# Linear Mappings are Lipschitz Continuous .... D&K Example 1.8.14 .... ....

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In summary, in the conversation about Duistermaat and Kolk's Example 1.8.14 on continuity in their book "Multidimensional Real Analysis I: Differentiation," the participants discuss proving the statement that any linear mapping A: R^n to R^p is Lipschitz continuous. They explain that this requires the inequality shown in (1), which can be rewritten as (2). The conversation concludes with a remark on the independence of the constant k in (2) from the variables x, x', and x''.
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I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of Example 1.8.14 ... ...

The start of Duistermaat and Kolk's Example 1.8.14 reads as follows:https://www.physicsforums.com/attachments/7753In the above example we read the following:

"... ... any linear mapping $$\displaystyle A \ : \ \mathbb{R}^n \rightarrow \mathbb{R}^p$$, whether bijective or not, is Lipschitz continuous and therefore uniformly continuous ... ... "I am somewhat unsure regarding proving this statement ... but I think the proof goes like the following:

For $$\displaystyle A \ : \ \mathbb{R}^n \rightarrow \mathbb{R}^p$$ to be Lipschitz continuous we require

$$\displaystyle \mid \mid Ax - Ax' \mid \mid \le k \mid \mid x - x' \mid \mid \text{ where } x, x' \in \text{dom} (A)$$ ... ... ... (1)now we have:

$$\displaystyle \mid \mid Ax \mid \mid \le k \mid \mid x \mid \mid$$ ... ... ... (2)Now ... maybe put $$\displaystyle x = x' - x''$$

... then (2) becomes $$\displaystyle \mid \mid A(x' - x'') \mid \mid = \mid \mid Ax' - Ax'' \mid \mid \le k \mid \mid x' - x'' \mid \mid$$

which is the required result ...Is that correct?

Peter

Yes, and you can be a bit more self-confident (Happy)
(Remark which may be superfluous: Your $k$ is independent of $x$, $x'$, $x''$.)

Krylov said:
Yes, and you can be a bit more self-confident (Happy)
(Remark which may be superfluous: Your $k$ is independent of $x$, $x'$, $x''$.)
Indeed, Krylov... you are probably right :)

Peter

## 1. What are linear mappings?

Linear mappings are mathematical functions that preserve linear combinations of vectors. In other words, they map one vector space to another in a linear fashion.

## 2. What does it mean for a linear mapping to be Lipschitz continuous?

A linear mapping is Lipschitz continuous if there exists a positive real number, called the Lipschitz constant, such that the distance between the images of any two vectors is less than or equal to the Lipschitz constant times the distance between the two original vectors.

## 3. How is Lipschitz continuity related to the D&K Example 1.8.14?

D&K Example 1.8.14 is a specific example that illustrates the concept of Lipschitz continuity. It shows that a linear mapping from a finite-dimensional vector space to itself is always Lipschitz continuous.

## 4. Why is Lipschitz continuity important?

Lipschitz continuity is important because it guarantees the stability of a function. In other words, small changes in the input will result in small changes in the output. This property is particularly useful in many mathematical and scientific applications.

## 5. Is every linear mapping Lipschitz continuous?

No, not every linear mapping is Lipschitz continuous. For example, the identity mapping on an infinite-dimensional vector space is not Lipschitz continuous. However, in finite-dimensional vector spaces, all linear mappings are Lipschitz continuous.

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