Convergence of operator I would like to know how to solve the following problem (since I didn't manage to solve it on today's exam):
Let $A_h:L^1(a,b)\to L^1(a,b)$ be defined: 
$$A_h f(x)=\frac{1}{h}\int_x^{x+h} g(t) dt,$$ where $g(t)=f(t)$ for $t\in(a,b)$ and $g(t)=0$ otherwise, for $0<h<\frac{b-a}{4}$. 
Find $\|A_h\|$. Does $A_h$ converge uniformly, strongly or weakly, when $h\to 0+$? 
Any hint is welcome. Thanks in advance.

This is too long for a comment, so I will post here as a answer. I will just give here some ideas and maybe you can finish it. I will use the notation $\tilde{f}$ (instead of your notation $g$), for the extension of $f$, which is zero outside $(a,b)$. With respect to the norm, note that 
\begin{eqnarray}
 \|A_hf\| &=& \int_a^b\left|\frac{1}{h}\int_{x}^{x+h}\tilde{f}(t)dt\right|dx      \nonumber \\
   &\le& \int_a^b\frac{1}{h}\int_{x}^{x+h}|\tilde{f}(t)|dtdx \nonumber \\
   &=& \frac{1}{h}\int_0^h\int_a^b|\tilde{f}(x+t)|dxdt \\
   &\le & \|f\|, \forall\ f\in L^1.
\end{eqnarray}
With respect to the convergence, I can prove it for at least strongly. Indeed, first assume that $f$ is continuous in $[a,b]$. For any $\epsilon>0$, choose $h>0$ such that $|\tilde{f}(t)-f(x)|\le \epsilon/(b-a)$ for $t\in (x,x+h)$ and $x\in (a,b)$. Note that 
\begin{eqnarray}
 \|A_h f-f\| &=& \int_a^b \left|\frac{1}{h}\int_x^{x+h}\tilde{f}(t)dt-f(x)\right|dx      \nonumber \\
   &\le& \frac{1}{h}\int_a^b\int_x^{x+h}|\tilde{f}(t)-f(x)|dtdx \nonumber \\
   &\le& \epsilon. \tag{1}
\end{eqnarray}
Now, given $\epsilon>0$, take $f\in L^1$ and choose $g\in C([a,b])$ with $\|f-g\|\le \epsilon$. First observe that 
\begin{eqnarray}
 \|A_h f-A_h g\|  &\le& \frac{1}{h}\int_a^b\int_x^{x+h}|\tilde{f}(t)-\tilde{g}(t)|dtdx      \nonumber \\
   &=& \frac{1}{h}\int_0^h\int_a^b |\tilde{f}(x+t)-\tilde{g}(x+t)|dxdt \nonumber \\
   &\le& \|f-g\|(b-a) \tag{2}
\end{eqnarray}
We combine $(1)$ and $(2)$, to conclude that for small $h$,
\begin{eqnarray}
 \|A_h f-f\| &\le& \|A_h f-A_h g\|+\|A_h g-g\|+\|g-f\|      \nonumber \\
   &\le& \epsilon (b-a)+\epsilon +\epsilon \nonumber \\
   &=& \epsilon (b-a+2)
\end{eqnarray}
It is worth to note that $A_h f(x)\to f(x)$ for a.e. $x\in (a,b)$, because of Lebesgue theorem.
