Can Laplace's transformation be equal to a Gaussian for any integer?

Question

Let $$\displaystyle M_n(f)=\int_0^1t^nf(t)dt, \quad \forall n\in\mathbb N$$ I ask if they exist a continuous function f on [0,1] such that $$M_n(f)=e^{-n^2}\quad \forall n\in\mathbb N$$ it seems obvious that such f does not exist, but I do not know how to prove it rigorously

My work: if we put $t=e^{-x}$ then $M_n(f)=\int_{0}^{\infty}e^{-nx}g(x)dx$. The function g is defined by $g(x)=f(e^{-x})e^{-x}$

The question amounts to looking for a continuous function g on $] 0, +\infty[$ such that $$\mathcal{L} (g) (n) =e^{-n^2}\quad n\in\mathbb N$$ with $\mathcal{L}$ : Laplace transform

Answer 1

We prove a more general claim:

Claim. Let $\mu$ be a signed finite Borel measure on $[0, 1]$ and write $M_n(\mu) = \int_{[0,1]} t^n \, \mu(\mathrm{d}t)$. Suppose $$\lim_{n\to\infty} r^n M_n(\mu) = 0 \tag{*} $$ holds for any $r > 0$. Then $\mu = c \delta_0$ for some constant $c$.

Note that OP's case corresponds to a signed measure of the form $\mu(\mathrm{d}t) = f(t) \, \mathrm{d}t$. Then the claim tells that there exists no such $\mu$ satisfying $M_n(\mu) = e^{-n^2}$ eventually. Indeed, any such $\mu$ would satisfy $\text{(*)}$, and then the claim leads to a contradiction that $M_n(\mu) = 0$ for all $n \geq 1$.

Proof of Claim. Assume that $\text{(*)}$ holds. For any $r > 0$ and $N \in \mathbb{N}_1$, we define

$$ S_N(r) := \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k!} r^{-Nk} M_{Nk}(\mu). $$

Then from the bound

$$\left| S_N(r) \right| \leq \sum_{k=1}^{\infty} \frac{1}{k!} r^{-Nk}\left| M_{Nk}(\mu)\right| \leq e \sup_{n \geq N} \left( r^{-n}\left| M_n(\mu) \right| \right), $$

we have $ \lim_{N\to\infty} S_N(r) = 0 $ for any $r > 0$. Moreover, by the Fubini's Theorem and the Dominated Convergence Theorem,

\begin{align*} 0 &= \lim_{N\to\infty} S_N(r)\\ &= \lim_{N\to\infty} \int_{[0,1]} \left( \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k!} (t/r)^{Nk} \right) \, \mu(\mathrm{d}t) \tag{$\because$ Fubini} \\ &= \lim_{N\to\infty} \int_{[0,1]} \left( 1 - e^{-(t/r)^N} \right) \, \mu(\mathrm{d}t) \\ &= \int_{[0,1]} \lim_{N\to\infty} \left( 1 - e^{-(t/r)^N} \right) \, \mu(\mathrm{d}t) \tag{$\because$ DCT} \\ &= \int_{[0,1]} \left( \mathbf{1}_{\{t > r\}} + (1-e^{-1})\mathbf{1}_{\{t=r\}} \right) \, \mu(\mathrm{d}t) \\ &= \mu([r,1])-e^{-1}\mu(\{r\}). \end{align*}

(When $r > 1$, we regard $[r, 1] = \varnothing$.) Consequently,

$$ \mu([r, 1]) = 0 $$

holds, initially when $r$ is not an atom of $\mu$, and then for all $r > 0$ by the limiting argument. Therefore $\mu$ must be concentrated at $0$. $\square$

Answer 2

Using the notation $$g(x)\risingdotseq G(s) \quad\text{if}\quad G(s)=\mathcal L(g(x)),$$ one can write $$M_{\large s}(f(t)) = \int\limits_0^1 t^{\large s} f(t)\,\mathrm dt = \int\limits_0^\infty e^{\large-sx}e^{\large-x}f(e^{\large-x})\,\mathrm dx =e^{\large-s^2},$$ $$g(x) = e^{\large-x}f\left(e^{\large-x}\right)\risingdotseq e^{\large-s^2},\tag1$$ $$\dfrac{\sqrt\pi}2\operatorname{erf} s = \int\limits_0^{\large s}M_{\large s}(f(t))\,\mathrm dt = \int\limits_0^1 \dfrac{t^{\large s}}{\ln t}f(t)\,\mathrm dt = \int\limits_0^\infty e^{\large-sx}\dfrac1x e^{\large-x}f(e^{\large-x})\,\mathrm dx,$$ $$\dfrac1x g(x)\risingdotseq \dfrac{\sqrt\pi}2\operatorname{erf} s.\tag2$$ Since $$e^{\large-\frac14x^2}\risingdotseq\sqrt\pi e^{\large s^2}\operatorname{erf}s,\tag3$$ then $$e^{\large-\frac14x^2}*g(x) = \dfrac2x g(x),$$ $$2g(x) = x\int\limits_0^{\large x} e^{\large-\frac14(x-t)^2}\, g(t)\,\mathrm dt = 2\int\limits_0^{\large x} g(t) \left(e^{\large-\frac14(x-t)^2}\right)'_t\,\,\mathrm dt + \int\limits_0^{\large x} e^{\large-\frac14(x-t)^2}\, tg(t)\,\mathrm dt \\ \overset{IBP}{=\!=\!=}\, 2 g(t) \left(e^{\large-\frac14(x-t)^2}\right)\bigg|_0^{\large x} - 2\int\limits_0^{\large x}e^{\large-\frac14(x-t)^2}g'(t)\,\,\mathrm dt + \int\limits_0^{\large x} e^{\large-\frac14(x-t)^2}\, tg(t)\,\mathrm dt,$$ $$\int\limits_0^{\large x} e^{\large-\frac14(x-t)^2}\, (tg(t)-2g'(t))\,\mathrm dt = g(0)e^{\large-\frac14x^2},$$ $$\int\limits_0^{\large x} e^{\large\frac12 xt}e^{\large-\frac14t^2}\, (tg(t)-2g'(t))\,\mathrm dt = g(0).\tag4$$ $RHS(4)=\mathrm{constant}(x),$ so the solution is defined via ODE task $$g'(x) = \frac t2 g(x),\quad g(0)= 0,\tag5$$ without reguar non-zero solutions.