I've been asked that if a real-valued random variable $X$ satisfies that for any $f\in C_c^{\infty}(\mathbb{R})=\{\text{smooth functions with compact support}\}$ one has $\mathbb{E}(f(X+1))=\mathbb{E}(Xf(X))$, then $X$ follows the Poisson($1$) distribution.
I know how to prove this if we assume $X$ is non-negative integer valued and we require $f$ to be bounded rather than smooth. Indeed, for each $A\subset\mathbb{Z}^+$, the equation $f(x+1)-xf(x)=1_{A}(x)-\sum_{k\in A}e^{-1}\frac{1}{k!}$ (here $x\in\mathbb{Z}^+\cup\{0\}$) has a unique solution $f:\mathbb{Z}^+\cup\{0\}\to\mathbb{R}$ with the initial value $f(0)=0$, and this solution is bounded. Then the condition applies to $f$ and thus $P(X\in A)=\sum_{k\in A}e^{-1}\frac{1}{k!}$, which gives the desired conclusion.
However, I can not figure out how this approach (which is standard for this type of question) extends to the problem I've encountered. Any help would be appreciated.
Let $\mu$ be the distribution of $X$. Then we have $$E(f(X + 1)) = \langle f(x + 1), \mu(x)\rangle_{x} = \langle f(x), \mu(x - 1)\rangle_{x}$$ and $$E(Xf(X)) = \langle xf(x), \mu(x)\rangle_{x} = \langle f(x), x\mu(x)\rangle_{x}.$$ Here $\langle g, T\rangle = T(g)$ is the pairing of compactly supported smooth functions $g$ and distributions $T$. Thus we have the equation $$\mu(x - 1) = x\mu(x).$$ Let $\phi(t) = \int e^{itx}\mu(dx)$ be the characteristic function of $\mu$. Take the characteristic function of both sides to get the following equality of tempered distributions: $$e^{it}\phi(t) = \frac{1}{i}\phi'(t).$$ Hence $$\phi'(t) = ie^{it}\phi(t).$$ Since $\mu$ is a probability measure, the right hand side is a continuous function, so $\phi'$ is continuous, so $\phi$ is a $C^1$ function. Moreover $\phi(0) = 1$. We have $$\frac{d\phi}{dt} = ie^{it}\phi,$$ $$\frac{1}{\phi}\,d\phi = ie^{it}dt,$$ $$\log(\phi) = e^{it} + C,$$ $$\phi = C\exp(e^{it}),$$ $$\phi = \exp(e^{it} - 1).$$ Hence $\phi(t) = \exp(e^{it} - 1)$, which is the characteristic function of Poisson(1).
If you didn't know the Poisson characteristic function, you can note that $\phi$ is $2\pi$ periodic, which implies by Fourier inversion that $\mu$ is supported on the Integers with $$\mu(j) = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-itj}\phi(t)\,dt = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-itj}\exp(e^{it} - 1)\,dt.$$ The integral will evaluate to the Poisson PMF.