Space of probability density function over compact set is compact or not Suppose $\Omega\subset\mathbb{R}^n$ is compact. Is the set of probability density functions over $\Omega$ compact with respect to weak convergence or $L^1(\Omega)$ convergence? Thank you!
 A: If $L^1(\Omega)$ is assumed infinite dimensional -i.e. $\Omega$ is not finite, this is not true with respect to the weak or strong topology on $L^1(\Omega)$. The set of probability density functions is just the unit sphere in $L^1(\Omega)$, but the unit sphere is never compact in an infinite dimensional space.
A: In order for the question to make sense, we must assume that $L^1(\Omega )\neq \{0\}$, which is the same as saying that the Lebesgue
measure of $\Omega $ is nonzero.  Otherwise there are no probability densities on $\Omega $,  whatsoever!
The set $\mathcal D$ of probability densities on $\Omega $ (which may be seen as probability measures that are absolutely
continuous with respect to Lebesgue measure) is,  by definition, the  set of all measurable functions
$$
  \varphi :\Omega \to {\mathbb R}
  $$
which are nonnegative a.e., and such that $\displaystyle \int_\Omega \varphi (x)\, dx=1$.
This set can be naturaly identified with a subset of $L^1(\Omega )$, which in turn has quite a few natural topologies.
Besides the norm topology, it may be equipped with the weak topology
$\sigma (L^1(\Omega ),L^\infty (\Omega ))$,
but there are still other weak
topologies of interest, namely
$\sigma (L^1(\Omega ),A)$,
where $A$ can be taken
to be various  natural subspaces of $L^\infty (\Omega )$, such as the set $C(\Omega )$  of all continuous functions on $\Omega $.
Since the weakest of the topologies so far mentioned is $\sigma (L^1(\Omega ),C(\Omega ))$, should we prove that $\mathcal D$ is not
compact relative to this topology, it will follow that $\mathcal D$ is not compact with the other topologies either.
This said,  let us prove that  $\mathcal D$ is not compact relative to  $\sigma (L^1(\Omega ),C(\Omega ))$.
With this topology, $L^1(\Omega )$ may be seen as a topological subspace of the dual space $C^*(\Omega )$, aka the space  $M(\Omega )$  of all
finite, signed,
Borel measures on $\Omega $, equipped with the weak$^*$-topology.
We will in fact prove that $\mathcal D$ is not closed in
$M(\Omega )$, from where it will follow that $\mathcal D$ is not compact.
Denoting by $\lambda $ the restriction of Lebesgue's measure to $\Omega $, let $x_0$ be any point in the support of $\lambda $, so  any
neigborhood of $x_0$ has nonzero measure.
For each positive integer $n$, let
$$
  V_n=\Omega \cap B_{1/n}(x_0),
  $$
and consider the normalized characteristic function
$$
  \varphi _n = \frac1{\lambda (V_n)}\mathbb 1_{V_n}.
  $$
One can now prove that
$$
  \int_\Omega f(x)\varphi _n(x)\, dx \to  f(x_0), \quad \forall f\in  C(\Omega ),
  $$
as $n\to \infty $,  which means that $\varphi _n$ converges to the Dirac measure $\delta _{x_0}$ in the weak$^*$-topology.  Since $\delta _{x_0}$ is not in
$\mathcal D$, we deduce that $\mathcal D$ is not closed, and hence also not compact.
