 Solution

The Lagrangian sufficiency theorem for an equality constraint states that if a feasible $x_{\ast }$ and a multiplier ${\lambda }_{\ast }$ exist such that $x_{\ast }$ globally maximizes $L(x,{\lambda }_{\ast })=f(x)-{\lambda }_{\ast }\cdot (g(x)-b)$ over the underlying domain, then $x_{\ast }$ globally maximizes $f$ over $g(x)=b$. Indeed $L=f$ at every feasible point, so global maximality of $L$ directly compares the feasible objective values. The underlying domain can already impose $x_i\ge 0$.

For $0<p<1$, choose $\lambda =p{d}^{1-p}$. Each function $t^p-\lambda t$ on $t\ge 0$ is a strictly concave function and attains its unique maximum at $t=1/d$, since $p{t}^{p-1}=\lambda$ there. The sum Lagrangian is therefore globally maximized when every $x_i=1/d$, and this point is feasible. For $p=1$ the objective is identically one on the feasible set. For $p>1$, each feasible $x_i$ lies in $[0,1]$, so $x_i^p\le x_i$, with equality only at $0$ or $1$; equality in the total is attained exactly at a simplex vertex. ConsequentlyFor $0<p<1$ the unique maximizer is $(1/d,\dots ,1/d)$; for $p=1$ every feasible point is a maximizer; for $p>1$ the maximizers are exactly the $d$ vertices with one coordinate one and all others zero. When $d=1$ these descriptions coincide.