Does the truncated Hausdorff moment problem admit absolutely continuous solutions? Let $\mu$ be a (Borel) probability measure on $[0,1]$ and define $m_j(\mu) = \int x^j\,\mu(dx)$. Let $k$ be a positive integer and consider the set $\mathcal C_{\mu,k}$ of probability measures $\nu$ on $[0,1]$ such that $m_j(\nu) = m_j(\mu)$ for $j = 1,\dotsc,k$.
We are interested in whether $\mathcal C = \mathcal C_{\mu,k}$ contains an absolutely continuous probability measure.
Some restrictions are obviously necessary: When $k=1$, evidently we must rule out $\mu \{0\} = 1$ or $\mu \{1\} = 0$, as either statement implies that $\mathcal C$ contains one point.  More generally, if we define the set 
$$
\mathcal M_k = \{ (m_1(\nu),\dotsc,m_k(\nu) ) \}
$$
of achievable moments,
where $\nu$ ranges over the probability measures on $[0,1]$, then it would seem (though we have not shown) that the boundary $\partial \mathcal M_k$ corresponds to discrete distributions supported by at most $k$ points.
We are aware of some literature on truncated Hausdorff moment problems, and have looked through it, but our particular question does not seem to be addressed.  When explicit representations (say, by densities that are sums of Bernstein polynomials) are used, the emphasis is usually on showing that the moments can be arbitrarily well approximated, but this does not rule out the possibility that the limit is no longer absolutely continuous.
 A: Here is a positive answer, for the interior points of the set $M_k:=\mathcal M_k$. Let indeed $c=(c_1,\dots,c_k)$ be any point in the interior of $M_k$. 
Let $P$ stand for the set of all probability measures on $[0,1]$. 
Let us show that then there is an absolutely continuous measure $\nu\in P$ such that $m_j(\nu)=c_j$ for $j=1,\dots,k$. 
For each $\mu\in P$ and each real $h\in(0,1/2)$, define $\mu_h\in P$ by the convolution-like condition that 
$$\int_{[0,1]} f\,d\mu_h=\int_{[0,1/2]}\mu(dx)\int_0^h\frac{du}h\,f(x+u)
+\int_{(1/2,1]}\mu(dx)\int_0^h\frac{du}h\,f(x-u) \tag{1}
$$
for all nonnegative Borel functions $f$ on $[0,1]$. Then $\mu_h$ is clearly absolutely continuous, with density 
$$[0,1]\ni y\mapsto\frac1h\,\mu\big((y-h,y)\cap[0,1/2]\big)+\frac1h\,\mu\big((y,y+h)\cap(1/2,1]\big).$$ 
Since $\mu\mapsto\mu_h$ is a linear operator, the set 
$$M_{k,h}:=\{(m_1(\mu_h),\dots,m_1(\mu_h))\colon\mu\in P\}
$$
is convex.  
It is enough to show that $c=(c_1,\dots,c_k)\in M_{k,h}$ for small enough $h$. Suppose the contrary. Then there is a hyperplane $H$ in $\mathbb R^k$ through the point $c$ such that the set $M_{k,h}$ is contained in a closed half-space $H_+$ whose boundary is $H$. Since $c$ is in the interior of $M_k$, there is a sphere $S_c(r)$ of positive radius $r$ centered at $c$ such that $S_c(r)\subseteq M_k$. Take the point $b=(b_1,\dots,b_k)\in S_c(r)$ that is at distance $r$ from $H_+$ and hence at distance $\ge r$ from the set $M_{k,h}$. 
Since $b\in S_c(r)\subseteq M_k$, there is some measure $\mu\in P$ such that $m_j(\mu)=b_j$ for all $j=1,\dots,k$. 
Since $|(x+u)^j-x^j|\le j|u|\le k|u|$ for $j=1,\dots,k$ and $x,x+u$ in $[0,1]$, by $(1)$ we have $|m_j(\mu_h)-b_k|=|m_j(\mu_h)-m_j(\mu)|\le kh/2$ for $j=1,\dots,k$, so that the distance from the point $(m_1(\mu_h),\dots,m_k(\mu_h))$ to $b$ is $\le k^{3/2}h/2<r$ if $h<2r/k^{3/2}$, which contradicts the condition that $b$ is at distance $\ge r$ from the set $M_{k,h}$. QED 
Addendum: For a point $c=(c_1,\dots,c_k)$ to be in the interior of $M_k$, it is enough that $c_j=m_j(\mu)$ for some measure $\mu\in P$ with at least $k+1$ points of support and all $j=1,\dots,k$; cf. e.g. the lemma in my answer to the question at [moment-matching]. 
This condition of $\mu$ having at least $k+1$ points of support is however not necessary; indeed, for large $k$, the measure $\mu$ representing the point $c$ may have only about $k/2$ points of support -- but those points need to be specially chosen; namely, the points need then to be the so-called roots of a principal representation of $c$; see e.g. Definitions 3.1 and 3.2 and Corollary 3.1 in Ch. II in [Karlin--Studden].
Addendum 2: In particular, it follows from the first sentence of the above Addendum that the condition that the point $c$ lie in the interior of $M_k$ is, not only sufficient, but also necessary for $c$ to be representable by an absolutely continuous probability measure. That is, $c$ is in the interior of $M_k$ if and only if $c_j=m_j(\mu)$ for some absolutely continuous $\mu\in P$ and all $j=1,\dots,k$. 
A: Yes, this is easy to do with a mild additional assumption (see below), if you are familiar with the connection between moment problems and spectral theory of Jacobi matrices.
An initial piece of the moments corresponds to that many coefficients $a_n,b_n$ of the associated Jacobi difference operator
$$
(Ju)_n=a_n u_{n+1} + a_{n-1}u_{n-1} + b_n u_n .
$$
In this terminology, you will now obtain what you want if you can continue the coefficients in such a way that $\sigma(J)\subseteq [0,1]$ and $J$ is purely absolutely continuous; here, $J$ is now an operator on $\ell^2(\mathbb N)$.
It's very easy in fact to make the spectral measure $\rho$ purely absolutely continuous on $[0,1]$; for example, we can just take $b_n=1/2$, $a_n=1/4$ for all large $n$.
So now we only need to make sure that $\rho$ is supported by $[0,1]$. We will certainly be able to do this if the measure $\rho_N$ that solves the finite moment problem and is supported by the corresponding number of points satisfies $\rho_N(\{ 0, 1\})=0$. This measure is the spectral measure of a finite Jacobi matrix, corresponding to a block in the upper left corner of $J$, written as an infinite matrix. We now take the next $a_n>0$ small, and then continue with $a_n=1/4$, $b_n=1/2$, as planned.
