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Solve the bank equation first:For $\mu >0$ the objective separates into pointwise strictly concave functions of positive $u$. Their derivative is $e^{-\beta t}/u-\mu e^{-\beta T}{e}^{r(T-t)}$. Setting it to zero gives the unique optimumLet $B={\int }_0^T{e}^{\beta s}ds$, equal to $(e^{\beta T}-1)/\beta$ when $\beta \ne 0$ and $T$ otherwise. Then $x^{\ast }(T;\mu )=W-B/\mu$ is strictly increasing from negative infinity to $W$ as $\mu$ increases from zero to infinity. Consequentlyis the unique positive multiplier with zero terminal balance.

For the constrained problem with no terminal reward, $x(T)\ge 0$ is exactly the weighted spending budget $\int e^{r(T-t)}u(t)dt\le W$. The candidate $u^{\ast }(\cdot ;{\mu }^{\ast })$ spends that budget exactly. Concavity gives $\log u-\log u^{\ast }\le (u-u^{\ast })/u^{\ast }$. Since $e^{-\beta t}/u^{\ast }={\mu }^{\ast }{e}^{-\beta T}{e}^{r(T-t)}$, integrating this inequality makes the utility function difference nonpositive for every feasible control. This proves the requested optimality of the logarithmic spending with a terminal multiplier solution directly, including uniqueness up to changes on measure-zero sets.