Take the set of all probability measures over $[0,1]$. Call this $\Delta [0,1]$. Consider an open ball of distributions in $\Delta [0,1]$, in the weak* topology. My question is, must this open ball contain a density with full support? In other words, for any distribution $\mu \in \Delta [0,1]$, does there exist another close to it which is absolutely continuous and has full support?

My approach: I think it definitely needs to have full support. If not, take its $\epsilon$-convex combination with the uniform distribution. For small enough $\epsilon$ this must lie inside the open ball. Now suppose the CDF is discontinuous (i.e. has an atom). It is an increasing function, so can only have jump discontinuities.

Now I think there is a result that says there is a continuous function close to every discontinuous function and a differentiable function close to every non-differentiable one. I should've known this result clearly already but I don't. Any reference for this one would be appreciated as well. (I'm a social sciences student, hence the gaps in basic knowledge simultaneously with the need to use results like these. :( )

Assuming the above result is true, this makes me think we indeed must have a density. But that result uses a different topology, and I'm confused if it is applicable here.

Any help is most appreciated.


1 Answer 1

  1. Differentiable functions are dense in the space of continuous functions; most students encounter this as Stone-Weierstrass.

    There are two good proofs, and many teachers prefer the conceptually-clear proof via Bernstein polynomials. But you should look for the one via mollification, which appears in (say) Rudin's Principles of Real Analysis.

    The latter proof also shows that continuous functions are dense in $L^p(\Omega)$ ($p<\infty$); a good (free, online) reference for that is Bass' Real Analysis for Grad Students.

  2. Mollification also lets you approximate (in the weak-$*$ topology) probability distributions by continuous functions.

    There are many different ways to write the argument. The easiest for you might be to mollify the CDF; continuous CDFs are dense in $L^2([0,1])$, and thus converge in the dual of $L^2([0,1])\supseteq C^1([0,1])$.

    Personally, I tend to prefer mollifying the measures directly, since then I don't have to rely on the "coincidence" that $C\subseteq L^2$ in spaces of finite measure. In that case, I define my density as $$(\phi_{\epsilon}*\mu)(x)=\int_{[0,1]}{\phi_{\epsilon}(x-t)\,\mu(dt)}$$ (where $\phi_{\epsilon}$ is a mollifier); Fubini-Tonelli implies this converges weak-$*$.

  3. You work too hard to show that full density is achievable. First, approximate by a continuous function; then it suffices to show that continuous function with full support are dense in the space of continuous functions in the sup norm, which follows from the first (and easy) part of your argument.

  • $\begingroup$ Thank you so much! This is very helpful. I didn't understand the 3rd point. what do you mean by "You work too hard to show that full density is achievable"? Also, if I understand #2 correctly, after mollifying with $\phi_{\epsilon}$ we get a continuous CDF which is close in weak-* to the original distribution $\mu$. But what about differentiability, i.e. having a density? Does that follow from denseness of differentiable functions in the space of continuous functions, i.e. #1? Thank you so much again! $\endgroup$
    – Canine360
    Jul 11, 2022 at 22:07
  • $\begingroup$ @Canine360: Ad #2, yes (or as I prefer to think of it, there was typo, $C([0,1])$ for $C^1([0,1])$). Ad #3, you spend some time in the question worrying about atoms. My point is that you can eliminate atoms with approximation and then worry about support. $\endgroup$ Jul 12, 2022 at 5:09

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