I'm trying to prove the following:

Show that $f$ grows fastest along path for which $\gamma'(t)=\nabla f(\gamma(t))$ than along any other path.

My reasoning is that of course $f$ along $\gamma$ must be growing the fastest since by definition the gradient at a point $x_0$ is the direction to which the function grows fastest and its magnitude gives the steepness of the growth.

But what about "any other path"?


Let $\eta$ be any other path. Then the growth of $f$ along $\eta$ is measured by the the derivative: $(f\circ \eta)'(t)=\nabla f(\eta(t))\cdot\eta'(t)$, which by the Cauchy-Schwarz inequality is $\leq || \nabla f(\eta(t)||\ || \eta'(t)||$. Therefore, the maximum value of the derivative will be attained when this is an equality, i.e. when the vectors are parallel. In other words, when $\eta'(t)=\lambda \nabla f(\eta(t))$, for some $\lambda$.

However, note that you can always reparametrize $\eta$ to get $\lambda=1$.

  • $\begingroup$ So how does this relate to $f(\gamma (t))$? $\endgroup$ – mavavilj Nov 24 '16 at 10:35
  • $\begingroup$ $f(\gamma(t))$ precisely satisfies that $(f\circ\gamma)'(t) = || \nabla f(\gamma(t)) ||^2$, or in other words, it makes the Cauchy-Schwarz inequality, an equality, therefore attaining the maximum value. $\endgroup$ – Sak Nov 24 '16 at 10:39
  • $\begingroup$ Why does it satisfy that? $\endgroup$ – mavavilj Nov 24 '16 at 10:40
  • $\begingroup$ Because of your hypothesis that $\gamma'(t)=\nabla f(\gamma(t))$. Just substitute in the inequality in my answer. Note that $\eta$ is any path, in particular you can choose $\gamma$ $\endgroup$ – Sak Nov 24 '16 at 10:41
  • $\begingroup$ Do you know if using C-S is the only way to solve this? $\endgroup$ – mavavilj Nov 24 '16 at 13:16

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