Does Lipschitz continuity imply α-Hölder continuity for all α∈(0,1)?

[bars]

Homework Statement

I only need help interpreting the following:
Show that every Lipschitz continuous function is α-Hölder continuous for
every α ∈ (0, 1
The definition of both is given in the homework so this seems trivial but it's a graduate level class. Am I mising something? Thanks for any help!

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

Well, what are those definitions? Why do you say this is "trivial"?

 

[bars]

Well by the definitions, Lipschitz is a special case of α-Hölder when α=1. Since α is contained in the interval (0,1] (which is the interval given for α-Hölder) then by def. every Lipschitz continuous function is α-Hölder continuous.

 

[christoff]

Well by the definitions, Lipschitz is a special case of α-Hölder when α=1. Since α is contained in the interval (0,1] (which is the interval given for α-Hölder) then by def. every Lipschitz continuous function is α-Hölder continuous.

Your question asked to show that it is α-Hölder continuous for every α ∈ (0, 1], not just for α=1. Unless this was a typo? Yes, Lipschitz implies Hölder of order 1. But does it imply this for all orders less than 1?

 

[bars]

Ahhh, great! yes your right I see it now. Funny how sometimes one can not see what is right in front of them. Thanks for the help, that's exactly what I needed.