I have the next exercice:
For $\epsilon \in ]0,1]$, we have the function
$$T_{\epsilon}(f)=\frac{1}{2\epsilon}\int_{-\epsilon}^{\epsilon}f(t)dt$$ and $T_0(f)=f(0).$
Show that for $\epsilon \in ]0,1]$ and $f\in E= C^1(]-1,1[)$ which are bounded as their derivative too, we have $$|T_{\epsilon}(f)-T_0(f)|\leq \frac{\epsilon}{2}||f||_E$$ where $||f||_E=\sup_{-1\leq t \leq 1}|f(t)|+\sup_{-1\leq t \leq 1}|f'(t)|$.
My ideas for this statement:
- I have the idea that I should do tend $\epsilon$ to $1$
- Rewrite in some convenient way $T_o(f)$ such that may have some integral of the form $\int_{-1}^{1}|f(t)|dt$ and after apply the $\sup$ for obtain $||f||_E$
Finally, I should use this result for preuve $L^p$ is not closed in $E'.$
Thanks a lot for your suggestions.
We have $$|T_0(f)-T_\epsilon(f)|\leq \frac{1}{2\epsilon}\int^\epsilon_{-\epsilon} |f(x)-f(0)|\,dx$$ Let $0<x<\epsilon$, then using The Mean Value Theorem we get the existence of $c_1\in (0,x)$ such that \begin{align*} |f(x)-f(0)|=|f'(c_1)|x\leq |x|\|f\|_E \end{align*} Applying MVT again for $-\epsilon<x<0$ gives us $c_2\in (x,0)$ such that $$|f(x)-f(0)|=|f'(c_2)||x| \leq |x| \|f\|_E $$ Therefore $$|T_0(f)-T_\epsilon(f)|\leq \frac{1}{2\epsilon}\int^\epsilon_{-\epsilon}|x|\|f\|_E\,dx =\|f\|_E\frac{1}{\epsilon}\int^\epsilon_0 x\,dx = \|f\|_E \frac 1 \epsilon \cdot \frac{\epsilon^2} 2 = \frac \epsilon 2 \|f\|_E$$