Let's $(X,\rho)$ be a metric space, where $X=L(0,1)$ and $\rho(x,y)$=$\int_{0}^{1}|x(t)-y(t)|$.
I need to prove that transformation $A:X\to X$ is a contraction mapping.
- $Ax(t)=\int_{0}^{1}e^{-(s+t)}x(s)ds+arctg(t)$,
- $Ax(t)=\int_{0}^{1}\frac{x(s)}{1+s+t}dt+f(t)$
I tried to do some transformations, but I am not sure if there are okay. Can you help me with finishing it?
$\int_{0}^{1}|\int_{0}^{1}e^{-(s+t)}x(s)ds+arctg(t)-\int_{0}^{1}e^{-(s+t)}y(s)ds-arctg(t)|dt=$
$=\int_{0}^{1}|\int_{0}^{1}e^{-(s+t)}x(s)-e^{-(s+t)}y(s)ds|dt=$
$=\int_{0}^{1}e^{-t}|\int_{0}^{1}e^{-s}(x(s)-y(s))ds|dt\leq \int_{0}^{1}e^{-t}\int_{0}^{1}e^{-s}|(x(s)-y(s))|dsdt=$
$(-e^{-1}+1)\int_{0}^{1}e^{-s}|(x(s)-y(s))|ds=\int_{0}^{1}e^{-s}|(x(s)-y(s))|ds$
That's all what I've got.
If it comes to the second one I do not have anything special so if you have any suggestions I will be really grateful.
To complete the first one (drop the last equality and) just note that $1-\frac 1 e <1$ and $\int_0^{1}e^{-s} |x(s)-y(s)|ds \leq \int_0^{1} |x(s)-y(s)|ds =\|x-y\|$.
Answer for the second transformation: $|Ax(t)-Ay(t)| \leq \frac 1 {1+t} \int_0^{1} |x(s)-y(s)| ds$ Now integrate this. $\|Ax-Ay\|=\int_0^{1}|Ax(t)-Ay(t)| \leq \int_0^{1} \frac 1 {1+t} dt \|x-y\|=\ln 2 \|x-y\|$ and $\ln 2 <1$.