Do equal integrals of $1/(1+x^a)$ imply equal measure?

Question

Suppose $\mu$ and $\nu$ are two probability measures on $[0,1]$ with the property that $\int_{0}^{1} \frac{1}{1+x^a} ~d\mu(x) = \int_{0}^{1} \frac{1}{1+x^a} ~d\nu(x)$ holds for every exponent $a > 0$. Does this imply that $\mu$ and $\nu$ are the same measure? If the answer is yes, is it sufficient to restrict to positive integer exponents $a$?

Answer 1

Yes, provided that you know a priori that $\int_{(0,1]} x^{-\kappa} d\mu(x) < \infty$ (and the same for $\nu$) for some $\kappa > 0$. By taking the limit $a \to \infty$, you can detect the size of the atom at the origin (if any), so we can assume without loss of generality that the measures are on $(0,1]$. Setting $f(z) = {1\over 2} (1+\tanh({z\over 2}))$, we can write $$ \int_0^1 {1\over 1+x^a}d\mu(x) = \int_0^\infty f(ay) d\hat \mu(y) $$ where $\hat \mu$ is the pushforward of $\mu$ under $x \mapsto -\log x$. (Our additional condition then guarantees that $\hat \mu$ has some finite exponential moment.) Taking derivatives in $a$ at $a=0$, we see that equality of the first expression between $\mu$ and $\nu$ implies that for every $k \ge 0$ $$ f^{(k)}(0) = 0 \quad\text{or}\quad \int_0^\infty y^k d\hat \mu(y) = \int_0^\infty y^k d\hat \nu(y)\;. $$ Since the $f^{(k)}(0)$'s are known to be proportional to the Bernoulli numbers for odd $k$, we conclude that the odd moments of $\hat \mu$ and $\hat \nu$ coincide. In particular, the first moments coincide, so we can normalise $\tilde \mu(dy) := y \hat \mu(dy)$ (and similarly for $\nu$) to be probability measures on $\mathbb{R}_+$ with coinciding even moments. Extend them evenly to all of $\mathbb{R}$ (so that odd moments vanish and therefore also coincide) and you're in the setting of the Hamburger moment problem. The fact that $\tilde \mu$ has some exponential moment implies the Carleman condition and we're done.

Answer 2

Let $m = \mu - \nu$. As in Martin Hairer's answer, we first consider the limits $a \to \infty$ and $a \to 0$ to see that $$ m(\{0\}) = m(\{1\}) = 0 $$ (only here we use the fact that $\mu$ and $\nu$ are probability measures, otherwise they can be arbitrary finite measures). Thus, $m$ is a measure on $(0, 1)$.

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We now substitute $a = e^{-b}$ and $$ t = \log(-\log x) , \qquad x = \exp(-e^t) . $$ If $M$ is the push-forward of $m$ under $x \mapsto t$, then we find that $$ \int_{-\infty}^\infty \frac{1}{1 + \exp(-e^{t - b})} M(dt) = 0 $$ for every $b \in \mathbb R$ (this is again similar to what Martin Hairer did in his answer). In other words, the convolution of a finite measure $M$ with the bounded function $$ \phi(t) = \frac{1}{1 + \exp(-e^{-t})} $$ is identically zero.

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By an appropriate variant of Wiener’s Tauberian theorem, the distributional Fourier transform of $M$ (a continuous function) is equal to zero on the spectrum of $\phi$ (that is, on the support of the distributional Fourier transform of $\phi$). Thus, it remains to prove that the spectrum of $\phi$ is all of $\mathbb R$.

Note that $\phi - 1 + \tfrac12 \mathbb 1_{(0,\infty)}$ decays exponentially fast at $\pm \infty$, and therefore its Fourier transform is an analytic function. The distributional Fourier transform of $1$ is just the Dirac measure. Finally, the distributional Fourier transform of $\mathbb 1_{(0,\infty)}$ is equal to an analytic function in $\mathbb R \setminus \{0\}$, and it has a singularity at zero. It follows that the distributional Fourier transform $\hat\phi$ of $\phi$ is equal to an analytic function in $\mathbb R \setminus \{0\}$. Since $\phi$ is real-valued, it is easy to see that $\hat\phi$ is not identically equal to zero neither in $(0, \infty)$ nor in $(-\infty, 0)$. We conclude that the support of $\hat\phi$ is indeed all of $\mathbb R$, as claimed.

(In fact, we can evaluate the distributional Fourier transform of $\phi$ explicitly: if I did not make a mistake, we have $$ \int_{-\infty}^\infty \phi(t) e^{-i z t} dt = \frac{3 \pi}{2} \delta_0(z) + \operatorname{P{.}V{.}} (1 - 2^{-1 + i z}) \Gamma(i z) \zeta(i z) $$ in the sense of distributions, and the support of the right-hand side is indeed $\mathbb R$.)