If characteristic function is $L^2$ does it mean that the distribution is absolutely continuos? Suppose, we have a characteristic function $\phi(t)$. 
Suppose that 
\begin{align}
\int_{-\infty}^\infty |\phi(t)|^2 dt <\infty
\end{align}
Does this mean the corresponding probability distribution is absolutely continuous?
I found that if the following condition  holds
\begin{align}
 \lim _{T \to \infty} \frac{1}{2T}\int_{-T}^T |\phi(t)|^2 dt=0
\end{align}
then the distribution is contionous. 
However, I think they mean continuous and I am not sure if this implies existence of density.     
 A: Yes, that's correct. In fact, the density of the probability distribution is square-integrable (with respect to Lebesgue measure).

Theorem Let $X$ be a random variable such that $$\int |\mathbb{E}e^{i \xi X}|^2 \, d\xi< \infty.$$ Then the distribution of $X$ has a density $f$ with respect to Lebesgue measure and $\int |f(x)|^2 \, dx <\infty$.

In fact, the converse holds also true, but this is a direct consequence of Plancherel's theorem.
References:


*

*Fournier, N., Printems, J.: Absolute continuity for some one-dimensional processes, Lemma 1.1.

*Kühn, F.: Probability and Heat Kernel Estimates for Lévy(-Type) Processes, Lemma 2.11.
Proof of the Theorem: Denote by  $\hat{\mu}(\xi)=\mathbb{E}e^{i \xi \cdot X}$ the characteristic function of $\mu := \mathbb{P}(X \in \bullet)$ and let $Y \in N(0,1)$ be independent of $X$. If wet set $Y_n := X+ \sqrt{\frac{1}{n}} Y$, then the distribution $\mu_n$ of $Y_n$ has a probability density $f_n$ with respect to Lebesgue measure and the characteristic function of $Y_n$ is given by \begin{equation*}
  \hat{\mu}_n(\xi) = \exp \left( - \frac{1}{n} \frac{|\xi|^2}{2} \right) \hat{\mu}(\xi), \qquad \xi \in \mathbb{R}^d.
 \end{equation*}
    This implies, in particular, \begin{equation*}
  \int_{\mathbb{R}^d} |\hat{\mu}_n(\xi)|^2 \, d\xi \leq \int_{\mathbb{R}^d} |\hat{\mu}(\xi)|^2 \, d\xi < \infty.
 \end{equation*}
    Applying Plancherel's theorem, we find \begin{equation*}
  \sup_{n \in \mathbb{N}} \int_{\mathbb{R}^d} |f_n(x)|^2 \, dx 
  = \frac{1}{(2\pi)^d} \sup_{n \in \mathbb{N}} \int_{\mathbb{R}^d} |\hat{\mu}_n(\xi)|^2 \, d\xi 
  \leq \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} |\hat{\mu}(\xi)|^2 \, d\xi < \infty.
 \end{equation*}
    Since the unit ball in $L^2(d\xi)$ is weakly (sequentially) compact (by the Banach-Alaoglu theorem), there exists a subsequence such that $f_{n_k} \to f \in L^2(d\xi)$ in $L^2(d\xi)$. In particular, there exists a subsequence of $(f_{n_k})_{k \in \mathbb{N}}$ which converges (Lebesgue)almost everywhere to $f$; thus, $f \geq 0$ a.e. Moreover, it follows easily that $\mu_n(dx) = f_n(x) \, dx$ converges vaguely to $f(x) \, dx$. On the other hand, we have $\hat{\mu}_n(\xi) \to \hat{\mu}(\xi)$ for all $\xi \in \mathbb{R}^d$, and therefore it follows from Lévy's continuity theorem that $\mu_n \to \mu$ in distribution. This implies in particular $\mu_n \to \mu$ vaguely. Now the uniqueness of vague limits gives $\mu(dx) = f(x) \, dx$, and this finishes the proof.
