 Solution

A descent direction $s$ satisfies $g^{T}s<0$, where $g=\nabla f(x)$; the first-order Taylor expansion then gives $f(x+\alpha s)<f(x)$ for sufficiently small $\alpha >0$. To compare directions one must fix their scale. Among Euclidean unit vectors, the Cauchy-Schwarz inequality givesThis is the unit steepest descent direction. The conventional unnormalized direction used in the algorithm is the same ray, $\boxed{s=-g}$. Equivalently it is the unique minimizer of $g^{T}s+\frac{1}{2}\Vert s{\Vert }_2^2=\frac{1}{2}\Vert s+g{\Vert }_2^2-\frac{1}{2}\Vert g{\Vert }_2^2$. If $g=0$ there is no strict first-order descent direction, and the conventional update is zero.