= Solution
Let $I=(U')^{-1}$ be the <inverse marginal utility>. The <derivative> $U'$ is continuous and strictly decreasing: concavity makes it nonincreasing, and equality at two distinct points would make $U$ affine between them, contradicting <strict concavity>. The <Inada conditions> therefore make $U'$ a bijection from $(0,\infty)$ onto $(0,\infty)$. For each $y>0$, the <derivative> of $U(x)-xy$ is positive below $I(y)$ and negative above it. Consequently its unique global maximum is finite and
$$
V(y)=U(I(y))-yI(y).
$$
There is no need to differentiate $I$ to find the first <derivative> of this <utility conjugate>. Comparing the maxima at $y$ and $y+h$ gives, for $h>0$,
$$
-hI(y)\leq V(y+h)-V(y)\leq-hI(y+h).
$$
Continuity of $I$, together with the corresponding inequalities for $h<0$, gives
$$
\boxed{V'(y)=-I(y),\qquad V'(0+)=-\infty,\qquad V'(\infty)=0.}
$$
Thus $V$ is strictly decreasing and <strictly convex>, since its <derivative> is strictly increasing. This proves all the asserted properties except the second <differentiability>, which does not follow from the printed hypotheses.
Here is an explicit <Inada utility with vanishing curvature>. Put
$$
w(s)=\frac{(s-1)^2}{s^2(s+1)^2},\qquad p(x)=\int_x^\infty w(s)\,ds=\frac1x+\frac4{x+1}+4\log\frac{x}{x+1},\qquad U(x)=\int_1^x p(r)\,dr.
$$
The positive integral defining $p$ makes $U$ strictly increasing. Also $p'=-w$ is negative except at one point, so $p$ is strictly decreasing and $U$ is <strictly concave>. Near zero $w(s)$ is asymptotic to $s^{-2}$, and at infinity it is asymptotic to $s^{-2}$; thus $p(0+)=\infty$ and $p(\infty)=0$. This smooth <utility function> satisfies every printed positive-domain assumption, but $U''(1)=0$. At $y_0=p(1)=3-4\log2$, differentiability of $I$ would imply, by differentiating $p(I(y))=y$,
$$
0\cdot I'(y_0)=1,
$$
which is impossible. Hence $V'=-I$ has no finite <derivative> there. More explicitly, $p(x)-p(1)=-(x-1)^3/12+o((x-1)^3)$, so its <inverse function> has a cube-root singularity. \b[The twice-differentiable conclusion is false as printed.] With the additional assumption $U''(x)<0$ everywhere, the <inverse function theorem> proves the intended <dual differentiability with nonvanishing utility curvature>:
$$
\boxed{V''(y)=-\frac1{U''(I(y))}>0.}
$$
For the financial assertions, use the positive-wealth <admissible trading strategy> class appropriate to the positive-domain <utility function>, and expectations for which the expressions are defined. A normalized terminal <state-price density> satisfies $Z_N>0$, $\mathbb EZ_N=1$, and defines an <equivalent martingale measure> $Q$ by $dQ=Z_N\,dP$. A <self-financing portfolio> obeys
$$
X_n=x+\sum_{k=0}^{n-1}\pi_k\cdot(S_{k+1}-S_k),
$$
where the holdings are measurable before the next price increment. Under $Q$ the gains are a <martingale> when integrable; for the usual lower-bounded admissible class, localization gives a <supermartingale>. In particular the <state-price budget constraint> is $\mathbb E[Z_NX_N]\leq x$. The positive-domain qualification matters: assumptions only on $U$ at positive arguments do not control its values at nonpositive wealth.
The definition of the <utility conjugate> gives the pointwise inequality
$$
U(X_N)\leq V(yZ_N)+yZ_NX_N.
$$
Taking expectations, using the <state-price budget constraint>, and then taking the supremum over strategies gives $u(x)\leq\mathbb EV(yZ_N)+yx$ for every $y>0$ and every <state-price density>. Taking the two infima proves the <utility duality with martingale deflators> upper bound:
$$
\boxed{u(x)\leq\inf_{y>0}\{v(y)+xy\}.}
$$
If an expectation is infinite, use the same inequality in the extended-real sense whenever the negative parts make it well defined; no subtraction of two infinities is needed.
A <complete market> is one in which every terminal <contingent claim> in the specified claim class is attainable by a <self-financing portfolio>. Bounded terminal claims suffice for the uniqueness argument. In the standard finite-horizon discrete-time market, absence of <arbitrage> gives existence of an <equivalent martingale measure> by the <fundamental theorem of asset pricing>. To prove uniqueness, let $Z_N$ and $\widetilde Z_N$ be two normalized <state-price densities>. Every bounded claim $H$ has a replicating price $p(H)$, and its replication has the same price under either <equivalent martingale measure>:
$$
\mathbb E[Z_NH]=p(H)=\mathbb E[\widetilde Z_NH].
$$
Take $H=\mathbf1_A$ for every terminal event $A$. The integrable difference $Z_N-\widetilde Z_N$ has zero integral over every event, hence vanishes almost surely: its positive and negative sets would otherwise give a nonzero integral. \b[The normalized terminal state-price density is unique up to null sets.] Its density process is then uniquely $Z_n=\mathbb E[Z_N\mid\mathcal F_n]$. This uses the usual pricing replication class, where bounded claims have bounded replicating value processes; arbitrary doubling strategies are not an admissible definition of completeness.
For <logarithmic utility>, direct maximization gives $V(y)=-\log y-1$. In the <complete market>, write $Z=Z_N$. The unique-density <utility conjugate> bound is minimized at $y=1/x$, giving
$$
u(x)\leq\log x-\mathbb E\log Z.
$$
The candidate terminal <portfolio wealth> is $H^*=x/Z$, whose cost is exactly $\mathbb E[ZH^*]=x$. If completeness includes nonnegative $Q$-integrable claims, it is replicable and attains the bound. There is also an approximation proof using only bounded-claim completeness. Define
$$
b_n=\mathbb E\frac{Z}{Z\vee n^{-1}},\qquad H_n=\frac{x}{b_n(Z\vee n^{-1})}.
$$
Then $0<b_n\leq1$, $b_n\uparrow1$, $H_n$ is positive and bounded, and $\mathbb E[ZH_n]=x$. Replicating $H_n$ therefore gives an admissible terminal claim of initial cost $x$. Since $\mathbb E(\log Z)^+\leq\mathbb EZ=1$, the negative parts of $-\log(Z\vee n^{-1})$ have an integrable common bound. <Monotone convergence> after adding this lower bound gives
$$
\mathbb E\log H_n=\log x-\log b_n-\mathbb E\log(Z\vee n^{-1})\longrightarrow\log x-\mathbb E\log Z.
$$
Consequently the <logarithmic terminal wealth in a complete market> is
$$
\boxed{u(x)=\log x-\mathbb E\log Z_N,\qquad H^*=\frac{x}{Z_N}.}
$$
The value may be $+\infty$ if $\mathbb E(\log Z_N)^-=\infty$; the displayed formula remains meaningful, while attainment by $H^*$ depends on the precise admissible claim class.
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