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Let $f, g : \mathbb{R} \rightarrow \mathbb{C}$ be Lebesgue measurable functions, and let $1 \leq p < \infty$.

If $f, g \in L^p$ and

$$ \large \lim_{y \rightarrow 0} \normalsize \left\| \frac{f(\bullet + y) – f(\bullet)}{y} – g(\bullet)\right\|_p= 0,$$

where $\|~\|_p$ is the $L^p$ norm, and we integrate with respect to the omitted variable, then we say that $f$ **is $L^p$ differentiable, and $g$ is the $L^p$ derivative of $f$**.

Now, suppose $f \in L^p$ and $f$ is differentiable **in the usual sense**. Moreover, assume that $f^{\prime} \in L^p$. My questions are:

- How to compute the $n_{th}$ derivative of a composition: ${\left( {f \circ g} \right)^{(n)}}=?$
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1)Does it follow that $f$ is also $L^p$ differentiable and $f^{\prime}$ is it’s $L^p$ derivative?

2)What if we assume that $f$ is $C^1$, in addition to the other hypotheses?

Here’s what I’ve tried: under the above hypotheses, we have that

$$\left| \frac{f(x + y) – f(x)}{y} – f^{\prime}(x)\right|^p \longrightarrow 0, ~\text{as}~y \rightarrow 0,$$

so we could apply the Dominated Convergence Theorem (switching from $y$ to an arbitrary sequence $(y_n)_n$ of course) if we could show that the above quantity is bounded by some fixed $L^p$ function of $x$ alone, that is, some function that doesn’t depend on $y$. This is where I got stuck; I can’t really eliminate the dependence on $y$.

Thank you.

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If $g=f’$ exists at every point, and is locally integrable, then the Fundamental theorem of calculus applies: $$f(x)= C+\int_0^x g(t)\,dt\tag1$$

For any $g\in L^p$ we have $\frac{1}{y}\int_{x}^{x+y}g(t)\,dt \to g$ in $L^p$. This is a standard fact about convolution with a mollifier (e.g., here). In terms of $f$, this means $\frac{1}{y}(f(\cdot+y)-f(\cdot))\to g$ in $L^p$, which was to be proved.

Personally, I would rather work with the assumption that (1) holds for some $g\in L^p$, instead of assuming everywhere differentiability of $f$. This gives a more general result ($f$ need not be everywhere differentiable) with less effort (no need to prove (1)). Pointwise-defined discontinuous derivatives brings much pain with little gain.

The functions $f$ for which (1) holds for some $g\in L^p$ belong to Sobolev space $W^{1,p}$ on every bounded interval. The result can be phrased as: for Sobolev functions, the $L^p$ derivative exists and agrees with the weak derivative (for $1\le p<\infty$).

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