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Have $f'(t) = f(t), f(0) = 1$ where $f \in C([0,c])$ where $0 < c < 1$. Using uniform norm, and Banach fixed point theorem with $(Tf)(x) = 1 + \int_0^x f(t) dt$. Show unique solution exists for the ODE.

I began by showing that $T$ is a contraction. As $||Tf_1 - Tf_2|| = ||\int_0^x (f_1 - f_2)(t)dt|| \le \int_0^x||f_1 - f_2||dt \le c \cdot ||f_1 - f_2||$ . Since $C([0,c])$ is complete, we then apply Banach fixed point theorem to get that there exists a unique $f^* \in C([0,c])$ such that $f^*(t) = 1 + \int_0^tf^*(s)ds$. And therefore, we can apply the fundamental theorem of calculus to see that $f^*(t)$ is differentiable on $(0,c)$ and satisfies the ODE on this interval.

My result is different from Picard's existence in that Picard's theorem works in an epsilon-interval around the initial condition and mine works further. Am I doing something wrong here? Or am I just able to get a stronger result because of more specific assumptions?

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You know what exactly the right hand side is, so you have an explicit bound on $\|Tf_1-Tf_2\|$. However, you still need $c<1$ in order to have a contraction. This means in some sense the epsilon-interval had already been adjusted from beginning in the exercise.

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