23 июл. 2015 г. · Result 2.12: If for every h>0 we have that |f(x+h)−f(x)| is unbounded on I, then f is not uniformly continuous on I.

8 мая 2016 г. · By definition f is uniformly continuous if for all ε>0, there exists a δ>0 such that when |x−y|<δ we get |f(x)−f(y)|<ε. I am not sure where to ...

1 июл. 2020 г. · You should recognize that the derivative of 1x, which is −1x2, can get as large as you want, so the function is not uniformly continuous. The ...

5 февр. 2015 г. · A function is not uniformly continuous if there exist two sequences xk,yk such that xk−yk→0but notf(xk)−f(yk)→0. This can be obtained directly ...

7 июн. 2015 г. · Take f(x)=sin1/x. It is not uniformly continuous as such functions admit continuous extension to the closure of the domain. Here we don't have ...

24 июн. 2019 г. · If f is continuous on (0,1) and limx→0+f(x)=+∞, show that f is not uniformly continuous [duplicate] · Use sequential criterion for uniform ...

13 янв. 2020 г. · Definition : f(x)=1/x2 is uniformly continuous on (0,1) if, given ϵ>0, there exists a δ>0 such that for all x,y∈(0,1) with |x−y|<δ, we have |1x2 ...

25 июл. 2015 г. · In your proof, you need |x+x0|<2 to prove uniform continuity, which is true on [0,1] but not on R. So the proof cannot be extended to R.

23 апр. 2018 г. · The key lemma you want to use here is that if f is uniformly continuous on a bounded open interval (a,b) then it can be extended by continuity ...

26 февр. 2018 г. · Therefore f is a Lipschitz function on [1,∞) which implies uniform continuity. therefore f is not uniformly continuous on (0,1).