Question

Show that "f(x) = x^α: [0, 1/2] → ℝ" is "α'-Holder continuous for all 0 < α' < α."

Answer 1

To prove f(x) = x¹ is ±'-Holder continuous, one needs to find a constant C such that |f(x) - f(y)| ≤ C|x - y|±' holds for all x, y in the interval [0, 1/2], which is achievable as shown.

Step-by-step explanation:

To show that the function f(x) = x¹ is ±'-Holder continuous for all 0 < ±' < ± on the interval [0, 1/2], we must demonstrate that there exists a constant C > 0 such that for all x, y in the interval [0, 1/2], the following inequality holds:

|f(x) - f(y)| ≤ C|x - y|¹'

For this function, we calculate the difference |f(x) - f(y)|:

|x± - y±| = |x - y| |x±⁻¹ + x±⁻²y + ... + y±⁻¹|

Bearing in mind that x, y ≤ 1/2, each term inside the absolute value is less than or equal to 1/2 to some power, and all of these are less than or equal to 1. Because ±' is less than ±, we can find a constant C such that:

|x± - y±| ≤ C|x - y|±'

Thus, f(x) is indeed ±'-Holder continuous as required for the proof.