# Characterization of Lipschitz Continuity on Shift of Finite Type by Variations

Jump to navigation
Jump to search

This article, or a section of it, needs explaining.In particular: If you're going to reference a concept in the title, you need to reference it also in the body of the proof. There is no indication anywhere here what "lipgloss continence" is.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

## Theorem

Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.

Let $f : X_\mathbf A \to \C$ be a continuous mapping.

Let $\theta \in \openint 0 1$.

Let $C > 0$.

Then:

- $\forall x, y \in X_\mathbf A : \size {\map f x - \map f y} \le C \map {d_\theta} {x, y}$

- $\forall n \in \N : \map {\mathrm {var}_n} f \le C \theta ^n$

where:

That is:

- $\ds \size f_\theta = \sup \set { \dfrac {\cmod {\map f x - \map f y} }{\map {d_\theta} {x,y} } : x,y \in X_\mathbf A, x \ne y}$

where $\size \cdot_\theta $ is the Lipschitz seminorm.

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: Maybe a separate page for this "That is"You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: ... and we still need to reference "Lipschitz continuity" somehowYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Proof

### Sufficient Condition

Suppose:

- $\forall x, y \in X_\mathbf A : \size {\map f x - \map f y} \le C \map {d_\theta} {x, y}$

Let $n\in\N$.

By definition of the metric:

- $\forall x, y \in X_\mathbf A:$
- $\forall i \in \openint {-n} n : x_i = y_i \implies \map {d _\theta} {x, y} \le \theta ^n$

Thus, by the assumption:

- $\forall x, y \in X_\mathbf A:$
- $\forall i \in \openint {-n} n :x_i = y_i \implies \size {\map f x - \map f y} \le C \map {d_\theta} {x, y} \le C \theta ^n$

Therefore:

- $\map {\mathrm {var}_n} f \le C \theta ^n$

$\Box$

### Necessary Condition

Suppose:

- $\forall n \in \N : \map {\mathrm {var}_n} f \le C \theta ^n$

Let $x, y \in X_\mathbf A$.

If $x = y$, then obviously:

- $\size {\map f x - \map f y} = 0 \le C \map {d_\theta} {x, y}$

So, let $x \ne y$.

Then:

- $\exists N \in \N : d_\theta (x,y) = \theta ^N$

In particular:

- $\forall i \in \openint {-N} N : x_i = y_i$

Therefore:

- $\size {\map f x - \map f y}\le \map {\mathrm {var}_N} f\le C \theta ^N = C \map {d_\theta} {x, y}$

$\blacksquare$