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? 

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 $x=y$. 
