Suppose I have a random sequence $X_n$ of cadlag functions on $[0,1]$ that converge weakly to $X$. In general this is meant with respect to the Skorkhod metric but suppose here I have that $X$ is continuous and $X_n$ converge weakly to $X$ with respect to the uniform metric on the space of cadlag processes $[0,1]$. (Under this metric this is a complete metric space however it is not separable.) I need that $X_n$ is well defined with respect to the uniform topology but suppose I have this as well.
Now suppose $X_n$ are piecewise constant and consider $\tilde{X}_n$, the linearized version of $X_n$. By this I mean the function I get by drawing straight lines between the jump discontinuities of $X_n$. Is it true that $\tilde{X}_n$ converges in law to $X$ again with respect to the uniform metric on continuous functions?
I'm pretty sure that this is true, and I believe I have a proof but I'd like confirmation. I thought this would be written down somewhere. I tried Billingsley, Convergence of Probability Measures, especially Chapter 3, Section 18 (The Uniform Topology), but I couldn't find it. Thanks for your help.
If I understand the question correctly, the linearization map $l$ have some parameters and can be chosen to satisfy some nice properties. For example, we can define it in such a way that $$ \|l(f) - l(g)\|\leq \|f-g\| $$ where $\|f\|:=\sup_{x\in [0,1]}|f(x)|$ denotes the uniform sup-norm. Hence, $l$ is a Lipschitz map and thus continuous. The result follows from the continuous mapping theorem.