Erwin Kreyszig's Introductory Functional Analysis With Applications, Section 2.7, Problem 9 Here is Problem 9 in the Problem Set following Section 2.7 in the book Introductory Functional Analysis With Applications by Erwine Kryszeg: 
Let $C[0,1]$ denote the set of all (real- or complex-valued) functions defined and continuous on the closed interval $[0,1]$ with the norm defined as follows: 
$$\Vert x \Vert_{C[0,1]} \colon= \max_{t \in [0,1]} \vert x(t) \vert \, \, \,  \forall x \in C[0,1]. $$ 
Let $T \colon C[0,1] \to C[0,1]$ be defined by 
$$Tx \colon= \int_{0}^t x(\tau) \, d \tau \, \, \,  \forall x \in C[0,1].$$ 
What is the range of $T$? 
Is $T$ injective? And if so, then what is the inverse $T^{-1}$? Is $T^{-1}$  bounded? 
I know that $T$ is bounded, for, given any $x \in C[0,1]$, we have 
$$ \Vert Tx \Vert_{C[0,1]} = \max_{t\in[0,1]} \vert \int_0^t x(\tau) \, d \tau \vert \leq \max_{t\in[0,1]} (\int_0^t \vert x(\tau) \vert \, d \tau ) = \int_0^1 \vert x(\tau) \vert \, d \tau  \leq \int_0^1 \max_{\tau\in[0,1]} \vert x(\tau) \vert \, d \tau = \int_0^1 \Vert x \Vert_{C[0,1]} \, d \tau = \Vert x \Vert_{C[0,1]}.$$ 
 A: Let $g$ be an element of the range of $T$.  Then there exists $f\in C[0,1]$ such that $g=Tf$.   As jflipp commented, this implies that for $t\in[0,1]$, $g'(t)=\frac{d}{dt}\int_0^t f(\tau)\,d\tau=f(t)$; this is one version of the Fundamental Theorem of Calculus you may look up in your Apostol text.  Thus $g$ is continuously differentiable on $[0,1]$ (take care of the special meaning this has at the endpoints), and $g(0)=0$.  
Conversely, suppose that $g$ is continuously differentiable on $[0,1]$ and $g(0)=0$. Then for $t\in[0,1]$, $\int_0^t g'(\tau)\,d\tau=g(t)-g(0)=g(t)$, by another version of the Fundamental Theorem of Calculus you may look up in your Apostol text.  Thus $g=Tg'$ is in the range of $T$.  
Thus the range is determined.  To show that $T$ is injective, suppose that $Tf=0$.  Then $0=(Tf)'=f$. 
Because $T$ is injective, we can define $T^{-1}: T(C[0,1])\to C[0,1]$, which is determined by the identity $T(T^{-1}g)=g$, where $g$ is continuously differentiable and $g(0)=0$.  As observed above, for this to be true requires $g'=T^{-1}g$, and sure enough, defining $T^{-1}g = g'$ yields $T(T^{-1}g))=Tg' = g$. Unboundedness of $T^{-1}$ can be seen by considering $T^{-1}f_n$ for $f_n(t)=t^n$.  
