Holder continuous of order 1/n: Let n be in N and define f(x

[dopey9]

Let n in N and define f : [−1, 1] -> R as f(x) = |x|^(1/n), for x in [−1, 1].

I want to show that f is Holder continuous of order 1/n

 

[marcmtlca]

you didnt comment as to my help on the other question. was it correct?

 

[dopey9]

you didnt comment as to my help on the other question. was it correct?

Sorry about that: Yes, it worked. Thank you!

 

[dopey9]

can anyone asnwer this?

is any one able to answer this question

 

[pka]

Re: can anyone asnwer this?

is any one able to answer this question

This may come as a great surprise to you but “Holder continuous” is not a widely known concept. In fact, I have spent a lifetime studying real analysis and have never seen it. I have probability seventy textbooks on analysis and it is in none of them I can find. Holder Inequality is widely used in integration theory.
That is why you have gotten no responses.

 

[dopey9]

Re: can anyone asnwer this?

is any one able to answer this question

This may come as a great surprise to you but “Holder continuous” is not a widely known concept. In fact, I have spent a lifetime studying real analysis and have never seen it. I have probability seventy textbooks on analysis and it is in none of them I can find. Holder Inequality is widely used in integration theory.
That why you have gotten no responses.

alright then no worries thankz