# A note on the completeness of Fourier-based metrics on measures

Evan Randles

###### Abstract

Resolving an open question of J. A. Carillo and G. Toscani, M. Stawiska
recently proved that certain metric spaces of probability measures equipped
with Fourier-based metrics are complete. In this note, we demonstrate the fine
nature of Stawiska’s result and show that, when such metrics are defined on
complex measures, the resulting metric spaces are not complete.

Keywords: Convergence of measures, Fourier-based metrics, moments.

Mathematics Subject Classification: Primary 28A33, 60B10; Secondary, 42B10,
46E27

We denote by $\mathcal{M}=\mathcal{M}(\mathbb{R}^{d})$ the set of complex-
valued finite Borel measures on $\mathbb{R}^{d}$. For each
$\mu\in\mathcal{M}$, we define its Fourier transform/characteristic function
$\widehat{\mu}:\mathbb{R}^{d}\to\mathbb{C}$ by

$\widehat{\mu}(\xi)=\int_{\mathbb{R}^{d}}e^{-ix\cdot\xi}\,d\mu(x)$

for $\xi\in\mathbb{R}^{d}$; here, $\cdot$ denotes the dot product. It is a
standard fact that measures are characterized by their Fourier transforms,
i.e., for $\mu,\nu\in\mathcal{M}$, $\mu=\nu$ if and only if
$\widehat{\mu}=\widehat{\nu}$ [Str94, Theorem 1.11]. Let us fix an integer
$m\geq 2$ and, for a pair of measures $\mu,\nu\in\mathcal{M}$, set

$d_{m}(\mu,\nu)=\sup_{\xi\neq
0}\frac{\absolutevalue{\widehat{\mu}(\xi)-\widehat{\nu}(\xi)}}{\absolutevalue{\xi}^{m}}$

where $\absolutevalue{\cdot}$ denotes the Euclidean norm on $\mathbb{R}^{d}$.
It is straightforward to see that $d_{m}$ is symmetric, non-negative, and
satisfies the triangle inequality. By virtue of the uniqueness property of the
Fourier transform mentioned above, we see that $d_{m}(\mu,\nu)=0$ if and only
if $\mu=\nu$. Thus $d_{m}$ is a metric on any subset of $\mathcal{M}$ for
which it is finite.

The metric $d_{m}$, when taken on the space of probability measures
$\mathcal{P}(\mathbb{R}^{d})\subseteq\mathcal{M}$, was introduced in the study
of asymptotics of solutions to the Boltzmann equation [GTW95]. We encourage
the reader to see the notes [CT07] which develop the theory for such Fourier-
based metrics on probability space and their applications. As pointed out in
[CT07] and [Sta20], the advantage of the metric $d_{m}$ is that it behaves
nicely on subsets of measures with (finite and) identical moments. To
introduces such subsets, let us first denote by $\mathcal{M}_{m}$ the subset
of $\mathcal{M}$ which have finite moments of order $m$, i.e.,
$\mu\in\mathcal{M}_{m}$ if and only if

$\int_{\mathbb{R}^{d}}\absolutevalue{x}^{m}\,d\mu(x)<\infty.$

Of course, because $1\in L^{1}(\mu)$, each $\mu\in\mathcal{M}_{m}$ has finite
moments of order up to and including $m$. For a specific choice
$M=\\{M_{\beta}\\}_{\absolutevalue{\beta}\leq m}\subseteq\mathbb{C}$ of
moments, let

$\mathcal{M}_{m}^{M}=\left\\{\mu\in\mathcal{M}_{m}\left|\,\,\int_{\mathbb{R}^{d}}x^{\beta}\,d\mu(x)=M_{\beta}\,\,\mbox{for}\,\,\absolutevalue{\beta}\leq
m\right\\}\right.;$

here, for a multi-index
$\beta=(\beta_{1},\beta_{2},\dots,\beta_{d})\in\mathbb{N}^{d}$,
$\absolutevalue{\beta}=\beta_{1}+\beta_{2}+\cdots+\beta_{d}$ and, for
$x=(x_{1},x_{2},\dots,x_{d})\in\mathbb{R}^{d}$,
$x^{\beta}=x_{1}^{\beta_{1}}\cdot x_{2}^{\beta_{2}}\cdots x_{d}^{\beta_{d}}$.
The set $\mathcal{M}_{m}^{M}$ is precisely the set of measures
$\mu\in\mathcal{M}_{m}$ with identical moments prescribed by
$M=\\{M_{\beta}\\}_{\absolutevalue{\beta}\leq m}$. When one considers the
analogous (sub) class of probability measures
$\mathcal{P}_{m}^{M}\subseteq\mathcal{M}_{m}^{M}$ (where, necessarily,
$\\{M_{\beta}\\}\subseteq\mathbb{R}$), M. Stawiska proved that
$(\mathcal{P}_{m}^{M}(\mathbb{R}^{d}),d_{m})$ is complete for every even
integer $m$. This result answered an open question posed by J. A. Carillo and
G. Toscani in [CT07]. The purpose of this note is to demonstrate the fine
nature of Stawiska’s completeness result, Theorem 1.1 of [Sta20]. Our result
is as follows.

######  Theorem 1.

Given any integer $m\geq 2$ and $M=\\{M_{\beta}\\}_{\absolutevalue{\beta}\leq
m}\subseteq\mathbb{C}$, $\mathcal{M}_{m}^{M}$ equipped with $d_{m}$ is a
metric space which is not complete.

In order the prove the theorem, we first present four lemmas. The first shows
that the Fourier transform exchanges the finiteness of moments for smoothness;
its proof is standard and omitted.

######  Lemma 1.

If $\mu\in\mathcal{M}_{m}$, then $\widehat{\mu}$ is $m$-times continuously
differentiable on $\mathbb{R}^{d}$, i.e., $\widehat{\mu}\in
C^{m}(\mathbb{R}^{d})$ and, for each multi-index $\beta\in\mathbb{N}^{d}$ for
which $\absolutevalue{\beta}\leq m$,

$(D^{\beta}\widehat{\mu})(\xi)=(-i)^{\absolutevalue{\beta}}\int_{\mathbb{R}^{d}}x^{\beta}e^{-ix\cdot\xi}\,d\mu(\xi)$

for $\xi\in\mathbb{R}^{d}$; here, for a multi-index
$\beta=(\beta_{1},\beta_{2},\dots,\beta_{d})\in\mathbb{N}^{d}$,
$D^{\beta}=\partial_{\xi_{1}}^{\beta_{1}}\partial_{\xi_{2}}^{\beta_{2}}\cdots\partial_{\xi_{d}}^{\beta_{d}}$.

######  Lemma 2.

Let $m$ and $\\{M_{\beta}\\}$ be as in the statement of the theorem and fixed.
Consider the multivariate polynomial

$P(\xi)=\sum_{|\gamma|\leq m}\frac{M_{\gamma}}{\gamma!}(-i\xi)^{\gamma}$

where $\gamma!=(\gamma_{1}!)(\gamma_{2}!)\cdots(\gamma_{d}!)$ and, for
$\delta\geq 0$, set

$\phi_{\delta}(\xi)=\left(P(\xi)+\absolutevalue{\xi}^{m+2}\sin\left(\frac{1}{\absolutevalue{\xi}^{m}}\right)e^{-\delta/|\xi|^{2}}\right)e^{-\absolutevalue{\xi}^{2m}}$

for $\xi\in\mathbb{R}^{d}\setminus\\{0\\}$ and $\phi_{\delta}(0)=P(0)$. If
$\delta>0$, then $\phi_{\delta}$ is the Fourier transform of a measure
$\mu_{1/\delta}\in\mathcal{M}_{m}^{M}$.

###### Proof.

For $\delta>0$, define

$f_{\delta}(\xi)=\begin{cases}\absolutevalue{\xi}^{m+2}\sin(1/\absolutevalue{\xi}^{m})e^{-\delta/\absolutevalue{\xi}^{2}}&\xi\neq
0\\\ 0&\xi=0\end{cases}$

so that
$\phi_{\delta}(\xi)=(P(\xi)+f_{\delta}(\xi))e^{-\absolutevalue{\xi}^{2m}}$ for
$\xi\in\mathbb{R}^{d}$. Observe that $f_{\delta}$ is smooth on
$\mathbb{R}^{d}\setminus\\{0\\}$. Further, for each multi-index
$\beta\in\mathbb{N}^{d}$, it is straightforward to see that there are
multivariate polynomials $Q_{1}$ and $Q_{2}$ (in the variables
$\xi_{1},\xi_{2},\dots,\xi_{d},$ and $\absolutevalue{\xi}$) and an integer
$k\geq 0$ for which

$(D^{\beta}f_{\delta})(\xi)=\left(Q_{1}(\xi)\sin(1/\absolutevalue{\xi}^{m})+Q_{2}(\xi)\cos(1/\absolutevalue{\xi}^{m})\right)\frac{e^{-\delta/\absolutevalue{\xi}^{2}}}{\absolutevalue{\xi}^{k}}$
(1)

for $\xi\neq 0$. We now prove that, for every $\beta\in\mathbb{N}^{d}$,
$(D^{\beta}f_{\delta})$ is continuous at $0$ with $(D^{\beta}f_{\delta})(0)=0$
by inducting along chains. The base case, i.e., that for $\beta=0$, is
immediate. Assuming the induction hypothesis, observe that, for $1\leq j\leq
d$,

$\displaystyle(D^{\beta+e_{j}}f_{\delta})(0)$ $\displaystyle=$
$\displaystyle\lim_{h\to 0}\frac{(D^{\beta}f_{\delta})(he_{j})-0}{h}$
$\displaystyle=$ $\displaystyle\lim_{h\to
0}\left(Q_{1}(he_{j})\sin(1/\absolutevalue{h}^{m})+Q_{2}(he_{j})\cos(1/\absolutevalue{h}^{m})\right)\frac{e^{-\delta/h^{2}}}{h^{k+1}}=0$

on account of the fact that $e^{-\delta/h^{2}}=o(h^{k+1})$ as $h\to 0$. A
similar computation making use of (1) guarantees continuity for
$D^{\beta+e_{j}}f_{\delta}$ at $0$ and so our proof by induction is complete.
Consequently, $f_{\delta}$ is smooth on $\mathbb{R}^{d}$ and, in view of (1),
for each multi-index $\beta$, there is a constant $C>0$ and an integer $l\geq
1$ for which

$\absolutevalue{(D^{\beta}f_{\delta})(\xi)}\leq C(1+\absolutevalue{\xi}^{l})$

for all $\xi\in\mathbb{R}^{d}$. Since
$\phi_{\delta}(\xi)=(P(\xi)+f_{\delta}(\xi))e^{-\absolutevalue{\xi}^{2m}}$
where $P(\xi)$ is a polynomial, it follows immediately from the above estimate
that $\phi_{\delta}$ is a member of the Schwartz class, i.e.,
$\phi_{\delta}\in\mathcal{S}(\mathbb{R}^{d})$. Using the fact that the
(inverse) Fourier transform is an isomorphism of
$\mathcal{S}(\mathbb{R}^{d})$, we conclude that

$\widecheck{\phi_{\delta}}(x)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\phi_{\delta}(\xi)e^{i\xi\cdot
x}\,d\xi$

defines a Borel measure $\mu_{1/\delta}$ via
$d\mu_{1/\delta}=\widecheck{\phi_{\delta}}(x)\,dx$ having finite moments of
all orders, c.f., [Str94, Note 1.1.1]. In particular,
$\mu_{1/\delta}\in\mathcal{M}_{m}$ and
$\widehat{\mu_{1/\delta}}=\phi_{\delta}$.

It remains to verify that $\mu_{1/\delta}\in\mathcal{M}_{m}^{M}$. To this end,
let $\beta$ be a multi-index for which $\absolutevalue{\beta}\leq m$. In view
of Leibniz’ rule, we have

$\displaystyle(D^{\beta}\phi_{\delta})(\xi)$ $\displaystyle=$ $\displaystyle
D^{\beta}\left(\left(P(\xi)+f_{\delta}(\xi)\right)e^{-\absolutevalue{\xi}^{2m}}\right)$
(2) $\displaystyle=$
$\displaystyle\sum_{\alpha\leq\beta}{\beta\choose\alpha}D^{\alpha}\left(P(\xi)+f_{\delta}(\xi)\right)D^{\beta-\alpha}\left(e^{-\absolutevalue{\xi}^{2m}}\right)$

for $\xi\in\mathbb{R}^{d}$; here $\leq$ is the standard partial ordering on
multi-indices, i.e., $\alpha\leq\beta$ provided $\alpha_{j}\leq\beta_{j}$ for
$j=1,2,\dots,d$. Now, for each multi-index $\alpha\leq\beta$, it is easy to
verify that

$D^{\beta-\alpha}\left(e^{-\absolutevalue{\xi}^{2m}}\right)\Big{|}_{\xi=0}=\begin{cases}1&\alpha=\beta\\\
0&\alpha<\beta\end{cases}$

in view of the fact that $\absolutevalue{\beta-\alpha}\leq m<2m$. With this
observation and that fact that $(D^{\alpha}f_{\delta})(0)=0$ for each
$\alpha$, (2) ensures that
$\left(D^{\beta}\phi_{\delta}\right)(0)=\left(D^{\beta}P\right)(0)$ and so it
follows that

$\left(D^{\beta}\phi_{\delta}\right)(0)=\sum_{\absolutevalue{\alpha}\leq
m}\frac{(-i)^{\absolutevalue{\alpha}}M_{\alpha}}{\alpha!}D^{\beta}(\xi^{\alpha})\Big{|}_{\xi=0}=(-i)^{\absolutevalue{\beta}}M_{\beta}.$

Thus, by virtue of the previous lemma, we have

$\int_{\mathbb{R}^{d}}x^{\beta}\,d\mu_{1/\delta}(x)=(i)^{\absolutevalue{\beta}}(D^{\beta}\phi_{\delta})(0)=M_{\beta}$

for each $\beta\in\mathbb{N}^{d}$ for which $\absolutevalue{\beta}\leq m$ and
therefore, $\mu_{1/\delta}\in\mathcal{M}_{m}^{M}$. ∎

######  Lemma 3.

For $\delta_{1},\delta_{2}>0$, we have

$d_{m}(\mu_{1/\delta_{1}},\mu_{1/\delta_{2}})\leq\absolutevalue{\delta_{2}-\delta_{2}}.$

###### Proof.

Set $\rho=\absolutevalue{\delta_{2}-\delta_{1}}$. Observe that, for $\xi\neq
0$, we have

$\displaystyle\frac{\absolutevalue{\phi_{\delta_{1}}(\xi)-\phi_{\delta_{2}}(\xi)}}{\absolutevalue{\xi}^{m}}$
$\displaystyle=$
$\displaystyle\absolutevalue{\xi}^{2}\absolutevalue{\sin(1/\absolutevalue{\xi}^{2})}\absolutevalue{e^{-\delta_{1}/\absolutevalue{\xi}^{2}}-e^{\delta_{2}/\absolutevalue{\xi}^{2}}}e^{-\absolutevalue{\xi}^{2m}}$
$\displaystyle\leq$
$\displaystyle\absolutevalue{\xi}^{2}\left(1-e^{-\rho/\absolutevalue{\xi}^{2}}\right).$

In view of the previous lemma,
$\widehat{\mu_{1/\delta_{k}}}=\phi_{\delta_{k}}$ for $k=1,2$, and so it
follows that

$\displaystyle d_{m}(\mu_{1/\delta_{1}},\mu_{1/\delta_{2}})$ $\displaystyle=$
$\displaystyle\sup_{\xi\neq
0}\frac{\absolutevalue{\phi_{\delta_{1}}(\xi)-\phi_{\delta_{2}}(\xi)}}{\absolutevalue{\xi}^{m}}$
$\displaystyle\leq$ $\displaystyle\sup_{\xi\neq
0}\absolutevalue{\xi}^{2}\left(1-e^{-\rho/\absolutevalue{\xi}^{2}}\right)$
$\displaystyle=$ $\displaystyle\sup_{\theta>0}g(\theta)$

where $g(\theta)=\theta(1-e^{-\rho/\theta})$. By a straightforward exercise in
single-variable calculus, it is easy to see that $g(\theta)\leq\rho$ for all
$\theta>0$. Thus,

$d(\mu_{1/\delta_{1}},\mu_{1/\delta_{2}})\leq\sup_{\theta>0}g(\theta)\leq\rho=\absolutevalue{\delta_{1}-\delta_{2}},$

as was asserted. ∎

######  Lemma 4.

For $\delta=0$, $\phi_{\delta}=\phi_{0}$ is not twice differentiable and
hence, for any $m\geq 2$, $\phi_{0}\notin C^{m}(\mathbb{R}^{d})$.

###### Proof.

Consider $f:\mathbb{R}\to\mathbb{R}$ defined by

$f(x)=e^{|xe_{1}|^{2m}}\phi_{0}(xe_{1})-P(xe_{1})=\absolutevalue{x}^{m+2}\sin(1/|x|^{m})$

for $x\in\mathbb{R}$. It is straightforward to verify that $f$ is continuously
differentiable on $\mathbb{R}$ with

$f^{\prime}(x)=\begin{cases}x\Big{(}(m+2)|x|^{m}\sin(1/\absolutevalue{x}^{m})-m\cos(1/\absolutevalue{x}^{m})\Big{)}&x\neq
0\\\ 0&x=0.\end{cases}$

With this, we see that the limit

$\lim_{h\to 0}\frac{f^{\prime}(h)-f^{\prime}(0)}{h}=\lim_{h\to
0}\left((m+2)\absolutevalue{h}^{m}\sin(1/\absolutevalue{h}^{m})-m\cos(1/\absolutevalue{h}^{m})\right)$

does not exist and hence $f$ is not twice differentiable at $0$. Of course, if
$\phi_{0}\in C^{m}(\mathbb{R}^{d})\subseteq C^{2}(\mathbb{R}^{d})$, the
construction of $f$ would imply that $f\in C^{2}(\mathbb{R})$ which is not the
case. Hence $\phi_{0}\notin C^{m}(\mathbb{R}^{d})$. ∎

###### Proof Theorem 1.

We first verify that $d_{m}$ is finite on $\mathcal{M}_{m}^{M}$. To this end,
let $\nu_{1},\nu_{2}\in\mathcal{M}_{m}^{M}$. In view of Lemma 1, an
application of Taylor’s theorem (with remainder in Peano form) guarantees
that, for $k=1,2$,

$\displaystyle\widehat{\nu_{k}}(\xi)$ $\displaystyle=$
$\displaystyle\sum_{\absolutevalue{\beta}\leq
m}\frac{D^{\beta}(\widehat{\nu_{k}})(0)}{\beta!}\xi^{\beta}+\mathcal{E}_{k}(\xi)\absolutevalue{\xi}^{m}$
$\displaystyle=$ $\displaystyle\sum_{\absolutevalue{\beta}\leq
m}\frac{(-i)^{\absolutevalue{\beta}}M_{\beta}}{\beta!}\xi^{\beta}+\mathcal{E}_{k}(\xi)\absolutevalue{\xi}^{m}$

where $\mathcal{E}_{k}$ is a continuous function with $\mathcal{E}_{k}(\xi)\to
0$ as $\xi\to 0$; here, we have used the fact that $\nu_{1}$ and $\nu_{2}$
have matching moments (given by $M=\\{M_{\beta}\\}$) up to order $m$). Thus

$\sup_{0<\absolutevalue{\xi}\leq
1}\frac{\absolutevalue{\widehat{\nu_{1}}(\xi)-\widehat{\nu_{2}}(\xi)}}{\absolutevalue{\xi}^{m}}=\sup_{0<\absolutevalue{\xi}\leq
1}\absolutevalue{\mathcal{E}_{1}(\xi)-\mathcal{E}_{2}(\xi)}=:C<\infty.$

Consequently,

$\displaystyle d(\nu_{1},\nu_{2})=\sup_{\xi\neq
0}\frac{\absolutevalue{\widehat{\nu_{1}}(\xi)-\widehat{\nu_{2}}(\xi)}}{\absolutevalue{\xi}^{m}}$
$\displaystyle\leq
C+\sup_{\absolutevalue{\xi}>1}\absolutevalue{\widehat{\nu_{1}}(\xi)-\widehat{\nu_{2}}(\xi)}\leq
C+\absolutevalue{\nu_{1}}(\mathbb{R}^{d})+\absolutevalue{\nu_{2}}(\mathbb{R}^{d})<\infty$

where we have used that, for $k=1,2$, the total variation
$\absolutevalue{\nu_{k}}$ of $\nu_{k}$ is a finite measure. Thus $d_{m}$ is
finite for each pair $\nu_{1},\nu_{2}\in\mathcal{M}_{m}^{M}$ and, since
$\mathcal{M}_{m}^{M}$ is non-empty by Lemma 2, we may conclude that
$(\mathcal{M}_{m}^{M},d_{m})$ is a metric space.

It remains to show that $(\mathcal{M}_{m}^{M},d_{m})$ is not complete. To this
end, for each positive integer $j$, define $\mu_{j}\in\mathcal{M}_{m}^{M}$ to
be the measure with Fourier transform $\phi_{1/j}$ guaranteed by Lemma 2. By
virtue of the Lemma 3,

$d(\mu_{j},\mu_{k})\leq\absolutevalue{\frac{1}{j}-\frac{1}{k}}$

and thus $\\{\mu_{j}\\}\subseteq\mathcal{M}_{m}^{M}$ is a Cauchy sequence in
the metric $d_{m}$. For suppose that, for some measure
$\mu\in\mathcal{M}_{m}^{M}$, $\lim_{j\to\infty}d_{m}(\mu_{j},\mu)=0$. By the
definition of $d_{m}$, it follows that
$\widehat{\mu}(\xi)=\lim_{j\to\infty}\phi_{1/j}(\xi)=\phi_{0}(\xi)$ for each
$\xi\in\mathbb{R}^{d}$. In view of Lemma 1, this implies that $\phi_{0}\in
C^{m}(\mathbb{R}^{d})$ which stands in contradiction to Lemma 4. Thus,
$(\mathcal{M}_{m}^{M},d_{m})$ is not complete. ∎

The example in the above proof can be used to illustrate that the conclusion
of Lemma 3.2 of [Sta20] fails in the context of complex measures. Through a
careful study of Section 2.9 of Chapter VI of [K04], it becomes clear that the
validity of Lemma 3.2 of [Sta20] is tied to the remarkable nature of positive
definition functions (arising as the Fourier transforms of positive measures
via Bochner’s theorem). Along similar lines, we point the reader to the
article [BD93] of L. Báez-Duarte which illustrates that certain classical
results of convergence of probability measures must be modified to hold in the
complex setting. In particular, this is true of Levy’s continuity theorem
concerning the weak convergence, a theorem which is also used in Stawiska’s
proof. It would be interesting to understand the relationship between weak
convergence and convergence in $d_{m}$ in the context of complex Borel
measures.

## References

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