Conside Hölder continuous functions f 1 </msub> , &#x2026; , f m </

Jaime Coleman

Jaime Coleman

Answered question

2022-05-15

Conside Hölder continuous functions f 1 , , f m : R n R (with Hölder coefficient α). The claum is now that any rational function of f 1 , , f m with non vanishing denominator is also Hölder continuous (with coefficient α). The author says this is trivial, but I do not see why this should be trivial!? (I could not even prove the easiest case f(x)/g(x))
Do you have an idea by which it is easily proven?
Best regards.

Answer & Explanation

Jaylon Richmond

Jaylon Richmond

Beginner2022-05-16Added 10 answers

Such as stated, the result is not true. Take for instance f 1 ( x ) = e x 2 ,which are Lipschitz. Then f 1 ( x ) / f 2 ( x ) = e x 2 , which is not (globally) Lipschitz. Two conditions that will make the result true:
1. The f i are bounded
2. The denominator is bounded away from 0

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