# Equivalence of antiderivative in L1 sense and in the usual sense

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.

• I guess everything's happening on $\mathbb R$? – LSpice Apr 1 at 17:03
• This seems loosely related to Proposition 9.3 in "Haïm Brezis: Functional Analysis, Sobolev Spaces and Partial Differential Equations (2011)". – Jochen Glueck Apr 1 at 20:00

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 $$G_0(x) = c_0 + \int_0^x g(y)\, dy.$$