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We say that$\ f$ is differentiable w.r.t to $L_1$ if there exists a$\ g$ such that: $$ \lim_{h\to 0}\left\Vert\frac{f(x+h)-f(x)}{h} - g(x)\right\Vert_1 = 0 $$ where $\Vert \cdot \Vert_1$ is the $L_1$ norm. Since $f$ is in $L_1$, the corresponding$\ g$ must be in$\ L_1$ too, and so by Lebesgue, it has an antiderivative $G$ which is differentiable a.e, with $G'(x)=g(x)$.

My question is: does $f=G$ a.e?

Here is my line of thought: if $G$ is in $L_1$, it can be shown that $$ \hat{g}{(t)} = 2\pi it\hat{G}{(t)} = 2\pi it\hat{f}{(t)}, $$ which then implies that $f=G$ a.e. and so, in order to show that $f=G$ a.e, it is enough to show that$\ G$ is in$\ L_1$, and that's where i got stuck.

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    $\begingroup$ I guess everything's happening on $\mathbb R$? $\endgroup$ – LSpice Apr 1 at 17:03
  • $\begingroup$ This seems loosely related to Proposition 9.3 in "Haïm Brezis: Functional Analysis, Sobolev Spaces and Partial Differential Equations (2011)". $\endgroup$ – Jochen Glueck Apr 1 at 20:00
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Most antiderivative of $g$ are not in $L^1(\mathbb{R})$, in your case only one antiderivative $G_0$ will be in $L^1(\mathbb{R})$, the one actually equal to $f$. All the other antiderivatives $G$ are equal to $G_0 + c$, with $c \neq 0$, which is not in $L^1(\mathbb{R})$.

To proof that there exist one antiderivative $G_0$ in $L^1(\mathbb{R})$. You start by noticing that your $L^1(\mathbb{R})$ differentiability imply differentiability in the distributional sens so, for $\phi \in \mathcal{C}_{comp}^\infty(\mathbb{R})$, we have $$ \langle f',\phi \rangle = \langle g,\phi \rangle. \qquad (1) $$ Fix $G$ an antiderivative of $g$, you have $G' = g$ in the distributional sens. The equation $(1)$ become $$ \langle f',\phi \rangle = \langle G',\phi \rangle \implies \langle (f-G)',\phi \rangle = 0 $$ and than imply $f-G = c$ a constant. Choosing $G_0 = G + c$, we have $G_0 = f \in L^1(\mathbb{R})$.

You can further show that there exists a constant $c_0$ such that $$ G_0(x) = c_0 + \int_0^x g(y)\, dy. $$

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