I From a proof on directional derivatives

[Delta2]

TL;DR Summary

-

Given that the partial derivatives of a function $f(x,y)$ exist and are continuous, how can we prove that the following limit
$$\lim_{h\to 0}\frac{f(x+hv_x,y+hv_y)-f(x,y+hv_y)}{h}=v_x\frac{\partial f}{\partial x}(x,y)$$

I can understand why the factor $v_x$ (which is viewed as a constant ) appears in front of there, my difficulty in understanding is that inside the function the argument is $y+hv_y$ if it was just y, then everything would be fine.

 

[anuttarasammyak]

There seems all right iny+hv_y \rightarrow y

 

[Delta2]

There seems all right iny+hv_y \rightarrow y

Yes I can see that and it makes some sort of intuitive proof but I am looking for a rigorous proof.

 

[S.G. Janssens]

Summary:: -

Given that the partial derivatives of a function $f(x,y)$ exist and are continuous, how can we prove that the following limit
$$\lim_{h\to 0}\frac{f(x+hv_x,y+hv_y)-f(x,y+hv_y)}{h}=v_x\frac{\partial f}{\partial x}(x,y)$$

I can understand why the factor $v_x$ (which is viewed as a constant ) appears in front of there, my difficulty in understanding is that inside the function the argument is $y+hv_y$ if it was just y, then everything would be fine.

Are you allowed to use that it follows from your assumptions on existence of continuous partial derivatives that the total derivative $Df(x,y) = \Bigl(\frac{\partial f(x,y)}{\partial x} \, \frac{\partial f(x,y)}{\partial y}\Bigr)$?

If yes, can you see how that helps?

EDIT: To be more explicit, if yes, then
$$
\begin{aligned}
\frac{f(x+hv_x,y+hv_y)-f(x,y+hv_y)}{h} &=\frac{f(x+hv_x,y+hv_y)-f(x,y)}{h} - \frac{f(x,y+hv_y) - f(x,y) }{h}\\
&\to Df(x,y)(v_x,v_y) - \frac{\partial f(x,y)}{\partial y}v_y = \frac{\partial f(x,y)}{\partial x}v_x,
\end{aligned}
$$
as $h \to 0$.

 

[Delta2]

Are you allowed to use that it follows from your assumptions on existence of continuous partial derivatives that the total derivative $Df(x,y) = \Bigl(\frac{\partial f(x,y)}{\partial x} \, \frac{\partial f(x,y)}{\partial y}\Bigr)$?

If yes, can you see how that helps?

Sorry I can't understand how the total derivative helps here. What I am trying to prove is that the directional derivative (with respect to a vector $\vec{v}=(v_x,v_y)$) is equal to the dot product of gradient and the vector $\vec{v}$.

 

[S.G. Janssens]

Sorry I can't understand how the total derivative helps here. What I am trying to prove is that the directional derivative (with respect to a vector $\vec{v}=(v_x,v_y)$) is equal to the dot product of gradient and the vector $\vec{v}$.

The total derivative $Df(x,y)$ of $f : \mathbb{R}^2 \to \mathbb{R}$ at the point $(x,y)$ is a linear map from $\mathbb{R}^2$ to $\mathbb{R}$. I regard the gradient as the coordinate representation of $Df(x,y)$ with respect to the standard basis of $\mathbb{R}^2$.

(By tradition abuse of notation (of which I am also regularly guilty, for instance in post #4), the distinction between $Df(x,y)$ and its coordinate representation is ignored, but this is not always good practice.)

If $v$ is a direction vector in $\mathbb{R}^2$ with standard coordinate representation $(v_x,v_y)$, then application of the linear map $Df(x,y)$ to $v$ is identical to matrix-vector multiplication (in this case: the dot product) of the gradient with $(v_x,v_y)$. (That is a fact from linear algebra, more so than from calculus.)

 

[S.G. Janssens]

If, on the other hand, you are asking why application of the coordinate-free total derivative $Df(x,y)$ to the coordinate-free vector $v$ gives you the directional derivative of $f$ at $(x,y)$ in the direction of $v$, then note that
$$
\|f((x,y) + hv) - f(x,y) - Df(x,y)hv\| = o(\|hv\|)
$$
by total differentiability of $f$ at $(x,y)$. So,
$$
\lim_{h \to 0}\frac{\|f((x,y) + hv) - f(x,y) - hDf(x,y)v\|}{h} = 0
$$
as well. (This is just "Fréchet differentiability implies Gateaux differentiability".)

 

[Delta2]

@S.G. Janssens I admit I am a bit lost. I haven't heard before about Frechet and Gateaux derivatives, but anyway let me ask this
What is the primary definition of the total derivative and when you say:

If v is a direction vector in R2 with standard coordinate representation (vx,vy), then application of the linear map Df(x,y) to v is identical to matrix-vector multiplication (in this case: the dot product) of the gradient with (vx,vy). (That is a fact from linear algebra, more so than from calculus.)

How do we prove the above

 

[S.G. Janssens]

I haven't heard before about Frechet and Gateaux derivatives

Fréchet = total, Gateaux = directional.

What is the primary definition of the total derivative

A function $f : \mathbb{R}^n \to \mathbb{R}^m$ is differentiable at a point $\mathbf{x} \in \mathbb{R}^n$ if there exists a linear map $Df(\mathbf{x}) : \mathbb{R}^n\to \mathbb{R}^m$ such that
$$
\|f(\mathbf{x} + \mathbf{z}) - f(\mathbf{x}) - Df(\mathbf{x})\mathbf{z}\| = o(\|\mathbf{z}\|)
$$
for $\|\mathbf{z}\| \to 0$.

How do we prove the above

This is linear algebra: Applying a linear map to a vector is equivalent to applying the representation of that map (in some chosen basis) to the representation of that vector (in the same basis).

 

[S.G. Janssens]

Also, it was not my intention to make this more difficult than necessary, and I am sorry if that happened anyway, but I cannot resist the coordinate-free definition, for a variety of reasons, such as: It keeps a clean separation between linear maps and their representations, and it generalizes directly from operators on $\mathbb{R}^n$ to operators on infinite-dimensional normed linear spaces. Here "directly" means that the proofs of the standard theorems carry over almost verbatim. (As long as these proofs do not rely on the local compactness of $\mathbb{R}^n$.)