Minimization of an entropy type functional with biased expectation constraint This question is a continuation of Minimization of an entropy type functional
Let $\mathcal P_c$ be the set of probability densities on $[0,1]$ with mean $c\in [0,1]$, i.e. $p\in \mathcal P_c$ iff
$$\int_0^1 p(x)dx=1,\quad \int_0^1 xp(x)dx=c\quad \mbox{and}\quad p(x)\ge 0, ~~\forall x\in [0,1].$$
Can we still find the minimizer of, or the characterization of the minimizer of
$$\min_{p\in\mathcal P_c}~ \left\{V(p) ~:=~ \int_0^1 \log\big(p(x)\big)p(x)dx + \int_0^1 \big(x\log(x)+(1-x)\log(1-x)\big)p(x)dx\right\}?$$
Any answer and comments are highly appreciated.
 A: For real $a$ and $x\in(0,1)$, let
\begin{equation*}
    p_a(x):=e^{ax+h(x)}/C(a),
\end{equation*}
where $h(x):=-x\ln x-(1-x)\ln(1-x)$ and $C(a):=\int_0^1 e^{ax+h(x)}\,dx$, so that $p_a$  is a pdf on $(0,1)$ and
\begin{equation*}
    V(p)=\int_0^1(p(x)\ln p(x)-h(x)p(x))\,dx \tag{1}
\end{equation*}
and
\begin{equation*}
    \int_0^1 xp_a(x)\,dx=L'(a), \quad L(a):=\ln C(a).
\end{equation*}
Also, the variance of the distribution with pdf $p_a$ is $L''(a)$, so that $L''>0$ and $L$ is strictly convex, and hence $L'(a)$ is strictly and continuously increasing, from $L'(-\infty+)=0$ to $L'(\infty-)=1$ (by Lemma 1 at the end of this answer).

For an illustration, here is the plot $\{(a,L'(a))\colon|a|\le30\}$:


So, for each $c\in(0,1)$, there is a unique real $a_c$ such that $L'(a_c)=c$ and hence
\begin{equation*}
    \int_0^1 xp_*(x)\,dx=c,
\end{equation*}
where
\begin{equation*}
    p_*:=p_{a_c}. 
\end{equation*}
(Note that $\int_0^1 xp(x)\,dx\in(0,1)$ for any pdf $p$ on $(0,1)$.)
So, by (1), for any pdf $q$ on $(0,1)$ with $V(q)<\infty$ and $\int_0^1 xq(x)\,dx=c$, the directional derivative of $V$ at $p_*$ in the direction of $q-p_*$ is
$$\begin{aligned}
&\frac d{dt}\,V(p_*+t(q-p_*))\Big|_{t=0} \\ 
&=\int_0^1 (1+\ln p_*(x)-h(x))(q(x)-p_*(x))\,dx \\ 
&=\int_0^1 (1+a_c x+h(x)-\ln C(a_c)-h(x))(q(x)-p_*(x))\,dx=0,
\end{aligned}\tag{2}$$
since $q$ and $p_*$ are pdf's on $(0,1)$, and $\int_0^1 x(q(x)-p_*(x))\,dx=0$.
The crucial point is that the function $V$ is convex (since $u\ln u$ is convex in $u\ge0$, with $0\ln0:=0$). So, $p_*$ is a minimizer of $V$.

Lemma 1: $L'(-\infty+)=0$ and $L'(\infty-)=1$.
Proof: Note that $0\le h(x)\le\ln2$ for all $x\in(0,1)$, and $h(0+)=h(1-)=0$. Therefore and by dominated convergence, if $0>a\to-\infty$, then, with the substitution $x=u/|a|$,
\begin{equation*}
    C(a)=\frac1{|a|}\,\int_0^{|a|}e^{-u+h(u/|a|)}\,du\sim\frac1{|a|}\,\int_0^{|a|}e^{-u}\,du\sim\frac1{|a|}, 
\end{equation*}
\begin{equation*}
    C'(a)=\frac1{a^2}\,\int_0^{|a|}ue^{-u+h(u/|a|)}\,du\sim\frac1{a^2}\,\int_0^{|a|}ue^{-u}\,du\sim\frac1{a^2}, 
\end{equation*}
so that
\begin{equation*}
    L'(a)=\frac{C'(a)}{C(a)}\sim\frac1{|a|}\to0. 
\end{equation*}
Similarly, if $0<a\to\infty$, then, with the substitution $x=1-u/a$,
\begin{equation*}
    C(a)=\frac{e^a}a\,\int_0^a e^{-u+h(1-u/a)}\,du\sim\frac{e^a}a, 
\end{equation*}
\begin{equation*}
    C'(a)=\frac{e^a}a\,\int_0^a(1-u/a)e^{-u+h(1-u/a)}\,du\sim\frac{e^a}a, 
\end{equation*}
so that
\begin{equation*}
    L'(a)=\frac{C'(a)}{C(a)}\to1. 
\end{equation*}
$\Box$
