## Modes of convergence on abstract (and not so abstract) sets

In how many ways can a series of real-valued functions on an abstract set converge? Having no measure on the domain eliminates the infinitude of modes of convergence based on integral norms. I can think of five modes of convergence of ${\sum f_n}$ where ${f_n\colon X\to \mathbb R}$:

• (P) Pointwise convergence: ${\sum f_n(x)}$ converges for each ${x\in X}$.
• (U) Uniform convergence: the partial sums ${S_n}$ of the series converge to some function ${f\colon X\to\mathbb R}$ uniformly, i.e., ${\sup |S_n-f| \to 0}$.
• (PA) Pointwise convergence of absolute values: ${\sum |f_n(x)|}$ converges for each ${x\in X}$.
• (UA) Uniform convergence of absolute values: like uniform, but for ${\sum |f_n|}$.
• (M) Weierstrass M-test convergence: ${\sum \sup |f_n| }$ converges.

Implications (all easy): (M) implies (UA), which implies both (U) and (PA). Neither (U) nor (PA) implies the other one, but each of them implies (P).

Perhaps (U) and (PA) deserve an illustration, being incomparable. Let ${X = [0, 1]}$. The constant functions ${f_n(x)=(-1)^n/n}$ form a series that converges uniformly but not in the sense (PA). In the opposite direction, a series of triangles with height 1 and disjoint supports converges (PA) but not (U).

Notably, the sum of the latter series is not a continuous function. This had to happen: by Dini’s theorem, if a series of continuous functions is (PA)-convergent and its sum is continuous, then it is (UA)-convergent. This “self-improving” property of (PA) convergence will comes up again in the second part of this post.

## From abstract sets to normed spaces

In functional analysis, the elements of a normed space ${E}$ can often be usefully interpreted as functions on the unit ball of the dual space ${E^*}$. Indeed, each ${x\in E}$ induces ${f_x(z) = z(x)}$ for ${z\in E^*}$. Applying the aforementioned modes of convergence to ${\sum x_n}$ with ${x_n\in E}$, we arrive at

• (P) ⇔ Convergence in the weak topology of E.
• (U) ⇔ Convergence in the norm topology of E.
• (PA) ⇔ Unconditional convergence in the weak topology of E.
• (UA) ⇔ Unconditional convergence in the norm topology of E.
• (M) ⇔ Absolute convergence, i.e., ${\sum \|x_n\|}$ converges.

The equivalences (P), (U), (M) are straightforward exercises, but the unconditional convergence merits further discussion. For one thing, there are subtly different approaches to formalizing the concept. Following “Normed Linear Spaces” by M. M. Day, let’s say that a series ${\sum x_n}$ is

• (RC) Reordered convergent if there exists ${x}$ such that ${\sum x_{\pi(n)} =x}$ for every bijection ${\pi:\mathbb{N}\to\mathbb{N}}$
• (UC) Unordered convergent if there exists ${x}$ such that for every neighborhood ${U}$ of ${x}$ there exists a finite set ${E\subset \mathbb{N}}$ with the property that ${\sum_{n\in F}x_n\in U}$ for every finite set ${F}$ containing ${E}$.
• (SC) Subseries convergent if for every increasing sequence of integers ${(n_k)}$ the series ${\sum x_{n_k}}$ converges.
• (BC) Bounded-multiplier convergent if for every bounded sequence of scalars ${(a_n)}$, the series ${\sum a_n x_n}$ converges.

In a general locally convex space, (BC) ⇒ (SC) ⇒ (UC) ⇔ (RC). The issue with reversing the first two implications is that they involve the existence of a sum for some new series, and if the space lacks completeness, the sum might fail to exist for no good reason. All four properties are equivalent in sequentially complete spaces (those where every Cauchy sequence converges).

Let’s prove that interpretation of (PA) stated above, using the (BC) form of unconditional convergence. Suppose ${\sum x_n}$ converges in the sense (PA), that is for every linear functional ${z}$ the series ${\sum |z(x_n)|}$ converges. Then it’s clear that ${\sum a_n x_n}$ has the same property for any bounded scalar sequence ${(a_n)}$. That is, (PA) implies bounded-multiplier convergence in the weak topology. Conversely, suppose ${\sum x_n}$ enjoys weak bounded-multiplier convergence and let ${z\in E^*}$. Multiplying each ${x_n}$ by a suitable unimodular factor ${a_n}$ we can get ${z(a_n x_n) > 0}$ for all ${n}$. Now the weak convergence of ${\sum a_n x_n}$ yields the pointwise convergence of ${\sum |z(x_n)|}$.

A theorem of Orlicz, proved in the aforementioned book by Day, says that (SC) convergence in the weak topology of a Banach space is equivalent to (SC) convergence in the norm topology. Thanks to completeness, in the norm topology of a Banach space all forms of unconditional convergence are equivalent. The upshot is that (PA) automatically upgrades to (UA) in the context of the elements of a Banach space being considered as functions on the dual unit ball.

## Weak convergence in metric spaces

Weak convergence in a Hilbert space is defined as pointwise convergence of functionals associated to the elements of the space. Specifically, ${x_n\rightarrow x}$ weakly if the associated functionals ${f_n(y) = \langle x_n, y \rangle}$ converge to ${f(y) = \langle x_n, y \rangle}$ pointwise.

How could this idea be extended to metric spaces without linear structure? To begin with, ${f_n(y)}$ could be replaced with ${\|x_n-y\|^2-\|x_n\|^2}$, since this agrees with original ${f_n}$ up to some constant terms. Also, ${\|x_n\|^2}$ here could be ${\|x_n-z\|^2}$ for any fixed ${z}$; to avoid introducing another variable here, let’s use ${x_1}$ for the purpose of fixed reference point ${z}$. Now we have a metric space version of the weak convergence: the functions
${f_n(y) = d(x_n,y)^2 - d(x_n,x_1)^2}$
must converge pointwise to
${f(y) = d(x,y)^2 - d(x,x_1)^2}$

The triangle inequality shows that strong convergence ${d(x_n,x)\rightarrow 0}$ implies weak convergence, as expected. And the converse is not necessarily true, as the example of a Hilbert space shows.

Aside: the above is not the only way to define weak convergence in metric spaces. Another approach is to think of ${\langle x_n , y\rangle}$ in terms of projection onto a line through ${y}$. A metric space version of this concept is the nearest-point projection onto a geodesic curve. This is a useful approach, but it is only viable for metric spaces with additional properties (geodesic, nonpositive curvature).

Also, both of these approaches take the Hilbert space case as the point of departure, and do not necessarily capture weak convergence in other normed spaces.

Let’s try this out in ${\ell^1}$ with the standard basis sequence ${e_n}$. Here ${f_n(y) = \|e_n-y\|^2 - 4 \rightarrow (\|y\| + 1)^2 - 4}$. Is there an element ${x\in \ell^1}$ such that

${\|x-y\|^2 - \|x-e_1\|^2= (\|y\|+1)^2 - 4}$ for all ${y}$?

Considering both sides as functions of one variable ${y_n}$, for a fixed ${n}$, shows that ${x_n=0}$ for ${n}$, because the left hand side is non-differentiable at ${y_n=x_n}$ while the right hand side is non-differentiable at ${y_n=0}$. Then the desired identity simplifies to ${\|y\|^2 - 1 = (\|y\|+1)^2 - 4}$ which is false. Oh well, that sequence wasn’t weakly convergent to begin with: by Schur’s theorem, every weakly convergent sequence in ${\ell^1}$ also converges strongly.

This example also shows that not every bounded sequence in a metric space has a weakly convergent subsequence, unlike the way it works in Hilbert spaces.