$W^{1,1}\subseteq AC$ and a certain property implies BV: why?

Question

Brézis states that the functions in $W^{1,1}(I)$, with $I$ a bounded interval, are absolutely continuous, and that, for $u\in L^1(I)$, if the following holds for some constant $C$:

$$\left|\int_Iu\phi'\right|\leq C\|\phi\|_{L^\infty},$$

for all $\phi\in\mathcal{C}^1_c(I)$, then $u$ is of bounded variation. I cannot quite see why this holds. How can I prove these statements? And are these characterizations or only implications? For example, is $W^{1,1}(I)=AC(I)$ or is it just $W^{1,1}(I)\subseteq AC(I)$?

Answer 1

There is a difference in that the elements of $W^{1,1}$ are equivalence classes of functions (up to equality a.e.), while the elements of AC and BV may (or may not) be understood as actual functions, for which changing the value at a point matters. That said, there is a close relation:

$u\in W^{1,1}(I)$ if and only if $u$ has an absolutely continuous representative.

A proof may be found in most Sobolev spaces books under "absolute continuity on lines", for example section 10.3 in Sobolev Spaces by Leoni. Usually one proves the $n$-dimensional case, but the case $n=1$ is what I stated above.

Summary: A function is in $W^{1,1}(I)$ iff its distributional derivative is an $L^1$ function, and the antiderivatives of $L^1$ functions are precisely absolutely continuous functions.

With BV it's more complicated, because there are two competing definitions (in one dimension):

1. Definition based on supremum of sums like $\sum |u(x_k)-u(x_{k-1})|$.
2. Definition based on distributional derivatives: the derivative must be a finite signed measure.

The first definition cares about pointwise values, the second does not (it operates with equivalence classes up to equality a.e.). The aforementioned book by Leoni uses notation BPV (bounded pointwise variation) for the former and BV for the latter. For example, $\chi_{\mathbb{Q}}$ is not in BPV, but is in BV (because it's the zero function as far as BV is concerned).

By the Riesz representation theorem, the property $\left|\int_Iu\phi'\right|\leq C\|\phi\|_{L^\infty}$ is equivalent to saying that the distribution $u'$ is a finite signed measure, and thus is equivalent to $u\in BV$.

Answer 2

Adding to @WORDS's answer, I thought up a way to prove $W^{1,1}\subseteq AC$. Here it is. Let $u\in W^{1,1}(a,b)$. Let $\rho_n$ be the standard mollifiers. We have:

\begin{align*} \rho_n\ast u,\rho_n\ast u'\in{}&\mathcal{C}^\infty(a,b)\,\,\forall n; \\ \rho_n\ast u\to{}&u\,\,\text{in }L^1(a,b); \\ \rho_n\ast u'\to{}&u'\,\,\text{in }L^1(a,b); \\ \rho_n\ast u'={}&(\rho_n\ast u)'. \end{align*}

Let $u_n:=\rho_n\ast u,u'_n:=\rho_n\ast u'$. $u_n'$ seeming to be the derivative of $u_n$ is no ambiguity, as the last equality above states. By FTIC we have:

$$u_n(x)-u_n(y)=\int_y^xu_n'(t)dt,$$

for all $x,y\in(a,b)$ and $n\in\mathbb{N}$. We know that convergence in $L^1$ implies convergence pointwise a.e. for a subsequence $u_{n_k}$. In particular, there exists $x_0\in(a,b)$ s.t.:

$$u_{n_k}(x_0)\to u(x_0).$$

Writing FTIC for $y=x_0$, we get:

$$u_{n_k}(x)-u_{n_k}(x_0)=\int_{x_0}^xu_{n_k}'(t)dt,$$

for all $n_k$ and $x$ (well, that would hold for $u_n$ for any $n$, but let's write what we will use). The LHS converges, by assumption, pointwise a.e. to $u(x)-u(x_0)$. The RHS converges to $\int_{x_0}^xu'(t)dt$, since:

$$\left|\int_{x_0}^x[u_{n_k}'(t)-u'(t)]dt\right|\leq\int_{x_0}^x|u_{n_k}'(t)-u'(t)|dt\leq\|u_{n_k}'-u'\|_{L^1(a,b)},$$

which is infinitesimal by assumption. Hence, we can say that, for a.e. $x\in(a,b)$:

$$u(x)-u(x_0)=\int_{x_0}^xu'(t)dt,$$

which implies $u$ is AC.

Now I will try concocting an argument showing the converse. Let $u$ be AC. Then we know $u$ is a.e. differentiable and:

$$u(x)-u(a)=\int_a^xu'(t)dt.$$

Of course we will need to show $u'$ is a weak derivative. So let $v\in\mathcal{D}(a,b)$. We want to show:

$$\int_a^bu'(t)v(t)dt=-\int_a^bu(t)v'(t)dt.$$

Letting $u_n,u_n'$ be mollifications as above, we certainly have:

$$\int_a^bu_n'(t)v(t)dt=-\int_a^bu_n(t)v'(t)dt,$$

by integration by parts and $u_n,v$ being smooth. Can we pass the limit under integral? I guess the argument is much the same: integral with $u$ minus integral with $u_n$, collapse to a single integral, bring the absolute value inside, $v'$ is test so it is bounded, which gives a constant, and what is left is $\|u_n-u\|_{L^1(a,b)}\to0$, and the same for the $u_n'$ with $v$ being test and hence bounded. Yeah, that works.

Addendum

To expand on @words's answer, call the following $(\ast)$:

$$\left|\int_Iu\phi'dt\right|\leq C\|\phi\|_{L^\infty}.$$

Consider the map $\mathcal{C}^\infty_c(I)\ni\phi\mapsto-\int_Iu\phi'$, the distributional derivative of $u$. $(\ast)$ is equivalent to saying said functional is continuous on $\mathcal{C}^\infty_c(I)$ viewed as a subspace of $L^\infty(I)$. Said subspace is not dense, as it would be in $L^p$ for $p\neq\infty$, but Hahn-Banach still extends the distributional derivative $u'$ of $u$ to an element of $(L^\infty(I))^\ast$. Sadly, this is not $L^1(I)$, but, as seems to be well-known, is the space of finitely additive signed measures on $I$. Thus, $(\ast)$ is equivalent to saying $u'$, the distributional derivative of $u$, extends to a finitely additive signed measure. This is one definition of BV. Unfortunately, this does not imply the second definition, which is the one I've been given in another course, since $\chi_{\mathbb{Q}}$ is not BV in my sense (BVP, as @words puts it), but is BV in the measure sense. Indeed:

1. For $n\in\mathbb{N}$, take $x_1=0,x_2=\frac1n,\dotsc,x_{n+1}=\frac nn=1$, and set $y_{2k}=x_{2k}$ for al suitable $k$'s, and for $y_{2k+1}$, take $x_{2k+1}$ and add $\dots110100100010000\dotsc$ at the end of its decimal expansion; this produces a partition of $[0,1]$ such that all terms $|\chi_{\mathbb Q}(x_i)-\chi_{\mathbb Q}(x_{i-1})|$ are 1 (either $|1-0|$ or $|0-1|$), hence the sum is $n$, so the total variation is the sup of a set that contains $\mathbb{N}$, hence $+\infty$, hence $\chi_{\mathbb{Q}}$ is not BVP;
2. Viewed as a functional on $\mathbb{C}^\infty_c(I)$ via integration, i.e. $\phi\mapsto\int_I\chi_{\mathbb{Q}}(x)\phi(x)dx$, $\chi_{\mathbb{Q}}$ is the zero functional, hence it is a finitely additive signed measure: the zero measure.

Indeed, in general, if $f=g$ a.e. on $I$, $f=g$ as a distribution, but the sums of the BVP definition can vary wildly.