I want to show that the solution of f'(x)=-af(x)+ag(x) depends continuous on its initial value. So given the differential equation $f'(x)=-af(x)+ag(x)$ with $f(0)=p$ , $x \geq 0$ , find a Constant C such that: $||f||_{C^1} \leq C ||g||_{C^0}$
Note that $||f||_{C^1}=||f||_{\infty}+||f'||_{\infty}$
and
$||g||_{C^0}=||g||_\infty$
My Approach: First I solved the equation above, $f(x)=e^{-ax}(p+a\int_0^x e^{as}g(s)ds)$
Using $||f||_{C^1}=||f||_{\infty}+||f'||_{\infty}$ und inserting the solution I get
$||f||_{C^1}=sup_{x \in [0,t]}|e^{-ax}(p+a\int_0^x e^{as}g(s)ds)|$+$sup_{x \in [0,t]}|-ae^{-ax}(p+a\int_0^x e^{as}g(s)ds+e^{-ax}+ae^{ax}g(x)|$
I split this into two parts:
First:
$sup_{x \in [0,t]}|e^{-ax}(p+a\int_0^x e^{as}g(s)ds)|=sup_{x \in [0,t]}|(p+a\int_0^x e^{as}g(s)ds)| \leq sup_{x \in [0,t]}(|p|+ |a\int_0^x e^{as}g(s)ds|)$
thats how far I got in the first part, I don't really know how to estimate the part with the Integral, especially because I don't know g.
Second:
$sup_{x \in [0,t]}|-ae^{-ax}(p+a\int_0^x e^{as}g(s)ds)+e^{-ax}+ae^{ax}g(x)|=sup_{x \in [0,t]}|-a(p+a\int_0^x e^{as}g(s)ds)+1+g(x)|$
$||g||_{C^0}=sup_{x \in [0,t]}|g(x)|$
I don't know how to find this C. Hope someone could show me how.