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I know that the question arose already in other contexts. However, I think this question might be different. If I have $f\in W^{1,1}$ then it is obvious that $\frac{f(x+t)-f(x)}{t}$ converges pointwise to the weak derivative $f'$. The question is: why does one have $\frac{f(\cdot+t)-f(\cdot)}{t}\to f'$ in $L^1$ or in other words, why $$\int{\frac{f(x+t-f(x)}{t}-f'(x)\, dt}\to 0$$ What I would need is a majoring function so that I can apply dominated convergence. Does anybody have an idea what such a function could be? Thanks in advance :)

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    $\begingroup$ Is the integration element meant to be $dx$? $\endgroup$
    – Willie Wong
    Commented Mar 12 at 16:04
  • $\begingroup$ Do you have a response to the answer below? $\endgroup$
    – Iosif Pinelis
    Commented Mar 26 at 21:24

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We have $$d(t):=\int dx\Big(\frac{f(x+t)-f(x)}{t}-f'(x)\Big) =\int dx\int_0^1 ds\,(f'(x+st)-f'(x)),$$ whence $$|d(t)|\le\int_0^1 ds\, J_t(s),$$ where $$J_t(s):=\int dx\,|f'(x+st)-f'(x)|\underset{t\to0}\longrightarrow0 \tag{10}\label{10}$$ for each $s$, since $f'\in L^1$ (see e.g. this answer). Also, $$|J_t(s)|\le\int dx\,(|f'(x+st)+|f'(x)|)=2\|f'\|_1<\infty.$$

So, by dominated convergence, $d(t)\underset{t\to0}\longrightarrow0$.

Instead of dominated convergence, one can use the fact that the convergence in \eqref{10} is uniform in $s\in[0,1]$.

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