I have used the assumption that a function, $f$ is everywhere Lipschitz to prove that another function, $\phi$, is a contraction. The function in question is \begin{equation} \phi(y,t) = y_0 + \int_{0}^{t} f(y(s)) ds \end{equation}
And the metric space in question, $X$, is the space of all bounded, continuous functions between $0$ and $T$.
The metric in use is \begin{equation} d(f,g) = \max_{t\in[0,T]}|f(t)-g(t)| \end{equation}
So I start with \begin{equation} |\phi(y_1)-\phi(y_2)| = |\;y_0 + \int_{0}^{t} f(y_1(s))ds \;-(y_0 + \int_{0}^{t} f(y_2(s))ds) \;| \end{equation}
Simplifying to \begin{equation} |\phi(y_1)-\phi(y_2)| = \;| \int_{0}^{t} f(y_1(s))-f(y_2(s))ds \;| \end{equation}
\begin{equation} = \int_{0}^{t} |\;f(y_1(s))-f(y_2(s))\;|\;ds \; \end{equation}
Which by the Lipschitz condition is \begin{equation} \le K \cdot \;\int_{0}^{t} |\;y_1(s)-y_2(s)\;|\;ds \; \end{equation} \begin{equation} \le K \cdot \;\int_{0}^{t} \max_{t\in[0,T]}|\;y_1-y_2\;|\;ds \; \end{equation} \begin{equation} = K \cdot \max_{t\in[0,T]}|\;y_1-y_2\;| \cdot \;\int_{0}^{t} 1\;ds \; \end{equation} \begin{equation} = K \cdot t\cdot \max_{t\in[0,T]}|\;y_1-y_2\;| \end{equation}
So if we have $c = Kt$, this meets the criteria that $\phi$ is a contraction.
I have now been asked to try and show that $\phi$ is a contraction if we assume that the function $f$ is LOCALLY Lipschitz. I believe this means that the Lipschitz condition only holds for restricted range of values of $y$.
I apologise if my work seems a bit disjointed or I have not fully explained a certain step in my proof. All I am asking is if anyone would understand how the change from everywhere Lipschitz to locally Lipschitz affects the proof. Does the proof still have a similar structure in general? What aspect of the proof changes? All help appreciated!