Using results
we have the inequality
\begin{equation}
\begin{split}
D_u f(x_0)
= \nabla f(x_0)^T u
\leq \Vert \nabla f(x_0) \Vert_2 \Vert u \Vert_2
= \Vert \nabla f(x_0) \Vert_2
\end{split}
\end{equation}
with equality if and only if there exists a constant $c \in [0, \infty)$ such that $\nabla f(x_0) = c u$ or $u = c \nabla f(x_0)$. If $\nabla f(x_0) = \boldsymbol{0}$, then $\nabla f(x_0) = c u$ is true with the choice of $c = 0$. If $\nabla f(x_0) \neq \boldsymbol{0}$, then the only proportionaly constant which satisfies the above condition and $\Vert u \Vert_2 = 1$ is given by $\Vert \nabla f(x_0) \Vert_2 u = \nabla f(x_0)$. $\square$