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Let $f:\mathbb{R}\to\mathbb{C}$ be a function Lebesgue-integrable on any finite interval and let $K$ be the space of infinitely differentiable equal to 0 outside a given finite interval. Be the generalised derivative of $f$ defined as the distribution $T':K\to\mathbb{C}$ defined by $$T'(\varphi)=-\int_{\mathbb{R}}f(x)\varphi'(x)d\mu$$where the integral is Lebesgue's. I read in Kolmogorov-Fomin's Элементы теории функций и функционального анализа that if $f$ is a function of bounded variation and its derivative as a function corresponds to its generalised derivative as a distribution, i.e. if $\int_{\mathbb{R}}f'(x)\varphi(x)d\mu=-\int_{\mathbb{R}}f(x)\varphi'(x)d\mu$ for any $\varphi\in K$, then $f$ is absolutely continuous.

I think I have been able to prove, by using the decomposition $f=H+\psi+\chi$ where $H$ is a step function, $\psi$ is absolutely continuous and $\chi$ is singular (i.e. with $\chi'(x)=0$ for almost all $x\in\mathbb{R}$), that, if the derivative of $f$ as a function corresponds to its generalised derivative as a distribution, then $f$ is identical to an absolutely continuous function almost everywhere, but I am not sure it holds everywhere and, if it did, I am not able to prove it. Does anybody know more about this fact? $\infty$ thanks!

I apologise in advance if I had used some scarcely formal and rigourous language, but the book I am following is quite informal and I fear I am assimilating not-so-good customs.

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Actually, this problem doesn't depend on knowing much about $f$. If $f$, $g$ are locally integrable on $\mathbb{R}$, and if $$ \int f\varphi'\,dx = -\int g\varphi\,dx,\;\;\; \varphi\in C_{0}^{\infty}(\mathbb{R}) $$ then $f$ is equal a.e. to a continuous function $\tilde{f}$, and $\tilde{f}$ is absolutely continuous with $\tilde{f}'=g$ a.e..

To see this, choose $\varphi'$ to converge to $\frac{1}{\epsilon}\chi_{[a-\epsilon,a]}-\frac{1}{\delta}\chi_{[b,b+\delta]}$ for $0 < \epsilon,\delta$ and $a \le b$, and $\varphi$ to converge to the integral of this function (which has $0$ total integral.) Then you get $$ \frac{1}{\epsilon}\int_{a-\epsilon}^{a}f\,dx-\frac{1}{\delta}\int_{b}^{b+\delta}f\,dx =-\int_{a-\epsilon}^{b+\delta}g\left[\int_{a-\epsilon}^{x}\frac{1}{\epsilon}\chi_{[a-\epsilon,a]}+\frac{1}{\delta}\chi_{[b,b+\delta]}\,dt\right]dx. $$ The inner integral on the right converges to $\chi_{[a,b]}$ as $\epsilon\downarrow$ and $\delta\downarrow 0$, and it remains uniformly bounded by $1$ in the process. So the limit of the expression on the right exists as $\epsilon \downarrow 0$, $\delta\downarrow 0$, whether one at a time, or together. That means that the limits on the left also exist at every $a$, $b$. We know that the limits on the left are left- and right-hand derivatives of $\int_{0}^{x}f\,dt$, and by the Lebesgue differentiation theorem, those limits are equal a.e. to $f$. So, we have the a.e. equality $$ f(a)-f(b) = -\int_{a}^{b}g\,dx. $$ The function on the right is continuous in $a$, $b$, which means that $f$ is equal a.e. to a continuous function $\tilde{f}$, and $\tilde{f}$ is absolutely continuous with $\tilde{f}'=g$ a.e..

Mollifier: You are concerned that the weak relation $$ \int_{\mathbb{R}}f'\varphi d\mu = -\int_{\mathbb{R}}f\varphi' d\mu,\;\; \varphi\in\mathcal{C}^{\infty}_{0}(\mathbb{R}) $$ cannot necessarily be strengthened to allow $\varphi$ to be a compactly supported absolutely continuous function instead. You can prove that such a thing can be done by finding $\eta \in \mathcal{C}^{\infty}_{0}(\mathbb{R})$ which is

  • non-negative, non-vanishing and constant in a neighborhood of $x=0$,
  • symmetric about $x=0$,
  • supported in $[-1,1]$,
  • bounded between $0$ and $\eta(0)$, and
  • normalized so that $\int \eta d\mu =1$.

Then one defines $$ \eta_{n}(x) = n\eta(nx) $$ so that $\int_{\mathbb{R}}\eta_{n}\,d\mu =1$ for all $n$. For any compactly supported absolutely continuous function $\varphi$, define $$ \varphi_{n} = \int_{\mathbb{R}}\eta_{n}(x-y)\varphi(y)\,d\mu(y). $$ The function $\varphi_{n}$ is in $\mathcal{C}^{\infty}_{0}(\mathbb{R})$. Because $\eta_{n}$ is supported in $[-1/n,1/n]$, and $\varphi$ is continuous, then $\varphi_{n}$ converges uniformly to $\varphi$ as $n\rightarrow\infty$. And, because $\varphi$ is absolutely continuous, then $$ \varphi_{n}' = \int_{\mathbb{R}}\eta_{n}'(x-y)\varphi(y)\,d\mu(y)= -\int_{\mathbb{R}}\eta_{n}(x-y)\varphi'(y)\,d\mu(y). $$ For my case $\varphi'$ is piecewise continuous, and the right side then converges pointwise everywhere to mean of the left- and right-hand limits of $\varphi'$, and it remains uniformly bounded by any bound for $\varphi'$. Thus, your weak equation $$ \int f'\varphi_{n}d\mu = -\int f\varphi_{n}'d\mu $$ becomes the following in the limit $$ \int f'\varphi d\mu = -\int f \varphi' d\mu. $$ For a general absolutely continuous $\varphi$, you can show that $\varphi_{n}$ converges uniformly to $\varphi$ and $\varphi_{n}'$ converges in $L^{1}(\mathbb{R})$ to $\varphi'$.

Mollifiers are very useful for extending integral equations for $\mathcal{C}^{\infty}_{0}$ functions to more general functions. If $\varphi$ has $k$ continuous derivatives that are compactly supported, then $\varphi_{n}$ and all $k$ derivatives converge uniformly to the corresponding derivatives of $\varphi$.

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    $\begingroup$ @DavideZena : The inner integral on the right is the piecewise linear function that is $0$ for $x \le a-\epsilon$, ramps linearly to $1$ at $x=a$, stays $1$ to $x=b$, and then ramps linearly back to $0$ at $x=b+\delta$, after which is stays $0$. Visualize what happens as either $\epsilon$ or $\delta$ tend down to $0$. You have bounded convergence. $\endgroup$ – Disintegrating By Parts Oct 31 '14 at 22:08
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    $\begingroup$ @DavideZena : The Bounded Convergence Theorem is a corollary of the Dominated Convergence Theorem: en.wikipedia.org/wiki/… . As the inner integral converges, it does so in a way that it remains uniformly bounded. But you can also argue directly without much trouble. $\endgroup$ – Disintegrating By Parts Nov 1 '14 at 0:03
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    $\begingroup$ @DavideZena : That's not what exists. It's the integral of that function $h_{a,b,\epsilon,\delta}(x)=\int_{-\infty}^{x}\left[\frac{1}{\epsilon}\chi_{[a-\epsilon ,a]}(t)+\frac{1}{\delta}\chi_{[b,b+\delta]}(t)\right]\,dt$ that has a limit as either or both $\delta$, $\epsilon$ tend to $0$. Therefore, $\int_{\mathcal{R}}g(x)h_{a,b,\epsilon,\delta}(x)dx$ has a limit as either or both of $\epsilon,\delta$ tend to $0$. Hence, the left side of the equation must have limits, which means that $\frac{1}{\epsilon}\int_{a-\epsilon}^{a}fdx$ has a limit as $\epsilon\downarrow 0$, etc. $\endgroup$ – Disintegrating By Parts Dec 31 '14 at 13:29
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    $\begingroup$ @DavideZena : it doesn't, but the integral with respect to it must converge, because the other side of the integral weak equation converges, which is how you conclude that $\frac{1}{\epsilon}\int_{a-\epsilon}^{a}f\,dx$ converges as $\epsilon\downarrow 0$. I've done nothing illegal in arriving at the equation; then I take limits of that legal equation, knowing that the right side converges, which forces the left side to converge. :) $\endgroup$ – Disintegrating By Parts Dec 31 '14 at 14:15
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    $\begingroup$ @DavideZena : Identify $\varphi$ and identify its derivative $\varphi'$. $\varphi$ is a piecewise linear function. $\endgroup$ – Disintegrating By Parts Dec 31 '14 at 15:51

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