# Missing argument in the proof of the Levy-Khintchine representation .

In one proof of the Levy Khintchine representation of the Laplace exponent of subordinators, the following argument is used:

"Assume $$f_n : [0,\infty) \to [0,\infty)$$ is a sequence of non-increasing functions such that

$$\int_{0}^{\infty} e^{-\lambda x} f_n(x) dx \; \text{ converges towards a finite value as } \; n \to \infty \text{ for each }\; \lambda \geq 0,$$

then necessarily the limit has the form

$$\int_{0}^{\infty} e^{-\lambda x} (\alpha \delta_0(x) + f(x) dx) = \alpha + \int_{0}^{\infty} e^{-\lambda x} f(x) dx$$

for some $$\alpha\geq 0$$ and some non-increasing function $$f$$."

How to prove this statement ?

It is quite unusual to be able to say anything about the "densities" under such a weak convergence statement. But the non-increasing assumption is crucial here.

## 1 Answer

Hopefully there is a more concise answer. I will prove much more than required, namely that $$f_n(x) dx$$ converges in distribution towards $$\delta_0(dx) + f(x) dx$$ with $$f$$ decreasing, and that $$f_n \to f$$ on every point of continuity of $$f$$. The idea will be to use the existence results for the moment problem.

Write $$g_n(\lambda) = \int_0^\infty e^{-\lambda x} f_n(x) \mathrm dx$$, then for every $$\lambda\geq 0$$, as $$n\to\infty$$ we have $$g_n(\lambda) \to g(\lambda)$$.

The derivatives of $$g_n$$.

Write $$h^{(k)}$$ for the $$k$$-th derivative of function $$h$$. We can easily check that $$g_n$$ is infinitely differentiable on $$(0,\infty)$$ with $$g_n^{(k)}(\lambda) = \int_0^\infty (-x)^k f_n(x) e^{-\lambda x} dx .$$

A bound on the moments of $$f_n(x) e^{-\lambda x}$$.

Since $$g_n(t) \to g(t)$$ for every $$t$$ we have $$A(t) := \sup_n g_n(t) < \infty$$. Fix $$\varepsilon>0$$. For every $$k$$, the maximum of $$x^k e^{-\varepsilon x}$$ is attained for $$x=k/\varepsilon$$ and equals $$e^{-1} k^k \varepsilon^{-k}$$, so that for every $$\lambda>\varepsilon$$ and every $$n$$, $$$$\int_0^\infty x^k f_n(x) e^{-\lambda x} dx \leq e^{-1} \varepsilon^{-k} g_n(\lambda-\varepsilon) \leq e^{-1} k^k \varepsilon^{-k} A(\lambda-\varepsilon) .$$$$ The fact that $$f_n$$ is non-increasing allows us to write $$tf_n(t) \leq \int_0^t f_n(x) dx \leq g_n(0) \leq A(0) .$$ Then, for $$0<\lambda\leq\mu$$, $$$$|g_n^{(k)}(\lambda) - g_n^{(k)}(\mu)| \leq \int_0^\infty x^k \frac{A(0)}{x} (e^{-\lambda x} - e^{-\mu x}) dx \leq (\lambda^{-k} - \mu^{-k}) A(0) k! .$$$$

The derivatives of $$g_n$$ converge to those of $$g$$.

We then proceed by induction. Assume that $$g_n^{(k)} \to g^{(k)}$$ for some $$k\geq 0$$ — this is the case for $$k=0$$. Since $$(-1)^k g_n^{(k+2)}$$ is non-negative, $$(-1)^k g_n^{(k+1)}$$ is increasing. Then for every $$0 and every $$n$$, we have $$\begin{multline} (-1)^k g_n^{(k+1)}(t) - A(0) k! (t^{-k} - u^{-k}) \leq (-1)^k g_n^{(k+1)}(u) \\\\ \leq (-1)^k\frac{g_n^{(k)}(u) - g_n^{(k)}(t)}{u-t} \leq (-1)^kg_n^{(k+1)}(t) \leq (-1)^k\frac{g_n^{(k)}(t)-g_n^{(k)}(s)}{t-s} \\\\ \leq (-1)^k g_n^{(k+1)}(s) \leq (-1)^k g_n^{(k+1)}(t) + A(0) k! (s^{-k} - t^{-k}) . \end{multline}$$ Taking the limit as $$n\to\infty$$, we first have $$(-1)^k\frac{g^{(k)}(u) - g^{(k)}(t)}{u-t} \leq (-1)^k\frac{g^{(k)}(t)-g^{(k)}(s)}{t-s}$$ as well as $$\left| \frac{g^{(k)}(u) - g^{(k)}(t)}{u-t} - \frac{g^{(k)}(t)-g^{(k)}(s)}{t-s} \right| \leq A(0) k! (s^{-k} - u^{-k}) ,$$ which together allow us to conclude that $$g^{(k)}$$ is differentiable in $$t$$. In addition, by the squeeze theorem, we have $$g_n^{(k+1)}(t) \to g^{(k+1)}(t)$$.

Existence of a limiting measure.

For every $$n,N$$, the $$N\times N$$ matrix $$H(n, \lambda)$$ where $$H(n,\lambda)_{i,j} = \int_0^\infty x^{i+j} f_n(x) e^{-\lambda x} dx = g_n^{(i+j)}(\lambda)$$ is positive semi-definite, see here. We have $$H(n,\lambda) \to H(\lambda)$$ as $$n\to\infty$$, where $$H(\lambda)_{i,j} = g^{(i+j)}(\lambda)$$. Since the eigenvalues of a matrix depend continuously on its coefficients, we deduce that $$H(\lambda)$$ is also positive semi-definite, thus that it is the Hankel matrix of a measure: hence, there exists $$\mu_\lambda$$ such that $$g^{(k)}(\lambda) = \int_0^\infty x^k \mu_\lambda(d x) .$$

Connexion between the $$\mu_\lambda$$.

The moments of $$\mu_\lambda$$ determine it uniquely. Indeed, by our previous estimates, for every $$0<\varepsilon<\lambda$$ $$\int_0^\infty x^k \mu_\lambda(dx) = \lim_{n\to\infty} \int_0^\infty x^k f_n(x) e^{-\lambda x} dx \leq e^{-1} k^k \varepsilon^{-k} A(\lambda-\varepsilon) .$$ Thus they satisfy Carleman’s condition.

We can then use the Fréchet-Shohat theorem to conclude that $$f_n(x) e^{-\lambda x} dx \to \mu_\lambda$$ in distribution.

In particular, since $$x \mapsto e^{-\mu x}$$ is continuous and bounded on $$[0,\infty)$$ for every $$\eta \geq 0$$, we have that for every $$h$$ continuous and bounded $$\int_0^\infty e^{-\eta x} h(x) f_n(x) e^{-\lambda x} dx \to \int_0^\infty e^{-\eta x} h(x) \mu_\lambda(dx)$$ as $$n\to \infty$$, but we also have that it converges to $$\int_0^\infty h(x) \mu_{\lambda+\eta}(dx)$$. Since this holds for every continuous and bounded test function $$h$$, we conclude that for every $$\lambda, \eta>0$$, $$\mu_{\lambda+\eta}(dx) = e^{-\eta x} \mu_\lambda(dx)$$.

This leads us to observe that for every $$t < \lambda$$, we have $$\int_0^\infty e^{+t x} \mu_{\lambda}(dx) = g(\lambda-t) < \infty$$. By the monotone convergence theorem, taking the limit as $$t\to\lambda$$, we get $$\int_0^\infty e^{\lambda x} \mu_\lambda(dx) = \lim_{t\to\lambda} g(\lambda-t) = g(0) < \infty .$$ It follows that there exists a finite measure $$\mu_0$$ with total mass $$g(0)$$ such that for every $$\lambda>0$$, $$\mu_\lambda(dx) = e^{-\lambda x} \mu_0(dx)$$. In addition, using the monotone convergence theorem we can also show that $$f_n(x) dx \to \mu_0(dx)$$ as $$n\to\infty$$.

Note that $$\mu_0$$ is not guaranteed to have any finite moment, so we cannot directly construct it like the other $$\mu_\lambda$$.

Towards the representation of $$\mu_0$$.

The dominated convergence theorem gives that $$\lim_{\lambda\to\infty} g(\lambda) = \mu(\{0\})$$.

Consider a non-negative, continuous and bounded function $$\rho$$ with support in $$[0,1]$$ and with integral $$1$$. For every $$0, by $$f_n(s) \leq \int_s^t f_n(x) \frac{1}{t-s} \rho\left(\frac{x-s}{t-s}\right) dx \leq f_n(t) \leq \int_t^u f_n(x) \frac{1}{u-t} \rho\left(\frac{x-t}{u-t}\right) dx \leq f_n(u)$$ and taking the limit as $$n\to\infty$$, $$\int_s^t \frac{1}{t-s} \rho\left(\frac{x-s}{t-s}\right) \mu_0(dx) \leq \liminf f_n(t) \leq \limsup f_n(t) \leq \int_t^u \frac{1}{u-t} \rho\left(\frac{x-t}{u-t}\right) \mu_0(dx) .$$ Taking $$\rho$$ that approaches the indicator function of $$[0,1]$$, then $$s\to t$$ and $$u\to t$$, we can deduce that $$x\mapsto \mu_0([x,\infty))$$ has a left and right derivative $$f$$ on $$(0,\infty)$$ that is non-negative and non-increasing, and $$f_n \to f$$ wherever $$f$$ is continuous.

• Thanks a lot for the wonderful answer Thomas ! Commented Jul 18 at 16:39