Weak convergence and tightness

A regulator produces a sequence of probability laws. We usually cannot compare individual random field configurations point by point, but we can compare the expectations of observables. Weak convergence makes this idea precise.

Prerequisites: probability measures and distributions.
Outcomes: recognize convergence through test observables; identify escape of probability; separate existence of subsequential limits, uniqueness, and physical identification.

A limit seen by observables

On $\mathbb R$, write $\mu_n\Rightarrow\mu$ when

$$\int f\,d\mu_n\longrightarrow\int f\,d\mu$$

for every bounded continuous real function $f$. Boundedness prevents the test itself from giving unlimited weight to a tiny distant tail. Continuity prevents an arbitrarily sharp detector at a moving boundary.

For $\mu_n=\delta_{1/n}$, the integral is $f(1/n)$, which tends to $f(0)$. Thus $\delta_{1/n}\Rightarrow\delta_0$. The event $\{0\}$ nevertheless has probability zero for each $n$ and probability one in the limit. Weak convergence does not promise convergence of probabilities for every set. Boundary mass matters.

For real random variables, the equivalence between distribution-function convergence at continuity points and this test-function formulation is Durrett, version 5, Theorem 3.2.9. Theorems 3.2.12–3.2.14 state the corresponding real-line selection and tightness results. Their role here is compactness, not a QFT reconstruction theorem.

Probability must not run away

The sequence $\delta_n$ has no weakly convergent subsequence to a probability measure on $\mathbb R$. Every bounded interval eventually contains none of its probability.

A family is tight if, for every $\epsilon>0$, there is one compact set $K_\epsilon$ with

$$\mu_n(K_\epsilon)\geq1-\epsilon \quad\hbox{for all }n.$$

On the real line a sufficiently large closed bounded interval is a compact set. The same interval must work for the whole family. Tightness excludes $\delta_n$ but includes $\delta_{1/n}$.

Here is a useful sufficient estimate that can be checked directly. If $\sup_n\mathbb E[X_n^2]\leq M$, then

$$\mathbb P(|X_n|>R)\leq \frac{M}{R^2}.$$

Choose $R>\sqrt{M/\epsilon}$. The compact interval $[-R,R]$ contains at least $1-\epsilon$ of every law. This is a tail estimate uniform in the sequence index. It does not say that the individual variables are almost surely bounded.

Two failures that compactness cannot repair

First, a tight sequence need not converge without choosing a subsequence. Let $\mu_{2n}=\delta_0$ and $\mu_{2n+1}=\delta_1$. All laws are supported in $[0,1]$, but a bounded continuous function that separates 0 from 1 has alternating expectations. There are two different subsequential limits.

Second, weak convergence need not preserve unbounded moments. Set

$$\mu_n=(1-n^{-2})\delta_0+n^{-2}\delta_n.$$

For bounded $f$, the difference between $\int f\,d\mu_n$ and $f(0)$ is at most $2\|f\|_\infty/n^2$, so $\mu_n\Rightarrow\delta_0$. Yet $\int x^2d\mu_n=1$ for every $n$, while the limiting second moment is zero. Uniform integrability, or stronger suitable tail estimates, is needed to justify passing such moments through the limit.

This is the reason field-theory arguments distinguish convergence of bounded observables from convergence of correlation functions.

What changes for fields?

A continuum field often takes values in a distribution space. The configuration space and its topology must be stated before the word “compact” has a meaning. A bounded set in an infinite-dimensional normed space need not be compact. Bounds for each separately chosen smeared variable also need not establish tightness of the entire field in the intended topology.

A typical construction therefore has separate tasks:

1. Place all regulated objects in a common space.
2. Prove estimates that give compactness in a specified topology.
3. Extract a subsequential law and define its observables.
4. Show all relevant subsequences describe the same theory, when uniqueness is claimed.
5. Establish symmetry, positivity, nontriviality, and physical reconstruction for that law.

This is a map of tasks, not a universal theorem that any moment estimate completes. The uniform-estimates bridge supplies the physical-units check that also belongs in the argument.

Exercise: what exactly is missing?

Suppose lattice correlators converge along two subsequences to $C^{(1)}$ and $C^{(2)}$, and both decay exponentially. Can you conclude that there is one continuum theory with a mass gap?

Solution. No. One must identify or relate the two limits, show that their full compatible observable families define the intended QFT, and connect sufficiently rich correlation estimates to the entire physical spectrum. Exponential decay of a single channel does not supply those missing statements. Even the decay constants must be interpreted in common physical units.

Return to lesson 8 and limits and uniformity. The official problem description, §6 places regulator control within the larger construction problem.