How Does Definition 8.9 Imply Differentiability Near Point p?

[Math Amateur]

I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ...

I need help in fully understanding Browder's comments on Definition 8.9 ... ...

Definition 8.9 including Browder's remarks reads as follows:View attachment 7469
In the above text from Browder, we read the following:

" ... ... We say that \(\displaystyle T \ : \ \mathbb{R}^n \mapsto \mathbb{R}^m\) is an affine map if it has the form \(\displaystyle x \mapsto c + Lx\) for some \(\displaystyle c \in \mathbb{R}^m\) and linear map \(\displaystyle L \ : \ \mathbb{R}^n \mapsto \mathbb{R}^m\). Thus Definition 8.9 says roughly that a function is differentiable at \(\displaystyle p\) if it can be approximated near \(\displaystyle p\) by an affine function. ... ..."I cannot see exactly how Definition 8.9 implies that a function is differentiable at \(\displaystyle p\) if it can be approximated near \(\displaystyle p\) by an affine function ... that is a function of the form \(\displaystyle x \mapsto c + Lx\) ...

Can someone please demonstrate rigorously that Definition 8.9 implies that a function is differentiable at \(\displaystyle p\) if it can be approximated near \(\displaystyle p\) by an affine function ... that is a function of the form \(\displaystyle x \mapsto c + Lx\) ... ?
Help will be much appreciated ...

Peter

 

[HallsofIvy]

It's a direct calculation. "If function F(x) can be approximated by an affine function near p" then there exist c and L such that $$F(x)= c+ L(x- p)+ \epsilon(x-p)$$ where $$\epsilon(x-p)$$ is the "error" in the approximation that goex to 0 as x goes to p ($$\epsilon(0)= 0$$).

Then $$F(p)= c+ \epsilon(0)= c$$ and $$F(p+ h)= c+ L(h)+ \epsilon(h)$$.
So $$F(p+ h)- F(p)= Lh+ \epsilon(h)$$

$$\frac{1}{|h|}\left(F(x+p)- F(p)- Lh\right)= \frac{1}{|h|}\left(Lh+ \epsilon(h)\right)= L\frac{h}{|h|}+ \frac{\epsilon}{|h|}$$.

We only need that we can choose $$\epsilon(x)$$ goes to 0 faster than x.

 

[math771]

Sure, I can help explain this concept to you. First, let's break down Definition 8.9 and understand its components.

The definition states that a function T: \mathbb{R}^n \mapsto \mathbb{R}^m is an affine map if it has the form x \mapsto c + Lx, where c \in \mathbb{R}^m and L: \mathbb{R}^n \mapsto \mathbb{R}^m is a linear map. This means that for any input x in \mathbb{R}^n, the output of T will be c + Lx, where c is a constant vector and L is a linear transformation.

Now, let's consider the concept of differentiability. A function is said to be differentiable at a point p if there exists a linear map L: \mathbb{R}^n \mapsto \mathbb{R}^m such that the following limit exists:

\lim_{h \to 0} \frac{f(p+h)-f(p)-L(h)}{\|h\|} = 0

This limit represents the rate of change of the function f at the point p, and the linear map L is the best linear approximation of f at that point.

So, how does this relate to Definition 8.9? Well, if a function T is an affine map of the form x \mapsto c + Lx, then it is automatically a linear map (since L is a linear transformation). This means that at any point p, the best linear approximation of T will simply be L itself. Therefore, T is differentiable at p and the linear map L is its best linear approximation.

In other words, if a function can be approximated near p by an affine function (which is a linear function), then it is differentiable at p and the best linear approximation is simply the linear map L that defines the affine function.

I hope this helps clarify the concept for you. Let me know if you have any further questions. Happy reading!