# Show that $|T_{\epsilon}(f)-T_0(f)|\leq \frac{\epsilon}{2}||f||_E,\; \epsilon\in]0,1]$

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 $$0, then using The Mean Value Theorem we get the existence of $$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 gives us $$c_2\in (x,0)$$ such that $$|f(x)-f(0)|=|f'(c_2)||x| \leq |x| \|f\|_E$$ Therefore $$|T_0(f)-T_\epsilon(f)|\leq \frac{1}{2\epsilon}\int^\epsilon_{-\epsilon}|x|\|f\|_E\,dx =\|f\|_E\frac{1}{\epsilon}\int^\epsilon_0 x\,dx = \|f\|_E \frac 1 \epsilon \cdot \frac{\epsilon^2} 2 = \frac \epsilon 2 \|f\|_E$$