Let fk01 r be a sequence of functions such that 1 fkxm1 for...

Let f_(k):[0,1]->R be a sequence of functions such that (1) |f_(k)(x)|<=M_(1) for all kinN and xin[0,1], (2) |f_(k)^(')(x)|<=M_(2) for all kinN and xin[0,1]. for some positive M_(1),M_(2). (a) Prove that there exists a subsequence of {f_(k)} which converges uniformly on [0,1.] (b) If the assumption (1) is omitted, can {f_(k)} still have a convergent subsequence? If yes, prove it; If not, give an counterexample. (c) Show that the assumption (1) can be replaced by f_(k)(0)=0 for all kinN. Let f : [0,1] -> R be a sequence of functions such that (1)|fz(x)|< Mi for all k e N and x E[0,1] (2)f{(c)|< M2 for all k e N and x E[0,1] for some positive Mi, M2 (a) Prove that there exists a subsequence of{fk}-i which converges uniformly on [0,1]. (b) If the assumption (1) is omitted, can {fi}-, still have a convergent subsequence? If yes, prove it; If not, give an counterexample. (c) Show that the assumption (1) can be replaced by fk(0) = 0 for all k e N.

Transcript

00:01 Okay, in this question, we are given a sequence of functions.

00:04 Let's see fk is a function from the closed interval 0, 1 to r.

00:12 We assume they are continuous because we have a prime of f is less or equal to m2 and fk itself is less or equal to m1.

00:31 Those are for any x in this closed interval.

00:37 Now we want to show, or go to show, there exists a subsequence, let's call it kn instead of ak.

00:57 Okay, there exists subsequence such that f and k converges to some f uniformly.

01:12 Okay, so for sequence of continuous function on the closed interval on the compact set with some boundless assumption to prove some uniform convergence we want to use the so -called arzano -oscaldi theory.

01:40 Okay, from this a -a theory we know if we can prove fk is is uniformly bounded and equally continuous, then this statement is true.

02:07 The uniform boundness has been given here.

02:12 I mean for any k we have a uniform bound, we have a uniform bound m1 for any k.

02:20 Now we want to show its equal continuity we know f minus f as the function is differentiable.

02:42 We know by the mean value theory, which is less than or equal to theta times, oh this is actually equal to sorry this has some number between x and y but by our first assumption for any x we have a uniform bound m2 that means this is less or equal to m2 times x minus 1 so you can see the difference of f k is bounded by a constant times the difference of f x minus 1 this is just a definition of the equilibrium.

03:39 So this is right, that means our aa theory has been checked.

03:45 That means there must be some subsequence of fk such that fkn, i mean let's call this subsequence as fkn, such that fkn converges to some f uniformly.

04:00 The second part, i mean, if this is 1, this is 2.

04:13 If the assumption 1 has been omitted, we want to show this statement is not true.

04:24 It's very simple.

04:26 The assumption 1 is the uniform bounded condition.

04:32 So to break this assumption, we only need to find some unbounded sequence.

04:38 For example, let's define fk in x to be equal to the constant k times the indicated function on 0, 1.

04:48 That means for f1, it is a function which is equal to 1 from 0 to 1.

04:56 This is our f1, and 0 otherwise.

05:01 And this is our f2...