 Solution

Put $c=e^{-\gamma }$, where $\gamma$ is the Euler--Mascheroni constant, and let $f(u)=w(u)-c$. The Buchstab function equation becomesThe given Gamma function bound implies $uf(u)\to 0$. Indeed,so $u/\Gamma (u+1)\to 0$.

Suppose first that $f\ge 0$ throughout an interval $[u,u+1]$, with $u\ge 1$. The delay differential equation and integration successively on $[u+1,u+2]$, $[u+2,u+3]$, and so on propagate nonnegativity to the entire tail $[u,\infty )$. In particular, $vf(v)$ is nonnegative and nondecreasing for $v\ge u+1$. Its limit is zero, so it is identically zero there. Thus $f=0$ on that tail.

The delay differential equation then propagates this equality backwards: if $f=0$ for $v\ge a$, then $f(v-1)=0$ for $v>\max (a,2)$. Repeating finitely many times would give $f=0$ on $(1,2)$, contradicting $f(v)=1/v-c$ there. Hence every interval $[u,u+1]$ contains a point where $f<0$.

If instead $f\le 0$ on $[u,u+1]$, the same propagation makes $vf(v)$ nonpositive and nonincreasing on the tail. Its limit zero again forces it to vanish, giving the same contradiction. Thus every such interval also contains a point where $f>0$. The oscillation of the Buchstab function is therefore strict on every unit interval: