Injectivity of $T:C[0,1]\rightarrow C[0,1]$ where $T(x)(t):=\int_0^t x(s)ds $ Prob. 2.7-9 in Erwin Kreyszig's "Introductory Functional Analysis with Applications": Is this map injective?
Let $C[0,1]$ denote the normed space of all (real or complex-valued) functions defined and continuous on the closed interval $[0,1]$ on the real line, with the maximum norm given by 
$$
\Vert x \Vert_{C[0,1]} = \max_{t \in [0,1]} \vert x(t) \vert \ \ \ \mbox{ for all } \ x \in C[0,1].
$$ 
Let $T \colon C[0,1] \to C[0,1]$ be defined as follows: for each $x \in C[0,1]$, let  $T(x) \colon [0,1] \to K$, where $K = \mathbb{R}$ or $\mathbb{C}$, be defined by
$$ 
\left( T(x) \right)(t) \colon= \ \int_0^t \ x(\tau) \ \mathrm{d} \tau 
\ \ \ \mbox{ for all } \ t \in [0,1].
$$ 
Then $T$ is a bounded linear operator with range consisting of all those continuously differentiable functions on $[0,1]$ that vanish at $t=0$. 
Am I right? 
Is $T$ injective? How to determine if $T$ is injective or not? 
 A: Suppose 
\begin{equation}
T(x_1)(t) = T(x_2)(t)
\end{equation}for all $t \in[0,1]$. As you have already figured out that the range of the operator is the set of all continuously differentiable functions (this follows from Fundamental Theorem of Calculus). In light of your observation just differentiate the two sides of the equation to get $x_1(t) = x_2(t)$, which shows that the operator is indeed injective.
A: Yes. Your answer about range is correct.
The operator is injective. To see it, since $x(t)$ is continuous at $[0,1]$, you can apply Fundamental theorem of calculus, given $F(t)\in R(T)$, $x(t)$ is uniquely determined by $x(t)=F'(t)$. Or you can argue directly that if $x_1(t_0)\not =x_2(t_0)$, for some  $t_0\in[0,1]$, by continuity $T(x_1)(1)\not =T(x_2)(1)$
A: If $\parallel x\parallel =M$, then $$ |T(x)(t)|=\bigg|\int_0^t x(s)ds \bigg|
\leq \int_0^t M \leq M $$ Hence bounded.
And $\frac{d}{dt} T(x)(t)=x(t)$ is continuous. And $T(x)(0)=0$.
If $T(x)=T(y)$ then $
\parallel T(x)- T(y)\parallel =0$ So $$ \forall
t,\ \int_0^t (x-y)(s) ds =0
$$
Assume that $t_0\in (0,1)$ with $(x-y)(t_0) >0$. Then $x-y \geq c> 0
$ on $ [t_0-\delta, t_0+\delta] \subset [0,1]$. Then $$
\int_0^{t_0+\delta } (x-y) = 0 + \int_{t_0-\delta }^{t_0+\delta }
(x-y)\geq 2\delta c > 0 $$ Contradiction.
Hence $x\leq y$ on $(0,1)$. Similarly $y\leq x$. Hence $x=y$ on
$(0,1)$. By continuity we have $x=y$. 
