= Solution
Let the one-period riskless gross return be $R>0$, with $0<d<R<u$, and let the physical up probability be $\pi\in(0,1)$. The <binomial market> has <risk-neutral probability>
$$
q=\frac{R-d}{u-d}\in(0,1),\qquad qu+(1-q)d=R.
$$
Under its <equivalent martingale measure> $Q$, successive multipliers $J_k$ are independent, taking $u,d$ with probabilities $q,1-q$. The <risk-neutral pricing> function has the extension
$$
g_r(s)=R^{-(n-r)}\mathbb E_Q\left[f\left(s\prod_{k=r+1}^nJ_k\right)\right],\qquad s>0.
$$
For every fixed positive multiplier $A$, the function $s\mapsto f(As)$ is <convex> when $f$ is <convex>. Taking a positive weighted sum preserves <convexity>. More explicitly, for $0\leq\lambda\leq1$,
$$
g_r(\lambda s+(1-\lambda)t)\leq\lambda g_r(s)+(1-\lambda)g_r(t).
$$
Thus the pricing extension is <convex>, and so are its values on the attainable stock-price grid.
The <replicating portfolio in a binomial market> holds
$$
\Delta_r(s)=\frac{g_{r+1}(us)-g_{r+1}(ds)}{(u-d)s}
$$
shares over the next interval. If $r+1<n$, put $A=\Delta_{r+1}(us)$ and $D=\Delta_{r+1}(ds)$. These are the secant slopes of $g_{r+2}$ on $[uds,u^2s]$ and $[d^2s,uds]$, respectively. For a <convex function>, the slope of a secant on a later adjacent interval is no smaller than that on the earlier interval: apply the defining <convexity> inequality to the middle point $uds$, then rearrange. Hence $A\geq D$. Using <backward option pricing> twice gives
$$
\begin{aligned}
g_{r+1}(us)-g_{r+1}(ds)
&=R^{-1}\left[q\{g_{r+2}(u^2s)-g_{r+2}(uds)\}+(1-q)\{g_{r+2}(uds)-g_{r+2}(d^2s)\}\right],\\
\Delta_r(s)&=\frac{qu}{R}A+\frac{(1-q)d}{R}D.
\end{aligned}
$$
The two coefficients are positive and sum to one. Therefore <convex-payoff delta monotonicity in a binomial market> yields
$$
\boxed{\Delta_{r+1}(us)\geq\Delta_r(s)\geq\Delta_{r+1}(ds).}
$$
In particular an up move cannot reduce the <stock> holding. “Increases” here means nondecreases: an affine payoff gives equality, so strict increase cannot be claimed for every <convex> payoff.
For the <expected utility maximization> problem, let $P$ be the physical measure and $Z=dQ/dP$ on terminal paths. If a path has $j$ up moves, its <binomial-market probability density> is
$$
Z_j=\left(\frac q\pi\right)^j\left(\frac{1-q}{1-\pi}\right)^{n-j}.
$$
Every terminal payoff is replicable in this <complete market>. Buying a positive payoff $H$ with initial wealth $w_0$ therefore imposes the <state-price budget constraint> $\mathbb E_QH=R^nw_0$. Since the <utility function> is increasing, the full budget is used. Its derivative and second derivative are
$$
v'(x)=x^{-(\gamma-1)/\gamma},\qquad
v''(x)=-\frac{\gamma-1}{\gamma}x^{-(2\gamma-1)/\gamma}<0.
$$
Put $a=\gamma/(\gamma-1)$. The first-order condition $v'(H_*)=\lambda Z$ for a positive multiplier $\lambda$ gives $H_*=C Z^{-a}$. The budget determines $C$ uniquely. Independence of the successive moves gives
$$
D:=\mathbb E_QZ^{-a}
=\left[q\left(\frac q\pi\right)^{-a}+(1-q)\left(\frac{1-q}{1-\pi}\right)^{-a}\right]^n.
$$
Thus the optimal claim, as a function of the number of up moves, is
$$
\boxed{H_*(j)=\frac{R^nw_0}{D}
\left(\frac\pi q\right)^{aj}
\left(\frac{1-\pi}{1-q}\right)^{a(n-j)}.}
$$
Because $S_n=S_0u^jd^{n-j}$ uniquely determines $j$, this is also a terminal-stock-price payoff, attainable by <backward option pricing> and <claim replication>. It rewards states relatively more probable under the investor's physical measure than under the pricing measure. If $\pi=q$, the optimum is the riskless payoff $R^nw_0$.
To prove global optimality rather than merely assert a stationary point, use the tangent inequality for the strictly <concave function> $v$:
$$
v(H)\leq v(H_*)+v'(H_*)(H-H_*)=v(H_*)+\lambda Z(H-H_*).
$$
Taking physical <expectations> makes the last term zero for any fully financed competing claim, and nonpositive for any claim costing less. Therefore $\mathbb E_Pv(H)\leq\mathbb E_Pv(H_*)$, with equality only when $H=H_*$ almost surely. Finiteness causes no difficulty on the finite binomial tree, and $H_*>0$ in every terminal state.
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