Ytukyg

Let $X$ be a real-valued random variable with $varphi_Xin L^2(mathbb{R})$, where $varphi_X$ denotes the characteristic function of $X$. Prove that $P(X=x)=0, forall xinmathbb{R}$.

Since $varphi_Xin L^2(mathbb{R})$, I believe that implies that $X$ must be continuous, then we use that to conclude that $P(X=x)=0$ for all $xinmathbb{R}$. However, I am having a hard time formalizing this idea. Any help would be appreciated.

Hints: Show the following auxiliary statements:

1. Let $chi$ be the characteristic function of a real-valued random variable $Y$ with distribution $mu$, i.e. $$chi(xi) = mathbb{E}e^{i xi Y} = int_{mathbb{R}} e^{i xi y} mu(dy).$$ Apply Fubini's theorem to show that $$frac{1}{2T} int_{-T}^T chi(xi) e^{-ix xi} dxi = mu({x}) + int_{mathbb{R} backslash {x}} frac{sin(T(y-x))}{T(y-x)} mu(dy).$$ Use the dominated convergence theorem to let $T to infty$: $$lim_{T to infty} frac{1}{2T} int_{-T}^T chi(xi) e^{-ix xi} , dxi = mu({x}) = mathbb{P}(Y=x).$$
   


2. Let $phi$ be the characteristic function of a real-valued random variable $X$. Show that $|phi|^2$ is the characteristic function of the random variable $Y:=X-X'$ where $X' sim X$ is an independent copy of $X$.

Now let $X$ be a random variable such that its characteristic function $phi$ is square-integrable. By Step 2, we know that $|phi|^2$ is the characteristic function of $Y=X-X'$ where $X' sim X$ is an independent copy of $X$. Since

$$lim_{T to infty} left| frac{1}{2T} int_{-T}^T |phi(xi)|^2 e^{-ix xi} dxi right| leq lim_{T to infty} frac{1}{2T} int_{mathbb{R}} |phi(xi)|^2 d xi =0$$

it follows from Step 1 (applied to $chi=|phi|^2$) that

$$mathbb{P}(Y=x) = 0$$

for all $x in mathbb{R}^d$. Since $Y=X-X'$ for $X sim X'$ independent, this implies that $mathbb{P}(X=x)=0$ for all $x in mathbb{R}^d$.