ is invertible. To see this note that since $\|T\|<1$, the series $\sum_{n=0}^\infty\|T\|^n$ converges (a geometric series). Also, it is well-known (and easy to prove) that $\|T^n\|\leq\|T\|^n$. Thus the series $\sum_{n=0}^\infty\|T^n\|$ converges. The operators over $C[0,1]$ form a Banach space, so the absolutely convergent series are convergent, thus the series $\sum_{n=0}^\infty T^n$ converges to a bounded linear operator $S$. I leave it up to you to verify that $S$ is the inverse of $(I-T)$ (multiply the partial sums of $S$ with $(I-T)$, see what you get and take limits).
(c) No, it is not surjective as you said: the constant function $1$ has no preimage. If $f\in C[0,1]$ such that $\int_0^xe^{-y}f(y)dy=1$ for all $x$, then, differentiating, we get $e^{-x}f(x)=0$ for all $x$, thus $f=0$, but then $\int_0^xe^{-y}f(y)dy=0$, a contradiction.
(d) It is obvious that $\|I+T\|\leq 1+\|T\|$, but that's as far as I can go. I think one strategy would be trying to compute the norm over polynomials, since they are uniformly-dense in $C[0,1]$. We can compute $T(x\mapsto x^n)$ with integration by parts but I'm not sure about this.
