Cauchy problem and operator theory problem Would you mind giving only hint to me in order to start to solve???

Let 
  $T : C([0, 1]) \to C([0, 1]) $ 
  be the linear operator defined by 
  $T(f) := y,$ 
  where y is the solution of the Cauchy problem
  $$y'(x)=y(x)+f(x),\;\;\; y(0)=0$$
  (a) Prove that $T$ is compact.
(b) Prove that $T$ has no eigenvalues.

 A: Hints: Using the product rule with integrating factors, we find that $$\frac{d}{dx}(y(x)e^{-x}) = y'(x)e^{-x}-y(x)e^{-x}=f(x)e^{-x}$$ So integrating both sides from $0$ to $u$ and then multiplying by $e^{u}$ we find $$y(u) = \int_0^uf(x)e^{u-x}dx=Tf(u)$$
To prove that $T$ is compact, I suggest using the fact that $|f(x)|\leq \|f\|_{\infty}$ in order to show that the collection $\{Tf:\|f\|_{\infty}<1\}$ is equicontinuous and pointwise bounded in $C[0,1]$.
To show that $T$ has no eigenvalues, then suppose that $$\lambda f(u) = \int_0^u f(x)e^{u-x}dx$$ Now differentiate both sides of this equation with respect to $u$ and you will see that $f$ satisfies a differential equation which is easily solved, with the initial data $f(0)=0$. Can the resulting function be an eigenvector?
A: The solution to the differential equation
$y'(x) = y(x) + f(x), \; y(0) = 0, \tag 1$
defined on the unit interval $[0, 1]$, is given by the formula
$y(x) = e^x \displaystyle \int_0^x e^{-s}f(s)ds, \tag 2$
as may readily be seen by direct differentiation; from (2):
$y'(x) = (e^x)'\displaystyle \int_0^x e^{-s}f(s)ds + e^x (\int_0^x e^{-s}f(s)ds)'$
$= e^x \displaystyle \int_0^x e^{-s}f(s)ds + e^x ( e^{-x}f(x)) = y(x) + f(x); \tag 3$
we corroborate that $y(x)$ satisfies the requisite initial condition:
$y(0) = e^0 \displaystyle \int_0^0 e^{-s}f(s)ds = 0. \tag 4$
We will show the map $T:C([0, 1]) \to C([0, 1])$ given by 
$T(f) =  e^x \displaystyle \int_0^x e^{-s}f(s)ds \tag 5$
is compact via the Arzela-Ascoli Theorem.  Let $f_n(x) \in C([0, 1])$ be a sequence such that
$\Vert f_n(x) \Vert \le 1 \tag 6$
for every $n$; that is, $f_n(x)$ lies in the unit ball of $C([0, 1])$; then setting
$y_n(x) = T(f_n(x)), \tag 7$
we have
$\Vert y_n(x) \Vert = \Vert e^x \displaystyle \int_0^x e^{-s}f_n(s)ds \Vert \le \Vert e^x \Vert \Vert \displaystyle \int_0^x e^{-s}f_n(s)ds \Vert$
$\le e \displaystyle \int_0^x \Vert e^{-s} f_n(s) \Vert ds \le  e \displaystyle \int_0^x \Vert e^{-s} \Vert \Vert f_n(s) \Vert ds \le e \displaystyle \int_0^1  1ds = e, \tag 8$
which shows the sequence $y_n(x) = T(f_n(x))$ is uniformly bounded.  Now suppose $x_1, x_2 \in [0, 1]$ with $\vert x_1 - x_2 \vert < \delta$, and say, without loss of generality, $x_2 \ge x_1$; then we have the following somewhat gory presentation of an estimate for $\vert y_n(x_2) - y_n(x_1) \vert$:
$\vert y_n(x_2) - y_n(x_1) \vert = \vert e^{x_2} \displaystyle \int_0^{x_2} e^{-s}f_n(s)ds \ -  e^{x_1} \displaystyle \int_0^{x_1} e^{-s}f_n(s)ds \vert$
$= \vert e^{x_2} \displaystyle \int_{x_1}^{x_2} e^{-s}f_n(s)ds + (e^{x_2} -  e^{x_1}) \displaystyle \int_0^{x_1} e^{-s}f_n(s)ds \vert$
$\le \vert e^{x_2} \vert \vert \displaystyle \int_{x_1}^{x_2} e^{-s}f_n(s)ds \vert + \vert (e^{x_2} -  e^{x_1}) \vert \vert \displaystyle \int_0^{x_1} e^{-s}f_n(s)ds \vert$
$\le e \vert \displaystyle \int_{x_1}^{x_2} e^{-s}f_n(s)ds \vert + \vert e^{x_1}(e^{x_2 - x_1} - 1) \vert \vert \displaystyle \int_0^{x_1} e^{-s}f_n(s)ds \vert$
$\le e \displaystyle \int_{x_1}^{x_2} \vert  e^{-s}f_n(s) \vert ds + \vert e^{x_1} \vert \vert e^\delta - 1 \vert \vert \displaystyle \int_0^{x_1} e^{-s}f_n(s)ds \vert$
$\le e \displaystyle \int_{x_1}^{x_2} \vert  e^{-s} \vert \Vert f_n(s) \Vert  ds + e \vert e^\delta - 1 \vert \vert \displaystyle \int_0^{x_1} e^{-s}f_n(s)ds \vert$
$\le e \displaystyle \int_{x_1}^{x_2} \Vert f_n(s) \Vert  ds + e \vert e^\delta - 1 \vert \displaystyle \int_0^{x_1} \vert e^{-s}f_n(s) \vert ds$
$\le e(x_2 - x_1) \Vert f_n(x) \Vert + e \vert e^\delta - 1 \vert \displaystyle \int_0^{x_1} \vert e^{-s} \vert \Vert f_n(s) \Vert ds$
$\le e \delta + e \vert e^\delta - 1 \vert \displaystyle \int_0^{x_1} \vert e^{-s} \vert \Vert f_n(s) \Vert ds$
$\le  e \delta + e \vert e^\delta - 1 \vert \displaystyle \int_0^{x_1} \Vert f_n(s) \Vert ds \le e \delta + e \vert e^\delta - 1 \vert \Vert f_n(x) \Vert x_1 \le e(\delta + \vert e^\delta - 1 \vert); \tag 9$
now, given $\epsilon$, we may choose $\delta$ sufficiently small that
$\delta + \vert e^\delta - 1 \vert < \dfrac{\epsilon}{e}; \tag{10}$
then we have
$\vert y_n(x_2) - y_n(x_1) \vert \le e(\delta + \vert e^\delta - 1 \vert) < e\dfrac{\epsilon}{e} = \epsilon, \tag{11}$
independently of $n$, $x_1$ or $x_2$; thus the sequence $y_n(x)$ is in fact equicontinuous; thus the hypotheses of Arzela-Ascoli are met, and we conclude that $y_n(x)$ contains a uniformly convergent subsequence; we conclude that $T$ is indeed compact.
We show that $T$ has no eigenvalue.  If $\lambda \in \Bbb R$ were an eigenvalue of $T$, then by (5) we would have
$e^x \displaystyle \int_0^x e^{-s}f(s)ds = T(f(x)) = \lambda f(x) \tag {12}$
for some $0 \ne f(x) \in C([0, 1])$; setting
$g(x) = e^{-x}f(x), \tag {13}$
we see that
$\displaystyle \int_0^x g(s) ds = \lambda g(x), \tag {14}$
and since $f(x) \in C([0, 1])$, the integral $\int_0^x e^{-s}f(s) ds = \int_0^x g(s)ds$ is differentiable in $x$, so (14) yields
$g(x) = \lambda g'(x); \tag{15}$
if now $\lambda = 0$, we have
$f(x) = e^x g(x) = \lambda e^x  g'(x) = 0, \tag{16}$
forbidden for an eigenvector of $T$; if, on the other hand, $\lambda \ne 0$, by (15) $g(x)$ satisfies the differential equation
$g'(x) = \lambda^{-1} g(x) \tag{17}$
with $g(0) = 0$ from (14).  We thus must have, by uniqueness of solutions, $g(x) = 0$ on $[0, 1]$, and
$f(x) = e^x g(x) = 0 \tag{18}$
for all $x \in [0, 1]$, again forbidden by the definition of eigenvectors.  This contradiction implies there is no legitimate  $f(x) \in C([0, 1])$ satisfying (12); $T$ is a compact operator without eigenvalues.
