Weak derivative in $L^1$ Consider $f:(0,1)\mapsto L^1(0,1)$ defined as $f(t)(x)=t\chi_{[0,t]}(x)$. I want to show that $\frac{f(t+h)-f(t)}{h}$ does not converge in $L^1(0,1)$. I have $\frac{f(t+h)-f(t)}{h}=\frac{(t+h)\chi_{[0,t+h]}-t\chi_{[0,t]}}{h}=\frac{t\chi_{(t,t+h]}+h\chi_{[0,t+h]}}{h}$ and therefore
\begin{equation}
\int_0^1 |\frac{t\chi_{(t,t+h]}+h\chi_{[0,t+h]}}{h}|\leq\frac{\int_t^{t+h} |t|dx}{h}+\int_0^{t+h}dx
\end{equation}
Clearly the second integral converge to $t$ as $h\to0$ and the first integral should converge to $t$ as $h\to 0$. Is it correct ? How  to get the result from this ?
 A: When $h>0$,
\begin{aligned}
 \frac{f_{t+h}-f_t}{h}(x) &= \frac{t}{h}\mathbb{1}_{(t,t+h]} +\mathbb{1}_{([0,t+h]}\\
\frac{f_{t-h}-f_t}{-h}(x) &= \frac{t}{h}\mathbb{1}_{(t-h,t]} +\mathbb{1}_{([0,t-h]}
\end{aligned}
The terms $\mathbb{1}_{[0,t-h]}$ and $\mathbb{1}_{[0,t+h]}$ converge to $\mathbb{1}_{(0,t]}$ in $L_1$ and also a.s. The terms $x\mapsto\frac{t}{h}\mathbb{1}_{(t-h,t]}(x)$ and $x\mapsto\frac{t}{h}\mathbb{1}_{(t,t+h]}(x)$ converge to $0$ a.s. So, if the differential quoatient is to converge in $L_1$, the limit should be $x\mapsto\mathbb{1}_{0,t-h]}(x)$. However
$$\left\|\frac{f_{t\pm h}-f_t}{\pm h}\right\|_1\xrightarrow{h\rightarrow0}2t$$
while $\|\mathbb{1}_{[0,t]}\|_1=t$.
Another way to see the lack of convergence is by computing
\begin{aligned}
\left\|\frac{f_{t+h}-f_t}{h}-\frac{f_{t-h}-f_t}{-h}\right\|_1&=\int\frac{t}{h}\big(\mathbb{1}_{(t,t+h]}(x)-\mathbb{1}_{(t-h,t]}(x)\big) + \mathbb{1}_{(t-h,t+h]}(x)\,dx\\
&=2t>0
\end{aligned}
for all $0<t<1$.
A: First let us write out the function
$$ g_h = \frac{f(t+h) - f(t)}{h} = \frac{(t + h) 1_{(0,t+h]} - t 1_{(0,t]}}{h} = \frac{t}{h} 1_{(t,t+h]} + 1_{(0,t+h]} $$
You want to show that the limit as $h \to 0$ does not exists in $L^1$. 
The second function quite "obviously" (we'll write down the proof later) converge to $1_{(0,t]}$ as $h\to 0$. The first function, however, does not converge in $L^1$: if you fix $t$ and plot it for $h \searrow 0$ you will see that it gets taller and taller but skinnier and skinnier. 
In order to show the non-convergence, we can use Cauchy's criterion: 


*

*If as $h\to 0$ this converges, then so does it when $h\searrow 0$. 

*... and so does it if we set $h = 2^{-n}$ and take limit $n \to \infty$...

*... and so $g_{2^{-n}}$ should form a Cauchy sequence. 


Take the difference (assuming $n < m$ WLOG)
$$ g_{2^{-n}} - g_{2^{-m}} = 2^{n}t 1_{(t,t+2^{-n}]} - 2^m t 1_{(t,t+2^{-m}]} + 1_{(t+2^{-m}, t+2^{-n}]} = (2^{n} - 2^m) t 1_{(t,t+2^{-m}]} + (1+2^{n}t) 1_{(t+2^{-m},t+2^{-n}]} $$
Its $L^1$ norm is equal to 
$$ 2(1 - 2^{n-m})t + (2^{-n} - 2^{-m}) \geq t$$
And hence Cauchy's criterion is violated and we don't have convergence. 
