Proposition

If $f: \mathbb{R}^n \rightarrow \mathbb{R}$, define a vector field $\grad f$ by

\[\begin{align*} (\grad f)(p) = D_1f(p) \cdot (e_1)_p + \cdots + D_nf(p) \cdot (e_n)_p. \end{align*}\]

For obvious reasons we also write $\grad f = \nabla f$. If $\nabla f(p) = w_p$, prove that $D_vf(p) = \ev{v, w}$ and conclude that $\nabla f(p)$ is the direction in which $f$ is changing fastest at $p$.

Solution

\[\begin{align*} D_vf(p) &= \ev{v, Df(p)} & \text{(Problem 2.29)} \\ &= \ev{v, D_1f(p)e_1 + \cdots + D_nf(p)e_n} & \text{(Theorem 2.7)}\\ &= \ev{v_p, (D_1f(p)e_1)_p + \cdots + (D_nf(p)e_n)_p} \\ &= \ev{v_p, \grad f(p)} \\ &= \ev{v_p, w_p} \\ &= \ev{v, w}. \end{align*}\]

The rate at which $f$ is changing in the direction $v$ is $\frac{\ev{v, w}}{\ev{v, v}}$, which is bounded by $\ev{w, w}$ by the Cauchy–Schwarz inequality. Therefore, $w$ is the direction in which $f$ is changing fastest at $p$.