Various kinds of derivatives

Solution 1:

Let's begin with some preliminaries.

For a locally integrable function $g$, let $\iota_g$ denote the distribution $\varphi \mapsto \int g(x)\varphi(x)\,dx$ (when we want to distinguish between the function and the induced distribution, otherwise we also denote the distribution with $g$). Let $D$ be the distributional derivative operator, i.e. $DT[\varphi] = -T[\varphi']$. For an integrable function $h$, let $I(h) = \int_{-\infty}^\infty h(x)\,dx$. Let $u$ be your favourite test function with integral $1$. Then we can write every test function $\varphi$ as $\varphi = I(\varphi)\cdot u + \eta'$, where $\eta(x) = \int_{-\infty}^x \varphi(t) - I(\varphi)u(t)\,dt$. We have the

Lemma: If $T$ is a distribution such that $DT = \iota_g$ for a locally integrable function $g$, then $T = \iota_{G+c}$ where $G(x) = \int_0^x g(t)\,dt$ and $c$ is an appropriate constant.

Proof: $G$ is a continuous (absolutely continuous, even) function; its distributional derivative is $\iota_g$:

$$\begin{align} D\iota_G[\varphi] &= - \int_{-\infty}^\infty G(x)\varphi'(x)\,dx\\ &= -\int_0^\infty \left(\int_0^xg(t)\,dt\right) \varphi'(x)\,dx + \int_{-\infty}^0 \left(\int_x^0g(t)\,dt\right) \varphi'(x)\,dx\\ &= \int_0^\infty \left(\int_t^\infty (-\varphi'(x))\,dx\right) g(t)\,dt + \int_{-\infty}^0 \left(\int_{-\infty}^t \varphi'(x)\,dx\right) g(t)\,dt\\ &= \int_0^\infty\varphi(t)g(t)\,dt + \int_{-\infty}^0 \varphi(t)g(t)\,dt\\ &= \iota_g[\varphi]. \end{align}$$

Let $c := T[u] - \iota_G[u]$. Then $T = \iota_{G+c}$:

$$\begin{align} \iota_{G+c}[\varphi] &= \iota_{G+c}[I(\varphi)\cdot u + \eta']\\ &= I(\varphi)\iota_{G+c}[u] - D\iota_{G+c}[\eta]\\ &= I(\varphi)(\iota_G[u] + c) - \iota_g[\eta]\\ &= I(\varphi)T[u] - DT[\eta]\\ &= I(\varphi)T[u] + T[\eta']\\ &= T[I(\varphi)\cdot u + \eta']\\ &= T[\varphi]. \end{align}$$

Together with the fact that for a locally integrable function $g$ we have $\iota_g = 0 \iff g = 0\: [\text{a.e.}]$, that will give us a stronger form of 4. and it's also useful in 2. and 5.

1. We have the somewhat stronger result that it suffices that $f'_c$ exists almost everywhere and $f$ be absolutely continuous. The proof tomasz gave covers that situation.

2. See the first part of the proof of the lemma. The integral of $f_c'$ has distributional derivative $\iota_{f_c'}$. But under the hypotheses that $f_c'$ exists everywhere, $f$ is the integral of $f_c'$ (plus a constant), so $f_c' = f_d'$.

3. Also true. Let $\displaystyle q_h(f)(x) =\frac{f(x+h)-f(x)}{h}$. The assumption is that $q_h(f) \to f_p'$ in $L^p$, but then $q_h(f) \to f_p'$ in $\mathscr{D}'$. But $q_h(f) \to Df$ in $\mathscr{D}'$, so $\iota_{f_p'} =D\iota_f$. We can also see that without theory: $$\begin{align} \int_{-\infty}^\infty f_p'(x)\varphi(x)\,dx &= \lim_{h\to 0} \int_{-\infty}^\infty q_h(f)(x)\varphi(x)\,dx\qquad \left(\varphi \in L^q\right)\\ &= \lim_{h\to 0} \frac1h \int_{-\infty}^\infty \left(f(x+h) - f(x)\right)\varphi(x)\,dx\\ &= \lim_{h\to 0} \frac1h \int_{-\infty}^\infty f(x)\left(\varphi(x-h) - \varphi(x) \right)\,dx\\ &= \lim_{h\to 0} \int_{-\infty}^\infty f(x) \frac{\varphi(x-h)-\varphi(x)}{h}\,dx\\ &= - \int_{-\infty}^\infty f(x)\varphi'(x)\,dx\qquad \left(q_{-h}(\varphi) \to \varphi' \text{ in } L^q\right) \end{align}$$

4. By the lemma, if $f_d'$ is locally integrable, then $f$ is (almost everywhere) the integral of $f_d'$ (plus a constant), so $f$ is differentiable at least in all Lebesgue points of $f_d'$ with $f_c'(x) = f_d'(x)$ there. If $f$ and $f_d'$ are continuous, then $f$ is everywhere differentiable with derivative $f_c' = f_d'$.

5. Of course $f_d' \in L^p$ does not imply that $f\in L^p$, so that needs to be an additional assumption. But if $f \in L^p$ and $f_d' \in L^p$, then, by 4., $f$ is - possibly after modification on a null set - the integral of $f_d'$ and $f$ is almost everywhere differentiable, so the more general version of point 1. yields that $f_p'$ exists and equals $f_c' = f_d'$.

Solution 2:

Partial answer:

1. is true: \begin{multline} \int_{\bf R} \left\lvert \frac{f(x+h)-f(x)}{h}-f'_c(x)\right\rvert^p\, dx\leq\frac{1}{h}\int_{\bf R}\int_0^h \left\lvert f'_c(x+t)-f'_c(x)\right\rvert^p\,dt \, dx=\\ =\frac{1}{h}\int_0^h\int_{\bf R} \left\lvert f'_c(x+t)-f'_c(x)\right\rvert^p\,dx \, dt \end{multline} And the last expression tends to zero as $h\to 0$ (because translations are continuous in $L^p$).
2. is true -- this is a simple application of product rule and fundamental theorem of calculus. It is applicable, because an everywhere differentiable function with integrable derivative is absolutely continuous.
3. I don't really know.
4. is true (up to a modification on a measure zero set). Consider $g(x)=\int_0^x f'_d$. Then $g$ is a $C^1$ function, and $f'_d$ is its classical derivative, and hence it's also its distributional derivative, so $f-g$ has weak derivative zero, so it must be constant (up to a set of measure zero), so $f$ is a $C^1$ function up to a set of measure $0$.