contestada

Suppose K⊆Rn is compact, f:K→R is continuous, and ϵ>0. Show that there is a number A>0 such that
|f(x)−f(y)|≤A∥x−y∥+ϵ,∀x,y∈K.

Respuesta :

By using the concept of compact set, it can be proved that

|f(x)−f(y)|≤A∥x−y∥+ϵ,∀x,y∈K.

What is compact set?

A set K is said to be compact if every open cover of K has a finite subcover.

Let K⊆Rn is compact  f:K→R is continuous, and ϵ>0

Let there exist [tex]x_n, y_n[/tex] ∈ K such that |f([tex]x_n[/tex])−f([tex]y_n[/tex])| > n∥[tex]x_n[/tex]−[tex]y_n[/tex]∥+ϵ,

Since K is compact there is a subsequence [tex]x_{nk}[/tex] and [tex]y_{nk}[/tex] of [tex]x_n, y_n[/tex] respectively such that [tex]x_{nk}[/tex] converges to x and [tex]y_{nk}[/tex] converges to y.

So,  |f([tex]x_{nk}[/tex])−f([tex]y_{nk}[/tex])| > [tex]n_k[/tex]∥[tex]x_{nk}[/tex]−[tex]y_{nk}[/tex]∥+ϵ,

Since f is continuous,

We can write

|f(x)−f(y)| > [tex]n_k[/tex]∥x - y∥+ϵ,

This is true for infinite many [tex]n_k[/tex]

So ||x - y|| = 0

|f(x) - f(y)| > ϵ, a contradiction since f is continuous

So,  there is a number A>0 such that

|f(x)−f(y)|≤A∥x−y∥+ϵ,∀x,y∈K.

To learn more about compact set, refer to the link-

https://brainly.com/question/17175854

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