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.

2 thoughts on “Weak convergence in metric spaces”

  1. I suggest an alternative: if $X$ is the metric space, then $x_n \to x$ weakly iff $f(x_n) \to f(x)$ for every short map $f : X \to \mathbb R$.

    1. This does not seem different from ordinary convergence in the metric, since $f$ could be the distance function to $x$, $f(z) = d(z,x)$.

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