# Show that f'(x) grad f(x) is positive

1. Aug 16, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I'm preparing a maths exam so I run through a lot of problems, and I am unsure whether I tackled this one correctly or not:

Prove that for every differentiable function $f: \mathbb{R}^3 \to \mathbb{R}$ and for every point $x \in \mathbb{R}^3$ the following statement is true:

$\lim\limits_{t \to 0} \frac{f(x+t \nabla f(x)) - f(x)}{t} \geq 0$

2. Relevant equations

I have the following definitions:

$f'(x) = \frac{\partial (f_1, ..., f_m)}{\partial (x_1, ..., x_n)} (x)$

Directional derivative: $f'(x)v = \lim\limits_{t \to 0} \frac{1}{t} (f(x + tv) - f(x))$

3. The attempt at a solution

So what I did was pretty straightforward:

$\lim\limits_{t \to 0} \frac{f(x + t \nabla f(x)) - f(x)}{t} = f'(x) \nabla f(x)$
$= [\partial_1 f(x) \partial_2 f(x) \partial_3 f(x)] \cdot [\partial_1 f(x), \partial_2 f(x), \partial_3 f(x)]$
$= (\partial_1 f(x))^2 + (\partial_2 f(x))^2 + (\partial_3 f(x))^2 \geq 0$

Is that legit? Not sure about my handling from $f'(x)$. Should I have used the total differential definition? Would I obtain the same result? I'm still unclear about total derivatives :)

Julien.

2. Aug 16, 2016

### Ray Vickson

For a vector $\vec{u} \in R^3$ we have
$$\lim_{t \to 0} \frac{f(\vec{x} + t \vec{u}) - f(\vec{x})}{t} = \langle \vec{u}, \vec{\nabla} f(\vec{x}) \rangle,$$
where $\langle \vec{p}, \vec{q} \rangle$ denotes the inner product of the vectors $\vec{p}, \vec{q}$. You are taking $\vec{u} = \vec{\nabla} f(\vec{x})$.

Last edited: Aug 16, 2016
3. Aug 16, 2016

### Staff: Mentor

It looks right to me. I would have written the limit as $(\nabla_{(\nabla f)(x)}f)(x) \overset{(*)}= (\nabla f (x))\cdot (\nabla f (x)) = |\nabla f (x)|^2$. The total differentiability of $f$ is hidden in $(*)$ which allows you to write the directional derivative in direction $v$ ($=(\nabla f)(x)$ in this case) as $D_v f(x) = \nabla f(x) \cdot v$. Other than directional differentiability and partial differentiability (which are simply special directional derivatives), the total differentiability allows us to choose any direction. In this case $v=(\nabla f)(x)$ is a linear combination of partial derivatives.

4. Aug 17, 2016

### JulienB

@Ray Vickson Hi and thanks for your answer! I like the way you write the definition with the inner product. At the end it gives the same result that I wrote, right?

@fresh_42 Hello and thank you for your answer as well. :) Of course I realised the equality holds only because the problem states that the function is differentiable. I am still unsure about the meaning of the total derivative. So far I understand it like that: the partial differentials hold for analytical use of a function, while the total differential becomes useful when wanting to approximate the slope at a given point. Would you support that statement? If not, could you maybe give me an example of partial and total differentiation so that I can visualise it? My interpretation is based on the following observation (for the case $f: \mathbb{R}^n \to \mathbb{R}$):

$f' (\vec{x}) = \frac{\partial (f_1, ... , f_m)}{\partial (x_1, ... , x_n)} (x) = \mbox{vector}$
$f' (\vec{x}) = \sum \frac{\partial f}{\partial x_i} dx_i = \mbox{scalar}$

First of all, is that correct? And if so, would you say that "the total differentiability allows us to choose any direction" when we plug in arbitrary $dx_i$ for a given point $\vec{x}$?

Julien.

5. Aug 17, 2016

### Staff: Mentor

Not really. Partial derivatives also approximate the slope at a given point. Just in the direction of the coordinates. A total differential approximates a function $f$ as a whole by a linear function $L$ (the total differential), i.e. you may write $f(\vec{x}+\vec{v}) = f(\vec{x}) + L(\vec{v}) +r(\vec{v})$ such that $\lim_{\vec{v} \rightarrow 0} \frac{r(\vec{v}}{||\vec{v}||} = 0$. This means $f$ can be approximated by $L$ such that the error terms in $r$ run faster towards zero than $L$ does, or as it often occurs (e.g. in Taylor expansions), $r$ has terms in $\vec{v}$ of higher orders than $1$.

I have difficulties with naming everything $x$. You should try to get used to distinguish when you are speaking of a function in the variable $x$, coordinates $x_i$ or $x^i$ and eventual evaluation points $x_0=p,$ e.g. where the derivative is evaluated at. I'm not sure if I got you right. E.g. you may denote a gradient better by $$grad(f) = \nabla f = \sum_{i=1}^{n}\frac{\partial f}{\partial x_i} \vec{e}_i$$ or more physical $$\nabla f = \sum_{i=1}^{n}\frac{\partial f}{\partial x_i} dx_i$$ and $$\nabla_p f = \sum_{i=1}^{n}\left(\frac{\partial f}{\partial x_i}\right)_{x=p} dx_i$$

In addition, how can $f'(\vec{x})$ simultaneously be a vector and a scalar? Only by evaluation of $\nabla_p f$ at a certain point, which should not be denoted by the variable name.

An example for the difference between partial and total differentiation is the following:
$$f(\vec{x}) = \begin{cases} \frac{2x_1x_2}{x_1^2+x_2^2} & \text{if } (x_1,x_2) \neq (0,0) \\ \\ 0 & \text{if } (x_1,x_2) = (0,0) \end{cases}$$
$f$ is partially differentiable at $(0,0)$ but not continuous at $(0,0)$. And you might want to calculate what happens along the line $x_1 = x_2$. The latter is an example of a direction that is not along a coordinate.

6. Aug 18, 2016

### JulienB

@fresh_42 Thank you again for your answer, it is very complete and I am slowly starting to understand. I will study the matter for some more time on my own and probably come back here for more questions later on :)

Julien.