Let's say we have a sequence of functions $f_n \in L^1(\Omega)$ for some nice bounded open set $\Omega \subset \mathbb{R}^N$. In general, even along a subsequence, we don't expect a weak limit (in the sense of testing against $L^{\infty}$ functions), so we often work with the weaker notion of weak$^*$ convergence, that is, a Radon measure $\mu$ such that

$$ \int_{\Omega} f_n \phi dx \rightarrow \int_{\Omega} \phi d \mu$$

for every $\phi \in C_b (\Omega)$.

My question is this: If we know through some other means that $\mu$ is absolutely continuous with respect to the Lebesgue measure and therefore given by some density $f \in L^1$, so that the above becomes

$$ \int_{\Omega} f_n \phi dx \rightarrow \int_{\Omega} f \phi dx$$

then how far is this from being able to say that the $f_n$ are actually weakly converging to $f$? In other words, does the above hold with any $L^{\infty}$ test function, not just $C_b$?

Obviously a simple density argument will not hold because $C_b(\Omega)$ is closed in the $L^{\infty}$ norm, but I am having trouble thinking of a counter-example.

It seems to me that if we could show for any Borel set $E$ that

$$ \int_{E} f_n dx \rightarrow \int_{E} f dx$$

then this would suffice- but obviously we do not have this in general for weak$^*$ convergence. Any thoughts?


Abstract counterexample

Let $E \subset \Omega$ be compact, nowhere dense, and have positive Lebesgue measure (like a fat Cantor set). Then $U = \Omega \setminus E$ is open and dense in $\Omega$. I claim that $L^1(U)$ is weak-* dense in $C_b(\Omega)^*$. This follows from Hahn-Banach, because $L^1(U)$ separates the points of $C_b(\Omega)$: suppose $\phi \in C_b(\Omega)$ is such that $\int \phi f = 0$ for all $f \in L^1(U)$. Take $f = \phi 1_U$; then this says $\int_U \phi^2 = 0$, so $\phi = 0$ almost everywhere on $U$. Since $U$ is open this means $\phi = 0$ everywhere on $U$, and since $U$ is dense we have $\phi = 0$ everywhere on $\Omega$.

In particular, this means you can choose a sequence $f_n \in L^1(U)$ converging weak-* to $f = 1_E$. But clearly $f_n$ does not converge weakly to $f$; take $\phi = 1_E$.

(I cheated a little: $C_b(\Omega)^*$ is not weak-* metrizable, so I cannot necessarily choose a sequence $f_n$ converging weak-* to $f$. You can fix this by choosing a compact regular $\Omega_0 \subset \Omega$ containing $E$ in its interior. Then do everything on $C(\Omega_0)^*$ where the bounded sets are weak-* metrizable.)

Hands-on counterexample

Let's work in one dimension. Let $\Omega = [0,1]$ (or any open set containing $[0,1]$, if you prefer). Let $E \subset [0,1)$ be a fat Cantor set which is closed, nowhere dense, and has positive Lebesgue measure $m(E) > 0$. Set $f = 1_E$. Now for each $n$, partition $[0,1)$ into the intervals $I_{n,j} = [(j-1)/n, j/n)$, $j=1,\dots,n$. Set $$f_n = \sum_{j=1}^n \frac{m(I_{n,j} \cap E)}{m(I_{n,j} \cap E^c)}1_{I_{n,j} \cap E^c}$$ so that $f_n$ is supported on $E^c \cap [0,1)$ and has the property that $\int_{I_{n,j}} f_n(x)\,dx = \int_{I_{n,j}} f(x)\,dx$ for each $j$.

Clearly we do not have $f_n \to f$ weakly (take $\phi = 1_E$).

But suppose $\phi$ is continuous on $\Omega$, hence uniformly continuous on $[0,1]$. Fix $\epsilon > 0$ and choose $N$ so large that the oscillation of $\phi$ is at most $\epsilon$ on every interval of length at most $1/N$. Then for any $n \ge N$ and any $I = I_{n,j}$ $$\begin{align*}\int_I \phi \cdot (f - f_n) &\le \max_I \phi \int_I f - \min_I \phi \int_I f_n \\ &= (\max_I \phi - \min_I \phi) \int_I f \\ &\le \epsilon \int_I f. \end{align*}$$ Summing over $j$ we have $\int \phi \cdot (f_n - f) \le \epsilon \int f = \epsilon m(E)$. A similar argument gives $\int \phi \cdot (f_n -f) \ge -\epsilon m(E)$. So we have $\left|\int \phi \cdot (f_n - f) \right| \le \epsilon m(E)$, and this shows that $\int \phi f_n \to \int \phi f$. So $f_n \to f$ in your weak-* sense.


Let us take $\Omega = (-1,1)$, and $f_n = n \, (\chi_{(0,1/n)} - \chi_{(-1/n,0)}$. It is easy to see that $$\int_\Omega f_n \, \varphi \, \mathrm d x \to 0$$ for all $C_b(\Omega)$, but $$\int_\Omega f_n \, \varphi \, \mathrm d x = 1$$ for $\varphi = \chi_{(0,1)} \in L^\infty(\Omega)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.