 Solution

For $g=\nabla f(x)\ne 0$, the unit steepest descent direction is $-g/\Vert g{\Vert }_2$ by the Cauchy-Schwarz inequality. The usual unnormalized direction is $\boxed{p=-g}$; choosing its scale separately as a line-search step gives the same ray. It satisfies $\boxed{g^{T}p=-\Vert g{\Vert }_2^2<0}$, so it is a descent direction. The first-order Taylor expansion then guarantees decrease for sufficiently small positive steps. The subsequent bounds use this unnormalized negative-gradient convention.