ThmDex – An index of mathematical definitions, results, and conjectures.
Tight upper bound to directional derivative
Formulation 0
Let $f : \mathbb{R}^{N \times 1} \to \mathbb{R}$ be a D5614: Differentiable real function at $x_0 \in \mathbb{R}^{N \times 1}$.
Let $u \in \mathbb{R}^{N \times 1}$ be a D5200: Real column matrix such that
(i) \begin{equation} \Vert u \Vert_2 = 1 \end{equation}
Then
(1) \begin{equation} D_u f(x_0) \leq \Vert \nabla f(x_0) \Vert_2 \end{equation}
(2) \begin{equation} D_u f(x_0) = \Vert \nabla f(x_0) \Vert_2 \quad \iff \quad \Vert \nabla f(x_0) \Vert_2 u = \nabla f(x_0) \end{equation}
Subresults
R5193: Maximal value for a directional derivative at a point
Proofs
Proof 1
Let $f : \mathbb{R}^{N \times 1} \to \mathbb{R}$ be a D5614: Differentiable real function at $x_0 \in \mathbb{R}^{N \times 1}$.
Let $u \in \mathbb{R}^{N \times 1}$ be a D5200: Real column matrix such that
(i) \begin{equation} \Vert u \Vert_2 = 1 \end{equation}
Using results
(i) R2863: Directional derivative is dot product of gradient and direction
(ii) R5192: Cauchy-Schwarz inequality for two real sequences

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 $\Vert \nabla f(x_0) \Vert_2 u = \nabla f(x_0)$. $\square$