Convergence of Functions when viewed as Distributions and other Convergence Conditions

Let $\Omega$ be a nonempty open subset of $\mathbb{R}^k$, $\mathfrak{B}$ the collection of all borel sets of $\mathbb{R}^k$ contained in $\Omega$, and $\mu$ the $k$-dimensional Lebesgue measure. Consider a sequence $\{ f_n \}$ of Borel measurable complex functions on $\Omega$ such that $f_n \in L_{loc}^{1}(\Omega)$ for all $n$, which means that for every compact $K \subseteq \Omega$ and every $n$, we have \begin{equation} \int_{K} \left| f_n \right| d \mu < \infty. \end{equation} Let $C_{c}(\Omega)$ be the set of all continuous functions $\phi:\Omega \rightarrow \mathbb{C}$ whose support is a compact subset of $\Omega$, and let $C_{c}^{\infty}(\Omega)$ be the set of all infinitely differentiable functions $\phi:\Omega \rightarrow \mathbb{C}$ whose support is a compact subset of $\Omega$. Let $\mathfrak{B}_{loc}$ be the set of all $E \in \mathfrak{B}$ such that the closure $\bar{E}$ of $E$ in $\mathbb{R}^k$ is a compact subset of $\Omega$. Finally, for every function $g:\Omega \rightarrow \mathbb{C}$, and every $T \subseteq \Omega$, we denote by $g_{| T}$ the restriction of $g$ to $T$.

Let $f$ be a Borel measurable complex function on $\Omega$ such that $f \in L_{loc}^{1}(\Omega)$, and consider the following four properties.

(A) For every compact subset $K$ of $\Omega$, we have \begin{equation} \lim_{n \rightarrow \infty} f_{n | K} = f_{| K} \end{equation} in $L^1(K)$.

\begin{equation} (B) \lim_{n \rightarrow \infty} \int_{E} f_n d\mu = \int_{E} f d\mu \qquad \forall E \in \mathfrak{B}_{loc} \end{equation}

\begin{equation} (C) \lim_{n \rightarrow \infty} \int_{\Omega} f_n \phi d\mu = \int_{\Omega} f \phi d\mu \qquad \forall \phi \in C_{c}(\Omega) \end{equation}

\begin{equation} (D) \lim_{n \rightarrow \infty} \int_{\Omega} f_n \phi d\mu = \int_{\Omega} f \phi d\mu \qquad \forall \phi \in C_{c}^{\infty}(\Omega) \end{equation}

It is immediate to prove that (A) implies (B), (C) and (D). Trivially, (C) implies (D). My question is: are there other implications among these three properties? Any proof or counter-example is welcome.

PS Clearly, by considering $g_n=f_n - f$, we can assume without loss of generality that $f=0$.

PSS For those who know distribution theory, note that, given a Borel measurable complex function $g$ on $\Omega$ such that $g \in L_{loc}^{1}(\Omega)$, if we set for every $\phi \in C_{c}^{\infty}(\Omega)=\mathcal{D}(\Omega)$ \begin{equation} T_{g}(\phi) = \int_{\Omega} g \phi d \mu, \end{equation} then (D) says that $T_{f_n} \rightarrow T_{f}$ in the weak*-topology of $\mathcal{D}^{'}(\Omega)$.

I have finally settled the problem. Since the solution is quite long, I will split it into four parts.

Let us start by proving that, surprisingly enough for me, (B) implies (C).

Let $K$ be a compact subset of $\Omega$. I will denote by $\mathfrak{B}_{K}$ the collection of all Borel sets of $\mathbb{R}^k$ contained in $K$. I will assume, without loss of generality, that $f=0$. Moreover, I will consider all the functions as restricted to $K$: in particular, with abuse of notation, I will denote by $f_n$ the restriction of the original $f_n:\Omega \rightarrow \mathbb{C}$ to $K$.

First of all, some terminology. Given a sequence of complex measures $\{ \lambda_n \}$ on $\mathfrak{B}_{K}$, we shall say that $\{ \lambda_n \}$ is uniformly absolutely continuous if for every $\epsilon > 0$, there exists $\delta > 0$, such that for every $E \in \mathfrak{B}_{K}$, with $\mu(E) < \delta$, we have $| \lambda_n(E) | < \epsilon$ for all $n$. We shall denote with $|\lambda_n|$ the total variation of $\lambda_n$. We need the following preliminary result.

Lemma

A sequence of complex measures $\{ \lambda_n \}$ on $\mathfrak{B}_{K}$ is uniformly absolutely continuous if and only if $\{ |\lambda_n| \}$ is uniformly absolutely continuous.

Proof. The ''if'' part is trivial. Let us prove the ''only if'' part. Assume by contradiction that for some $\epsilon > 0$ there exists a sequence $\{ E_m \}$ in $\mathfrak{B}_{K}$, such that $\mu(E_m) \rightarrow 0$ and for every $m$ there exists $n_m$ such that $|\lambda_{n_m}|(E_n) \geq \epsilon$. Fix $m$, and take a countable partition $\{ A_j \}$ of $E_m$, with $A_j \in \mathfrak{B}_{K}$ for every $j$, and such that $\sum_{j=1}^{\infty} |\lambda_{n_m} (A_j)| \geq \frac{\epsilon}{2}$. Choose $\bar{j}$ such that $\sum_{j=1}^{\bar{j}} |\lambda_{n_m} (A_j)| \geq \frac{\epsilon}{4}$. From Lemma 6.3 in [R], we deduce that $\left| \sum_{j=1}^{\bar{j}} \lambda_{n_m} (A_j) \right| \geq \frac{\epsilon}{4 \pi}$. Then put \begin{equation} D_m = \bigcup_{j=1}^{\bar{j}} A_j. \end{equation} You get $\mu(D_m) \leq \mu(E_m)$, so that $\mu(D_m) \rightarrow 0$, and $\left| \lambda_{n_m}(D_n) \right| \geq \frac{\epsilon}{4 \pi}$ for all $m$, a contradiction.

QED

Now, let us come back to our problem. Assume that (B) holds, and set for all $n$ \begin{equation} \lambda_n(E) = \int_{E} f_n d \mu \qquad (E \in \mathfrak{B}_K). \end{equation} From the converse of Vitali's Theorem (see [R], Exercise 6.10(g)), we deduce that $\{ \lambda_n \}$ is uniformly absolutely continuous, and by the previous lemma we deduce that $\{ \left| \lambda_n \right| \}$ is uniformly absolutely continuous. From [R], Theorem 6.13 we know that for any $n$: \begin{equation} \left| \lambda_n \right|(E) = \int_{E} \left| f_n \right| d \mu \qquad (E \in \mathfrak{B}_{K}). \end{equation} So, if $\eta > 0$, there exists $\delta > 0$ such that \begin{equation} \int_{E} \left| f_n \right| d \mu < \eta, \end{equation} for any $E \in \mathfrak{B}_{K}$ such that $\mu(E) < \delta$. Let $\{ E_{1},\dots, E_{m} \}$ be a finite subset of $\mathfrak{B}_{K}$, such that $\mu(E_j) < \delta$ for $j=1,\dots,m$, and \begin{equation} K = \bigcup_{j=1}^{m} E_j, \end{equation} To see that such a collection of sets exists, fix a positive integer $p$ such that $2^{kp} > \frac{1}{\delta}$, and consider the subdivision of $\mathbb{R}^{k}$ in dyadic $k$-cells \begin{equation} W = \left \{ (x_1,\dots,x_k) \in \mathbb{R}^k : \frac{j_i}{2^p} \leq x_i < \frac{j_i + 1}{2^p}, \quad i=1,\dots, k \right \}, \end{equation} where $( j_1, \dots, j_k )$ ranges in $\mathbb{Z}^{k}$, with $\mathbb{Z}$ denoting the set of all integer numbers. Take the intersections of these $k$-cells with $K$ to get the required collection.

Now, we have for any $n$ \begin{equation} \int_{K} \left| f_n \right| d \mu \leq \sum_{j=1}^{m} \int_{E_j} \left| f_n \right| d \mu \leq m \eta, \end{equation} so that $\{ f_n \}$ is bounded in $L^{1}(K)$ by $M= m \eta$.

Suppose that $\phi \in C_{c}(\Omega)$, that $\phi$ is real, with $\phi \geq 0$, and that the support of $\phi$ is contained in $K$. Let $\epsilon > 0$. Since $\phi$ is continuous, it is bounded, and from the construction in [R], Theorem 1.17, we deduce the existence of a simple function $s:K \rightarrow [0,\infty)$ such that $0 \leq \phi(x) - s(x) \leq \epsilon$ for all $x \in K$. From our hypothesis there exists $\nu > 0$ such that for $n > \nu$ we have \begin{equation} \left| \int_{K} f_n s d \mu \right| < \epsilon. \end{equation} We then have for any $n > \nu$ \begin{equation} \left| \int_{K} f_n \phi d \mu \right| \leq \left| \int_{K} f_n s d \mu \right| + \left| \int_{K} f_n (\phi - s) d \mu \right| < \epsilon + M \epsilon, \end{equation} and so \begin{equation} \lim_{n \rightarrow \infty} \int_{\Omega} f_n \phi d \mu = \lim_{n \rightarrow \infty} \int_{K} f_n \phi d \mu = 0. \end{equation} If now $\phi \in C_{c}(\Omega)$, $\phi$ is real, and the support of $\phi$ is contained in $K$, by considering the positive part $\phi^{+}$ and negative part $\phi^{-}$ of $\phi$ we get again \begin{equation} \lim_{n \rightarrow \infty} \int_{\Omega} f_n \phi d \mu = 0. \end{equation} Finally, if $\phi \in C_{c}(\Omega)$ and the support of $\phi$ is contained in $K$, by considering the real and imaginary part we get \begin{equation} \lim_{n \rightarrow \infty} \int_{\Omega} f_n \phi d \mu = 0. \end{equation}

QED

Now, we shall show by using a counterexample that (C) does not imply (B). Take $k=1$, $\Omega = \mathbb{R}$, and define for any positive integer $n$: \begin{equation} f_n(x) = \begin{cases} 2n^2 + 4n^{5} \left(x - \frac{2j+1}{2n} \right) & \text{if } x \in \left[ \frac{2j+1}{2n} - \frac{1}{2n^3}, \frac{2j+1}{2n} \right] \quad (j=0,1,\dots,n-1), \\ 2n^2 - 4n^{5} \left(x - \frac{2j+1}{2n} \right) & \text{if } x \in \left[ \frac{2j+1}{2n}, \frac{2j+1}{2n} + \frac{1}{2n^3} \right] \quad (j=0,1,\dots,n-1), \\ 0 & \text{otherwise}. \end{cases} \end{equation} Then $f_n$ is a piecewise linear function, $f_n \geq 0$, $f_n(x)=0$ if $x \notin I = [0,1]$, and \begin{equation} \int_{I} f_n d \mu = 1. \end{equation} Moreover, if $S_n$ is the support of $f_n$, we have \begin{equation} \mu(S_n) = \frac{1}{n^2}, \end{equation} so that \begin{equation} \sum_{n=1}^{\infty} \mu(S_n) < \infty. \end{equation} For any positive integer $m$, let $n_m$ be an integer such that \begin{equation} \sum_{n \geq n_m} \mu(S_n) < \frac{1}{m}, \end{equation} and put $E_m = \bigcup_{n \geq n_m} S_n$. Then for any $A \in \mathfrak{B}$, with $A \subseteq \mathbb{R} \setminus E_m$, we have \begin{equation} \int_{A} f_n d \mu \rightarrow 0. \end{equation} Assume there exists a Borel measurable $f:\mathbb{R} \rightarrow \mathbb{C}$ such that $f \in L_{loc}^{1}(\mathbb{R})$ and such that for every $E \in \mathfrak{B}_{loc}$ we have \begin{equation} \int_{E} f_n d \mu \rightarrow \int_{E} f d \mu. \end{equation} Then, from what we have proved above and [R], Theorem 1.39 we get that for any positive integer $p$, we have $f=0$ a.e. $[ \mu ]$ on $[-p,p] \setminus E_{m}$, so that $f=0$ a.e. $[ \mu ]$ on $\mathbb{R} \setminus E_{m}$. We deduce that $f=0$ a.e. $[ \mu ]$ on $\bigcup_{m=1}^{\infty} \left( \mathbb{R} \setminus E_{m} \right)$. If we set $E = \bigcap_{m=1}^{\infty} E_m$, then we have $\mu(E)=0$, so we conclude that $f=0$ a.e. $[ \mu ]$ on $\mathbb{R}$. But we have \begin{equation} \lim_{n \rightarrow \infty} \int_{I} f_n d \mu = 1, \end{equation} so (B) cannot hold.

We shall now show that instead (C) holds, with $f$ defined by

\begin{equation} f(x) = \begin{cases} 1 & \text{if } x \in \left[0, 1 \right], \\ 0 & \text{otherwise}. \end{cases} \end{equation}

Let $\phi \in C_{c}(\mathbb{R})$. For any positive integer $n$ define the function \begin{equation} \phi_{n}(x) = \begin{cases} \phi \left( \frac{2j+1}{2n} \right) & \text{if } \left[ \frac{j}{n}, \frac{j+1}{n} \right), \quad (j=0,1,\dots, n-1), \\ 0 & \text{otherwise}. \end{cases} \end{equation}

Choose $\epsilon > 0$, and let $\delta > 0$ be such that $| \phi(x) - \phi(y) | < \epsilon$ for every $x, y$ such that $|x - y | < \delta$. Let $\bar{n}$ be an integer such that $\bar{n} > \frac{2}{\delta}$. Then we have \begin{equation} \left| \int_{\mathbb{R}} f_n \phi d \mu - \frac{1}{n} \sum_{j=0}^{n-1} \phi \left(\frac{2j+1}{2n} \right) \right| = \left| \int_{I} f_n \phi d \mu - \int_{I} f_n \phi_n d \mu \right| \leq \int_{I} f_n \left| \phi - \phi_n \right| d \mu \leq \epsilon \int_{I} f_n d \mu = \epsilon. \end{equation} From [Ru], Theorem 6.7 we have \begin{equation} \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{j=0}^{n-1} \phi \left(\frac{2j+1}{2n} \right) = \int_{0}^{1} \phi(x) dx, \end{equation} where the right-hand side integral is the Riemann integral of $\phi$ over $I$. From [Ru], Theorem 11.33 we also know that \begin{equation} \int_{0}^{1} \phi(x) dx = \int_{I} \phi d \mu. \end{equation} So we conlude that \begin{equation} \lim_{n \rightarrow \infty} \int_{\mathbb{R}} f_n \phi d \mu = \int_{I} \phi d \mu = \int_{\mathbb{R}} f \phi d \mu. \end{equation}

QED

Finally, we show by means of a counterexample that (D) does not imply (C). Take $k=1$, $\Omega= \mathbb{R}$, and define for any positive integer $n$: \begin{equation} f_n(x) = \begin{cases} 2 n^{\frac{11}{4}} x & \text{if } x \in \left[ 0, \frac{1}{2n} \right], \\ n^{\frac{7}{4}}(2-2nx) & \text{if } x \in \left[ \frac{1}{2n}, \frac{3}{2n} \right], \\ n^{\frac{7}{4}}(-4+2nx) & \text{if } x \in \left[ \frac{3}{2n}, \frac{2}{n} \right], \\ 0 & \text{otherwise} \end{cases} \quad. \end{equation} If $\phi \in C_{c}^{\infty}(\mathbb{R})$, then for some $L >0$ we have $| \phi'(x) | \leq L$ for all $x \in \mathbb{R}$. So by using Lagrange Mean Value Theorem, we get \begin{multline} \left| \int_{\mathbb{R}} f_n \phi d \mu \right| = \left| \int_{\left[ 0, \frac{1}{n} \right]} f_n \phi d \mu + \int_{\left[ \frac{1}{n}, \frac{2}{n} \right]} f_n \phi d \mu \right| = \left| \int_{0}^{ \frac{1}{n}} f_n(x) \phi(x) dx - \int_{0}^{ \frac{1}{n}} f_n(x) \phi \left(x+\frac{1}{n} \right) dx \right| \leq \\ \leq \int_{0}^{ \frac{1}{n}} f_n(x) \left| \phi(x) - \phi \left(x+\frac{1}{n} \right) \right| dx \leq \frac{L}{n} \int_{0}^{ \frac{1}{n}} f_n(x) dx = \frac{L}{n} \frac{n^{\frac{7}{4}}}{2n} = \frac{L}{2 n^{\frac{1}{4}}}, \end{multline} so that (D) holds with $f=0$. To see that (C) does no hold, define now \begin{equation} \phi(x) = \begin{cases} \sqrt{x} & \text{if } x \in \left[ 0, 1 \right], \\ 2 - x & \text{if } x \in \left[ 1, 2 \right], \\ 0 & \text{otherwise} \end{cases} \quad. \end{equation} An easy computation shows that for $n \geq 2$ we have \begin{equation} \int_{\mathbb{R}} f_n \phi d \mu = \frac{n^{\frac{1}{4}}}{15 \sqrt{2}} \left(-68 + 36 \sqrt{3} \right), \end{equation} so that \begin{equation} \lim_{n \rightarrow \infty} \int_{\mathbb{R}} f_n \phi d \mu = - \infty. \end{equation}

QED

References

[R], Rudin, Real and Complex Analysis, Third Edition

[Ru], Principles of Mathematical Analysis, Third Edition

• Even though, the three parts of the solution are independent, to read them in the order I wrote them, select "oldest" among the ordering options. Sep 16 '16 at 17:44

Now, I show by means of a counterexample that (B) does not imply (A). To see this, consider $k=1$, $\Omega = \mathbb{R}$, and define for every positive integer $n$: \begin{equation} f_n(x)= \begin{cases} 1 & \text{if } x \in \left[\frac{2j}{2^n}, \frac{2j+1}{2^n}\right), \quad (j=0,1,\dots, 2^{n-1}-1), \\ -1 & \text{if } x \in \left[\frac{2j+1}{2^n}, \frac{2j+2}{2^n}\right), \quad (j=0,1,\dots, 2^{n-1}-1), \\ 0 & \text{otherwise}. \end{cases} \end{equation} Take $f=0$. Then by outer regularity of $\mu$, for every $E \in \mathfrak{B}$ and every $\epsilon > 0$, you can find an open set $V$ of $\mathbb{R}$ such that $E \subseteq V$ and $\mu(V \backslash E) < \epsilon$. Set $W = V \cap (0,1)$. $W$ is an at most countable union of open and pairwise disjoint intervals $\{ I_m \}$. Choose $\bar{m}$ such that \begin{equation} \sum_{m > \bar{m}} \mu(I_m) < \epsilon. \end{equation}

Since $f_n$ is periodic on each of the intervals $I_m$, $m=1,\dots,\bar{m}$, and $|f_n(x)| = 1$ for all $x \in (0,1)$, you can find $\bar{n}$ such that for all $n > \bar{n}$ \begin{equation} \left| \int_{I_m} f_n d\mu \right| < \frac{\epsilon}{\bar{m}} \end{equation} for $m=1,\dots,\bar{m}$. We then have for $n > \bar{n}$: \begin{equation} \left| \int_{W} f_n d \mu \right| = \left| \sum_{m} \int_{I_m} f_n d \mu \right| \leq \sum_{m \leq \bar{m}} \left| \int_{I_m} f_n d \mu \right| + \left| \sum_{m > \bar{m}} \int_{I_m} f_n d \mu \right| \leq \epsilon + \sum_{m > \bar{m}} \int_{I_m} |f_n| d \mu \leq 2 \epsilon. \end{equation}

If $D = E \cap (0,1)$, then $\mu(W \setminus D) < \epsilon$, so we have for $n > \bar{n}$: \begin{equation} \left| \int_{E} f_n d \mu \right| = \left| \int_{D} f_n d \mu \right| \leq \left| \int_{W} f_n d \mu - \int_{D} f_n d \mu\right| + \left| \int_{W} f_n d \mu \right| \leq \int_{W \setminus D} |f_n| d \mu + 2 \epsilon < 3 \epsilon. \end{equation}

This proves (B). But (A) is not satisfied with $f=0$ because we have for any $n$ \begin{equation} \int_{\left[0,1 \right]} |f_n| d \mu = 1. \end{equation} Actually, (A) cannot be satisfied by any $f \in L_{loc}^{1}(\Omega)$. Indeed, assume it is. Then, (B) would hold with the same $f$. So from we what we have proved we deduce that for any interval $[a,b]$ and any $E \in \mathfrak{B}$ contained in $[a,b]$, we would have \begin{equation} \int_{E} f d \mu = 0. \end{equation} So from [R], Theorem 1.39, we would have $f=0$ a.e. on $[a,b]$. And so $f=0$ a.e. on $\Omega$. But then $f$ cannot satisfy (A), as we have seen.

PS We know now that (B) implies (C), so our sequence $\{ f_n \}$ satisfies for sure (C). This can also be given a simple direct proof as follows. Now, if $\phi \in C_{c}(\Omega)$, then for every $\epsilon > 0$, there exists $\delta > 0$ such that $|\phi(x) -\phi(y)| < \epsilon$ for every $x, y \in \Omega$ such that $|x-y|< \delta$. Choose $\bar{n}$ such that $2^{\bar{n}} > 1/\delta$. For every $n > \bar{n}$, and every $j=0,1,\dots,2^{n-1}-1$, we have \begin{multline} \left| \int_{\left[\frac{2j}{2^n}, \frac{2j+2}{2^n}\right]} f_n \phi d \mu \right | = \left| \int_{\left[\frac{2j}{2^n}, \frac{2j+1}{2^n}\right]} \phi d \mu - \int_{\left[\frac{2j+1}{2^n}, \frac{2j+2}{2^n}\right]} \phi d \mu \right | \leq \\ \leq \int_{\left[\frac{2j}{2^n}, \frac{2j+1}{2^n}\right]} \left| \phi(x) - \phi ( x+ 1/2^{n}) \right| d \mu \leq \frac{\epsilon}{2^{n}}. \end{multline} So for $n > \bar{n}$ we have \begin{equation} \left| \int_{\Omega} f_n \phi d \mu \right | \leq \frac{\epsilon}{2}, \end{equation} which proves (C).