Csdrhrt

We denote by $D$ the space of the functions $C^{infty}(]-1,1[)$ and compact support and $F$ the space of the Lipschitzian functions on $[-1,1]$.

For $fin F$ and $varphiin D$, we note

$B(f,varphi)=int_{-1}^{1}f(t)varphi'(t)dt$

They ask me to show that:

1. 
   

   Let $varphiin D$ be zero outside of $[-1 + eta, 1 - eta] $ (with $0 <eta <1$). Show that for $epsilonin]0,frac{eta}{2}[$ we have

   
   
   

   $int_{-1+eta}^{1-eta}frac{f(t)-f(t+epsilon)}{epsilon}varphi(t)dt=int_{-1+eta/2}^{1-eta/2}f(t)frac{varphi(t)-varphi(t-epsilon)}{epsilon}dt$

   
   


2. 
   

   Prove that $|B(f,varphi)|le||f||_F||varphi||_1$.

   
   
   

   Where, $||f||_F=sup_{-1le tle 1}|f(t)|+sup_{-1le tle
   1}|f'(t)|$

   
   

For the second item, I must use the first item.

But in the first item I intuit that I should use the derivative definition or that $f $ be in the Lipschitz function space.

Could you please help me?

Sketch: Note first that $epsilon$ is a constant, so can be multiplied out from both integrals. Then, since $phi(t)$ is zero outside $[1-eta, 1-eta]$, the $f(t) phi(t)$ term can be ignored also. It then remains to show:

$$int_{-1+eta}^{1-eta} f(t+epsilon) phi(t) , dt = int_{-1+frac{eta}{2}}^{1-frac{eta}{2}} f(t) phi(t-epsilon) , dt.$$

Perform a substitution on the right hand side, then consider the intervals that both integrals are defined over, and use the range of $epsilon$ to conclude.