continious functions: Let f:[a,b] -> R be a real function

[dopey9]

Let f: [a,b] -> R be a real function

It is said that f satisfies the Holder condtion of order alpha >0 if there exists an M>0 and some tilda>0 such that for all x[0] in [a,b] and all x in [a,b] with 0<| x - x[0] | < tilda it holds |f(x) - f(x[0]\ < M |x-x[0]|^alpha


does any one know how to show that if f:[a,b] -> R satisfies the Holder condition of order aplha >0 at any x[0] in [a,b], then f is continious?

 

[marcmtlca]

Choose \(\displaystyle x_0\) some arbitrary point in [a,b]

Let \(\displaystyle \Large\Large \delta = \min((\frac{\epsilon}{M})^{\frac{1}{\alpha}},tilda)\)

Then for \(\displaystyle |x-x_0|<\delta\)

We use Holder's condition to conclude that:
\(\displaystyle \Large\Large |f(x)-f(x_0)|<M|x-x_0|^{\alpha}<M|(\frac{\epsilon}{M})^{\frac{1}{\alpha}}|^\alpha=\epsilon\)

This might work.. im sort of tired.

 

[dopey9]

correct

This might work.. im sort of tired.

Yes, it worked. Thank you!

 

[dopey9]

part 2 of question

Choose \(\displaystyle x_0\) some arbitrary point in [a,b]

Let \(\displaystyle \Large\Large \delta = \min((\frac{\epsilon}{M})^{\frac{1}{\alpha}},tilda)\)

Then for \(\displaystyle |x-x_0|<\delta\)

We use Holder's condition to conclude that:
\(\displaystyle \Large\Large |f(x)-f(x_0)|<M|x-x_0|^{\alpha}<M|(\frac{\epsilon}{M})^{\frac{1}{\alpha}}|^\alpha=\epsilon\)

This might work.. im sort of tired.

It is said that f satisfies the Holder condtion of order alpha >0 if there exists an M>0 and some tilda>0 such that for all x[0] in [a,b] and all x in [a,b] with 0<| x - x[0] | < tilda it holds |f(x) - f(x[0]\ < M |x-x[0]|^alpha

youve jus shown that f:[a,b] -> R satisfies the Holder condition of order aplha >0 at any x[0] in [a,b] whcih makes f continious

but now im stuck on the part where i want to show that f:[a,b] -> R satisfies the Holder condition of ORDER aplha >1 at any x[0] in [a,b]. then f is differentiable at any x[0] in [a,b]. and from this i need to find its derivative

im confused... do i cahnge apha to 1