$\lim_{t \rightarrow 0} \frac{1}{2t} \int_{b-t}^{b+t} |f- \frac{1}{2t} \int_{b-t}^{b+t} f| = 0$

Question

$\lim_{t \rightarrow 0} \frac{1}{2t} \int_{b-t}^{b+t} |f- \frac{1}{2t} \int_{b-t}^{b+t} f| = 0$

Can I use the Monotone convergence theorem besides the L'Hopital theorem to show this? with $t=1/n$

Answer 1

I consider the case when $f$ is continuous, then general case is covered in the answer below..

We have $$ \vert f(x)-\frac{1}{2t} \int_{b-t}^{b+t} f(y) dy \vert =\vert \frac{1}{2t} \int_{b-t}^{b+t} (f(x)-f(y)) dy\vert \leq \frac{1}{2t} \int_{b-t}^{b+t} \vert f(x)-f(y) \vert dy \leq \sup_{y\in (b-t,b+t)} \vert f(x)-f(y)\vert.$$ Thus, $$\frac{1}{2t} \int_{b-t}^{b+t} \vert f(x)-\frac{1}{2t} \int_{b-t}^{b+t} f(y) dy \vert dx \leq \frac{1}{2t} \int_{b-t}^{b+t} \sup_{y\in (b-t,b+t)} \vert f(x)-f(y)\vert \leq \sup_{x,y\in (b-t,b+t)} \vert f(x)-f(y)\vert.$$ However, if $f$ is continuous, then $f$ is uniformly continuous on compact sets and hence you obtain that the RHS of the last inequality goes to zero, which yields the claim.

Answer 2

Partial Solution:

Note that \begin{align*} &\dfrac{1}{2t}\int_{r-t}^{r+t}\left|f-\dfrac{1}{2t}\int_{r-t}^{r+t}f\right|\\ &\leq\dfrac{1}{2t}\int_{r-t}^{r+t}|f-f(r)|+\dfrac{1}{2t}\int_{r-t}^{r+t}\left|f(r)-\dfrac{1}{2t}\int_{r-t}^{r+t}f\right|\\ &=\dfrac{1}{2t}\int_{r-t}^{r+t}|f-f(r)|+\left|f(r)-\dfrac{1}{2t}\int_{r-t}^{r+t}f\right|. \end{align*} By Lebesgue's Differentiation Theorem one has for a.e. $r$ that \begin{align*} \dfrac{1}{2t}\int_{r-t}^{r+t}|f-f(r)|\rightarrow 0 \end{align*} and hence the proposition is true for a.e. $r$.

For a fixed $b$, I have no solution.