I am trying to follow a proof in the paper Wasserstein Generative Adversarial Networks by Arjovsky et al. (proof A in supplementary material).

They show that the convergenc of the total variation distance of two probability distributions corresponds to convergence in a strong topology induced by the norm and the convergence of the Wasserstein-1 distance of two distributions corresponds to convergence in a weak* topology.

If I understand that correctly, the convergence of a sequence in a topology induced by the norm always converges strongly (see here). And convergence in a weak* topology is weak* convergence by definition.

As far as I understand this, strong convergence induces weak* convergence. I did also find that in some lecture notes (e. g. here or here), but not in any textbook or scientific article. So does anyone know of literature I could cite in my thesis concerning this?

Or a proof that strong convergence induces weak* convergence (easily) understandable by electrical engineers would be fine, too, I guess.

Thank you very much in advance!


welcome to MSE.

The fact that strong convergence implies weak*-convergence in the dual space is so basic, that most books don't bother mentioning it. What might be a good thing to do is to reference "Functional Analysis" 2nd edition by Walter Rudin which is a good book on functional analysis containing a good introduction to the weak and weak* topology.

As for the proof, this consists of two parts, showing that strong convergence implies weak convergence and noting that weak convergence implies weak*-convergence.

Let $X$ be a normed vector space and let $X^{*}$ denote its dual space. By definition $X^{*}$ is the space of all continuous linear functional, so for any $\varphi\in X^{*}$ the operator norm $$\|\varphi\|=\sup\{|\varphi(x)|:x\in X,\|x\|\leq1\}$$ is finite. Now suppose $(x_{\alpha})$ is a net in $X$ (or $(x_{n})$ is a sequence if you don't want to use nets) converging strongly to $x\in X$. Then for all $\varphi\in X^{*}$ we find $$\lim_{\alpha}|\varphi(x_{\alpha})-\varphi(x)|=\lim_{\alpha}|\varphi(x_{\alpha}-x)|\leq\lim_{\alpha}\|\varphi\|\|x_{\alpha}-x\|=0.$$ As this holds for all $\varphi\in X^{*}$ by definition $(x_{\alpha})$ weakly converges to $x$.

For the next part it is important to note that the weak* topology is not defined on $X$ but $X^{*}$. So we are considering strong convergence on $X^{*}$ with respect to the operator norm, and a net $(\varphi_{\alpha})$ in $X^{*}$ weakly converges to $\varphi$ if for all $\psi\in (X^{*})^{*}$ we have $\lim_{\alpha}\psi(\varphi_{\alpha})=\psi(\varphi)$. Recall that $(\varphi_{\alpha})$ converges to $\varphi$ in the weak* topology if for all $x\in X$ we have $\lim_{\alpha}\varphi_{\alpha}(x)=\varphi(x)$. Note however that for $x\in X$ the evaluation map $\psi_{x}:X^{*}\rightarrow\mathbb{R},\varphi\mapsto\varphi(x)$ is a continuous linear functional on $X^{*}$ hence $\psi_{x}\in(X^{*})^{*}$. Therefore if $(\varphi_{\alpha})$ converges to $\varphi$ in the weak topology it also converges in the weak* topology. (For a mathematical audience I would just write that the weak convergence implying weak* convergence is trivial.)

Combining these two results gives that strong convergence on $X^{*}$ implies weak* convergence.

Let me know if any of these steps is still unclear.


As far as I understand this, strong convergence induces weak$^*$ convergence... does anyone know of literature I could cite in my thesis concerning this?

Yes: Proposition 3.13(ii) in Brezis book.


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