Metric on $M[0,1]$ the space of measures

Question

Let $B$ be the closed unit ball on the space $M[0,1]$ of the borel regular complex measures on $[0,1]$. For $\mu$ and $\nu$ $\in $ $M[0,1]$ define $d(u,v) = \sum_{n=0}^{\infty} \left|\int_{[0,1]}x^nd\mu - \int_{[0,1]}x^nd\nu \right|$. Show that $d$ is a metric on $M$ and that define the weak star topology on $B$ but not in $M$.

I know that $C_0([0,1])$ is separable and hence I know that $(K,w^*)$ is metrizable with $K$ $w^*$-compact on $M$. And even more if $(x_i)$ is a dense set on the space then $$\sum_{n=1}^{\infty}\frac{p_n(\varphi - \psi)}{1 + p_n(\varphi - \psi)}$$ where $p_n(\varphi) = \varphi(x_n)$, is a metric which defines the $w^*$ topology on $K$. I would like to use this result but the polynomials given are not dense on the space. Another problem I have is that I am not able of see why the series given in the statement converges.

Another way I would like to attack this is proving that the identity function $id:(K,w^*)\to (K,d)$ is continuos, and since $K$ is $w*$ compact the result will follow.

I am stuck on this lines, I would appreciate any idea.

ADDED:

Remember, Complex measures always have finite variation, by definition.
As was mentioned in the comments, $d$ can not be a metric since it can take an infinite value, for example, when you put $\nu = 0$ and $\mu$ the lebesgue measure then you will get $\int_{0}^1\frac{1}{1-x}$ which diverges. A simpler example is considering dirac measures. So the problem
I make a typo in the theorem I wanted to use, the metric associated with the dense subset $(x_i)$ is $$\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{p_n(\varphi - \psi)}{1 + p_n(\varphi - \psi)}$$

Maybe it is worth mentioning that the idea of the theorem I wanted to use is to observe that if I truncate the series then I have a pseudometric which is $w^*$ continuos, and since the series converges uniformly, then the series is $w^*$continuos. Last, since $(x_i)$ is dense one can prove that actually is a metric, and since is continuos one already has that $w^*$ topology includes the topology induced by the metric, the final inclusion follows using that $K$ is $w^*$ compact.

EDIT2:

I will write it to the author of the problem. But I think I have the essential key. I will expect that the correct $d$ on the problem is the one I give, if $p_n(\mu) = \int_{0}^1x^n d\mu$ then

$$d(\mu,\nu) = \sum_{n=1}^{\infty}\frac{1}{2^n}\frac{p_n(\mu - \nu)}{1 + p_n(\mu - \nu)}$$

in this case the truncated series is $w^*$ continuos, and since the series converges uniformly then $d$ is $w^*$ continuos, this implies that $\tau_d \subset w^*$. Now $d$ is a metric: if $d(\mu,\nu) = 0$ then each term of the sum is zero and then, by linearity, $\int p(x)d\mu - \int p(x)d\nu = 0$ with $p(x)$ being a polynomial, since the polynomials are dense on $C_0([0,1])$ and the integrals are linear functionals on $C_0[0,1]$ which are continuos, then this linear functionals are equal and then are equal the measures.

To finish, if we have that $id:(K,w^*) \to (K,\tau_d)$ is continuos, and since $K^*$ is $w^*$ compact then $id$ will be an homeomorphism.

I have not thought why this metric does not work on $M$

EDIT3:

Well finally, the author say he just forgot the factor $\frac{1}{2^n}$ in $d$, hence the metric is $$d(u,v) = \sum_{n=0}^{\infty}\frac{1}{2^n}\left|\int_0^1 x^n d\mu - \int_{0}^{1}x^n \right | d\nu$$

Whit this in mind I have a possible demonstration. I will propose it as answer to be discussed.

Why is $x^n$ $\mu$-integrable? Do you mean finite Borel measures? — amsmath, Oct 06 '19 at 03:28
@amsmath, As far as I understand, complex measures automatically have finite total variation. But clarity would be always welcomed... — Sangchul Lee, Oct 06 '19 at 03:34
@SangchulLee Gosh, you're right. I forgot about that. Sorry. — amsmath, Oct 06 '19 at 03:34
This distance is not well defined. For example, take the Lebesgue measure as $\mu$ and $\nu = 0$. Then $$d(\mu,\nu) = \sum_n\int_0^1 x^n,d\mu = \int_0^1\left(\sum_n x^n\right),dx = \int_0^1\frac 1{1-x},dx = \infty.$$ — amsmath, Oct 06 '19 at 03:41
I see no reason that $d$ is a metric on $B$, much less on $M[0,1]$. Indeed, $$\left| \int_{[0,1]}x^n ,\delta_1(\mathrm{d}x)-\int_{[0,1]}x^n ,\delta_0(\mathrm{d}x)\right|=\mathbf{1}{n\geq1},$$ and so, its sum over all $n\geq0$ will diverge. Perhaps the author wanted to point out that the weak-$*$ topology on $B$ is the same as the uniform topology given by pseudometrics $$d_n(\mu,\nu)=\left|\int{[0,1]}x^n,\mu(\mathrm{d}x)-\int_{[0,1]}x^n,\nu(\mathrm{d}x)\right|,$$ which in turn is the metric topology given by $$d(\mu,\nu)=\sum_{n=0}^{\infty} 2^{-n} d_n(\mu, \nu). $$ — Sangchul Lee, Oct 06 '19 at 03:42

score 1 · Answer 1 · answered Oct 06 '19 at 14:47

1

$M[0,1]$ cannot be $w^*$-metrizable because a dual space $X^*$ (of some Banach $X$) is $w^*$-metrizable if and only if $X^*$ is of finite dimension.

answered Oct 06 '19 at 14:47

Nick

2,769

Can you explain why? or give some bibliography? – HFKy Oct 06 '19 at 15:09
you may refer to https://math.stackexchange.com/questions/2486297/weak-topology-is-metrizable-iff-x-is-finite-dimensional/2487154 – Nick Oct 06 '19 at 15:20
Okey, it seems a nice answer. The problem is that they use a lot of theorems that I am not suppose to know. However, reading the answers give me some inspiration. Instead of using the result your propose, what do you think of this: – HFKy Oct 06 '19 at 15:36
If it were metrizable, the open ball of center 0 and radius 1 will be open (in fact this is always true because $d$ is weak star continuos). Then take a point distinct of zero in this ball, let call it $T$. Since the ball is open there exists a local neighborhood of the weak star topology containing $T$ inside the open ball. However, since $X$ is of infinite dimension, this neighborhood will contain an infinite dimension subspace, in particular a line. Let $H$ be a generator of such line, then $T+\alpha H$ will be in $B(0,1)$ for all $\alpha$. Since $H$ is not zero and $d$ is a metric, then – HFKy Oct 06 '19 at 15:41
$H(x^n)$ is not zero for some $n$, and hence if $\alpha$ is greater enough then is absurd that $T+\alpha H \in B(0,1)$. Am I right? – HFKy Oct 06 '19 at 15:42
why is that absurd? – Nick Oct 06 '19 at 15:45
because $\frac{|T(x^n)+\alpha H(x^n)|}{2^n}\geq \alpha\frac{|H(x^n)|}{2^n} - \frac{|T(x^n)|}{2^n}$ for $\alpha >0$ greater enough. Then eventually will be greater than 1, and then $d(T+ \alpha H,0) >1$ – HFKy Oct 06 '19 at 15:51

score 0 · Accepted Answer · edited Feb 21 '21 at 10:45

First, $$d(u,v) = \sum_{n=0}^{\infty}\frac{1}{2^n}\left|\int_0^1 x^n d\mu - \int_{0}^{1}x^n \right | d\nu$$ is metric on $M[0,1]$:

First the series is finite: this is because the total variation of the measures are finite and because $x^n$ is always less than 1, then this follows from triangular inequality.
The properties of a pseudometric are met: this is because the series is equal to $$d(u,v) = \sum_{n=0}^{\infty}\frac{1}{2^n} |p_n(\mu - \nu)|$$ and each $p_n$ are in fact seminorms.
Finally let's check that $d(\mu,\nu) = 0$ implies $\mu = \nu$. If this happens then the linear functionals associated $T_\mu, T_\nu \in C_0([0,1])'$ are equal on the polynomials of the form $x^n$, then by linearity are equal on the polynomials. Finally since the polynomials on $[0,1]$ are dense on $C_0([0,1])$ and the linear functionals continuous then they are equal on every function of $C_0([0,1])$, this implies $\mu = \nu$.

$d$ gives the weak star topology on B, the closed unit ball with the norm of the total variation:

$d$ is continuos: $d_k(\mu,\nu) = \sum_{n=0}^{k}\frac{1}{2^n}\left|\int_0^1 x^n d\mu - \int_{0}^{1}x^n \right | d\nu$ is $w^*$-continuous, and it converges uniformly to $d$, this is because here the variaton of the measures is less than 1, and since $x^n$ is less than 1, the series defining $d$ is less than $\sum_{n=0}^{\infty}\frac{1}{2^n}2 < \infty$ and hence uniform convergence follows from Weierstrass M-Test. Uniform convergence implies $w^*$ continuity of $d$.
Since $d$ is continuos we have that the identity map $id:(B,w^*) \to (B,\tau_d)$ is continuous. Now since $B$ is $w^*$ compact, this identity map is a homeomorphism. This follows because if $A$ is a closed set in $B$ then is compact, then $id(A) = A$ is compact, and since $(B,\tau_d)$ is Hausdorff then $A$ is closed. Hence the image of every closed set under the identity is closed.

This finishes the proof that $d$ defines the same topology.

Now this is not true in $M$. What is happening is that $d$ is not $w^*$ continuous. [Nick proposed another proof for this part that seems more efficient than this] Let's see: If $d$ defines the same topology on $M$ then $B_d(0,1)$ is $w*$-open. Since $X = C_0([0,1])$ has $dim = \infty$, then every neighborhood of zero in the weak star topology of $X'$ contains a subspace of infinite dimension, in particular $B_d(0,1)$ is not bounded and this is absurd. (In the comments of Nick's answer I give an argument more complicated using a linear functional $H$)

Metric on $M[0,1]$ the space of measures

2 Answers2