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Consider the following set up. Let $0 < a < b < \infty$ be fixed constants, and $G \subset \mathbb{R}^{d}$ be a compact set, for some fixed $d \in \mathbb{N}$.

Then define $$\mathcal{H}_{a}^{b} := \left\{h \colon G \to [a, b] \mid \int_{G} h d \mu = 1, h \text{ measurable} \right\},$$ to be the set of uniformly bounded probability densities on $G$.

Further let us endow $\mathcal{H}_{a}^{b}$ with the $L_{2}$-metric, i.e., for all $f_1, f_2 \in \mathcal{H}_{a}^{b}$ $$\|f_1 - f_2\|_{2} := \left(\int_{G} (f_1 - f_2)^{2} d\mu \right)^{1 / 2}$$

Let $\mathcal{E} \subset \mathcal{H}_{a}^{b}$, denote a closed and convex set. We will also consider $\mathcal{E}$ to be a restricted $L_{2}$-metric space of $\mathcal{H}_{a}^{b}$.

I want to understand basic topological properties of $\mathcal{E}$ and $\mathcal{H}_{a}^{b}$. All of these are with respect to the $L_{2}$-metric:

  1. Is $\mathcal{H}_{a}^{b}$ compact?
  2. If not compact, is $\mathcal{H}_{a}^{b}$ totally bounded?
  3. If not totally bounded, is $\mathcal{H}_{a}^{b}$ separable?
  4. Since $\mathcal{H}_{a}^{b}$ is bounded (by $2b$), $\mathcal{E}$ is bounded. Further $\mathcal{E}$ is also closed and convex by assumption. Do these 3 properties guarantee either compactness, total boundedness, separability of $\mathcal{E}$?

I am able to show that $\mathcal{H}_{a}^{b}$ is convex, and also bounded, but not the above properties. The properties can imply each other in the metric space setting, so if the most general property is proven, then it would solve many of these queries efficiently, e.g., compactness of $\mathcal{H}_{a}^{b}$ should imply total boundedness and thus separability of $\mathcal{H}_{a}^{b}$ (I believe).

Could anyone rigorously justify these, or provide counterexamples?

user4687531
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1 Answers1

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For simplicity, let us consider $G = [0,1] \subset \mathbb R$ and $(a+b)/2 = 1$. Let us consider the sequence $(h_k)$ defined via $$ h_k(x) = \begin{cases} a & \text{if } x \in [2n/2^k, (2n+1)/2^k] \text{ for some $n \in \mathbb N$},\\ b & \text{else}. \end{cases} $$ Then, $h_k \in H_a^b$, but $\|h_k - h_l\|_2 = \textrm{const}$ if $k \ne l$. Thus, $H_a^b$ cannot be compact and cannot be totally bounded.

Further, $H_a^b$ is a (closed) subset of $L^2(0,1)$ and $L^2(0,1)$ is separable and complete. This shows that $H_a^b$ is also separable.

A similar construction can be used in the general case.

gerw
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  • Thanks @gerw! I have a basic follow up here. How is your $h_k \in \mathcal{H}^{a}{b}$, since I don't believe it integrates to 1, for each $k$? That is isn't $\int{0}^{1} h_k(x) dx = \frac{a}{2^{k}} + (1 - \frac{1}{2^{k}})b$. I'm not sure how your sufficient condition $(a + b) / 2$ guarantees this integrates to 1? Could you please clarify? – user4687531 Sep 15 '22 at 22:10
  • Also, could you please add some details as to (A) why $\mathcal{H}^{a}{b}$ is closed in $L^{2}(G)$? And (B) is $\mathcal{H}^{a}{b}$ complete in the $L_2$ metric? I'm guessing if it is closed (i.e., property (A) holds) in $L^{2}(G)$, which is complete (per your claim) in $L_2$, then $\mathcal{H}^{a}_{b} \subset L^{2}(G)$ must also be complete in the L_2 metric (i.e., property (B) holds)? Is that correct? – user4687531 Sep 15 '22 at 22:17
  • $h_k \in H_b^a$: on one half of $[0,1]$ the function $h_k$ equals $a$, on the other half $b$. Thus, $\int h_k = (a+b)/2 = 1$. On (A): Convergence in $L^2$ implies pointwise a.e. convergence of a subsequence. This ensures that the limit lies between $a$ and $b$. The integral also converges. On (B): Yes, closedness and completeness of the surrounding space gives completeness of $H_a^b$. – gerw Sep 16 '22 at 07:54
  • thanks again for your kind clarifications. Since I'm new to these ideas, so a few more questions. (C) I'm still a bit unsure why $\int h_k = 1$. I see $h_k(x) = a \mathbb{1}{A_k}(x) + b a \mathbb{1}{[0, 1] \setminus A_k}(x)$, where $A_k := [2n/2^k, (2n+1)/2^k]$. This results in my expression for $\int h_k$ above. Since isn't $\lambda(A_k) = \frac{1}{2^{k}}$, for any fixed $n$. $\lambda$ is the Lebesgue measure here. – user4687531 Sep 16 '22 at 13:50
  • In short, I'm not sure how you get $1 / 2$. Could you please add some details here? It would really help my learning. I will def accept the answer once I understand this better. – user4687531 Sep 16 '22 at 14:06
  • Also for (A), could you please add some more details and citation for the result. I keep seeing counterexamples for $L^2$ convergence, and would like to appreciate the details of your claim to understand why they don't apply here. – user4687531 Sep 16 '22 at 14:07
  • Thanks for clarifying (B), that argument is clear now. Assuming I understand (A) better. – user4687531 Sep 16 '22 at 14:08
  • (C) I think you misunderstood my definition of $h_k$. The function $h_k$ oscillates (with period $2 /2^k$) between $a$ and $b$. (A) This is a standard result in measure theory, see, e.g., https://math.stackexchange.com/questions/611355/subsequence-convergence-in-lp – gerw Sep 19 '22 at 12:23
  • I thank you for your patience. I did not realize that $h_k$ also depended on $n$, I thought "some $n \in \mathbb{N}$", meant $n$ was fixed. This now makes $h_k$ to be a sequence in $h_kn$. I'm not sure how to understand that one half of $h_k$ takes value $a$. In short, could you please add a lot more details to your answer. It would help newcomers like me develop these skills better. – user4687531 Sep 20 '22 at 01:07
  • I've looked into this sequence https://math.stackexchange.com/questions/1412091/the-typewriter-sequence, but I still can't justify your claim above, since I don't know how you remove dependence on $n$ for each $h_k$. – user4687531 Sep 20 '22 at 01:08
  • For (A), could you please show how you use this type of subsequence argument to conclude that $H_a^b$ is closed in $L^2(G)$? And further how do you generalize your counterexample from $[0, 1]$ to an arbitrary compact set $G \subset \mathbf{R}^{d}$. I'm happy to mark this as correct, once these details are clear. Your help is appreciated here. – user4687531 Sep 20 '22 at 01:10
  • If we let $A_k^n := [2n/2^k, (2n+1)/2^k]$, then if we consider the sequence of sets $A_1^n$ (i.e. $k = 1$), don't we get $A_1^n = \left{[1, 3/2], [2, 5/2], [3, 7/2], \ldots \right}$, which has intersection of mass 0, with $G = [0, 1]$, right? Since all the intervals have a left most point of 1, and then are to the right of that point thereafter. So how does $\int_G h_1 = (a + b)/2$ here? I believe you claim this to be true for each $k \in \mathbf{N}$. I must be missing something. – user4687531 Sep 20 '22 at 01:25
  • You also take $n = 0$, i.e. $A_k = \bigcup_{n \in \mathbb Z} A_k^n \cap (0,1)$. – gerw Sep 20 '22 at 13:52
  • Aha - thanks, that clarifies! Are you able to describe how to extend this to general compact sets $G$? This is not obvious. – user4687531 Sep 20 '22 at 14:05
  • This can be done very similarly, since in $G$ every set of positive measure can be split into two subsets of equal measure. – gerw Sep 20 '22 at 14:53