# $L^p$ derivative vs normal derivative.

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:

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.

• This notion of $L^p$-differentiability is new to me, so just to make sure, the norm in the definition is taken as a function of $x$, correct? So that there's no reason to quantify $\forall x$ later? Aug 19 '13 at 21:39
• @JonathanY. Certainly, I apologize for that. I'll fix it right away.
– L..
Aug 19 '13 at 21:39

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 $1\le p<\infty$).