(Ex. 7.10. from Chapter 3 of "Advanced Calculus" by Loomis, Sternberg)
If f: R^n -> R is differentiable at a, then its differential dfa, being a linear functional in R^n, is given by its skeleton n-tuple L according to the formula
dfa(x) = (L, x) = Σ(LiXi) ( Σ from 1 to n)
In this context, we call the n-tuple L the gradient of f at a. Show from the Schwarz inequality that if we use vectors v of Euclidian length 1, then the directional derivate Dvf(a) is maximum when v points in the direction of the gradient of f.
I vaguely understand what I need to do here... I think that I have to show is that when the direcion v in
has the same direction as the gradient L (dfa(v) = (L, v) ? So that L is its basis?) and when the Euclidian norm of v is 1, then the directional derivate with such v is a uniform norm, and I'll show that by using the Schwarz inequality. (for any other direction, the directional derivate is either smaller of equal.)
I'm not sure if this idea is correct. And if it is, I'm not sure how to proceed. I'll be very grateful for help!
The gradient $$ \nabla f= \left( {\partial f\over\partial x}, {\partial f\over\partial y}, {\partial f\over\partial z}, ... \right). $$ At a particular point $a=(x,y,z,...)$ the gradient is a vector taking the values of the derivatives.
The directional derivative $$ \nabla_v f=\lim_{h->0} {f(x+hv)-f(x)\over h} =v\cdot\nabla f $$ as you can see from using the chain rule.
Supposing $v$ is some kind of unit vector swinging around at $a$, then what it's asking you to prove is that, at the point $a$, $v\cdot\nabla f$ is greatest when the two vectors $v$ and $\nabla f$ are parallel.
(This problem seems to be a test of whether you understand the definitions. That is to say it contains a lot of complicated-looking terminology, but the only thing that you're actually required to do is to apply the complicated terminology to solve a relatively straightforward problem.)