Let $T > 0$ and $L ≥ 0$. Consider $C[0, T]$, the space of all continuous real valued functions on $[0, T]$, with the metric $ρ$ defined by $ρ(x, y) = \sup\limits_{ (0≤t≤T)} e ^{−Lt}|x(t) − y(t)|$.
Consider the initial value problem
$ x' (t) = f(t, x(t))$ for $t > 0,$ $x(0) = x_0.$
where f : $\Bbb R × \Bbb R → \Bbb R$ is continuous and globally Lipschitz continuous with respect to x, i.e, there exists $L ≥ 0$ such that $|f(t, x) − f(t, y)| ≤ L|x − y|$ for all $t, x, y ∈ \Bbb R.$
Construct an integral operator and show that this operator is a contraction on $(C[0, T], ρ)$. Hence deduce that the initial value problem has a unique solution in $C^ 1 [0, T].$
To use the contraction mapping theorem, I have managed to prove that $(C[0, T], ρ)$ is a completes metric space. But I really don't know how to construct a suitable integral operator.
Let $L_f$ denote the Lipschitz constant of $f$ and choose $L := 2L_f$ in the definition of $\rho$. Define the standard Picard operator $$ Tx (t) := x_0 + \int_0^t f(s, x(s))\, ds, \qquad x\in C[0,T]. $$ Then it holds $$ (*) \qquad \rho(Tx, Ty) \leq \frac{1}{2} \rho(x,y), \qquad \forall x,y\in C[0,T]. $$ Namely, let $x,y\in C[0,T]$. By the very definition of $\rho$ we have that $$ |x(s) - y(s)| \leq \rho(x,y)\, e^{2 L_f s}, \quad \forall s\in [0,T]. $$ Since $$ \int_0^t e^{2 L_f s}\, ds = \frac{1}{2 L_f} (e^{2 L_f t} - 1) <\frac{1}{2 L_f} e^{2 L_f t}, $$ we get $$ |Tx(t) - Ty(t)| \leq \int_0^t L_f |x(s) - y(s)|\, ds \leq \int_0^t L_f \rho(x,y)\, e^{2L_f s}\, ds \leq \frac{1}{2} \rho(x,y) \, e^{2 L_f t}, $$ so that $$ e^{-2 L_f t} |Tx(t) - Ty(t)| \leq \frac{1}{2} \rho(x,y), \qquad \forall t\in [0,T] $$ and (*) follows.