I have some difficulties with solving that problem:
Let $X$ be a linear space and $X= \left \{u \in C([0,\alpha]):\|u\|=\sup \frac{|u(x)|}{x}\right \}$.
We have a differential equation:
$\left\{\begin{matrix} &y'=f(x,y) & \\ &y(0)=y_{0} & \end{matrix}\right.$
Where:
$|f(x,y)-f(x,z)|\leqslant \frac{q}{x}|y-z|$ for all x $\in [0,\alpha]$, all real $y,z$ and some $q<1$.
Prove that there exsists exactly one solution in $X$.
I started with saying that $X$ is a Banach space. Then, defining operator $T:X \rightarrow X$:
$T(u)(x)=\int_{0}^{x}f(t,u(t)+y_{0})dt$
I wanted to prove that for some $n\in \mathbb{N}$ , $T^n$ is a contraction. If it is, from Banach fixed point theorem I would have that $T$ has exactly one fixed point in $X$ and that would be the solution of the equation above. I just have no idea how to prove it. I tried to estimate $||T^n(u)(x)-T^n(v)(x)||$ but I couldn't come up with a good estimation. Thanks for any help!
I don't think you need to estimate $T^n$. It seems to me enough to prove $T$ is a contraction. For this we must use the particular norm of the space $X$. Note for all $t>0$, \begin{equation} \vert f(t,u(t)+y_0)-f(t,v(t)+y_0)\vert \leq \frac{q}{t}\vert u(t)-v(t)\vert\leq q\Vert u-v\Vert \end{equation} by the definition of the norm in $X$. Then for all $x>0$, \begin{eqnarray} \frac{\left\vert T(u)(x)-T(v)(x)\right\vert}x & = & \frac{\left\vert \int_0^{x}f(t,u(t)+y_0)-f(t,v(t)+y_0)dt \right\vert}{x}\\ & \leq & \frac{\int_0^x\left\vert f(t,u(t)+y_0)-f(t,v(t)+y_0)\right\vert dt}{x}\\ & \leq & \frac{\int_0^x\frac{q}{t}\left\vert u(t)-v(t)\right\vert dt}{x}\\ & \leq & \frac{q\int_0^x\left\Vert u-v \right\Vert dt}{x}\\ & \leq & q\Vert u-v \Vert \end{eqnarray} This implies $\Vert T(u)-T(v)\Vert \leq q\Vert u-v\Vert$, so $T$ is a contraction and by Banach fixed point theorem we conclude.