Question

Sequence of functions and uniform convergence.. I am having some troubles with this topic, so here it is...

\[f_{n}(x)\] = \[\large \log(1+\frac{1}{n(x-1)})\]

Now..

Limit for n->infinity is log1 so 0.

So it converges pointwise to 0. Now I have to find out if it uniformly converges, so here it is:

\[\large \sup|f_{n}(x) - f(x)| = \sup|f_{n}(x)|\]

Should go to 0.

Now I start to get confused. What I have to do exactly, take the first derivative respect to x and = 0 ?
Help!

Answer 1

x in R?

Answer 2

sorry, x belonging to ]1,+infinity[

Answer 3

Ok, you must be prove the following inequality :
\[\ln(1+u) \le u, \forall u \in (0,1)\]

Answer 4

I am not sure my calculus teacher actually used this method, you got any reference?

Answer 5

because we want to have a result : \[\sup_{x>1} |f_n (x)| \to 0\], and we see that \[\frac{1}{n} \to 0\] so we need take \[\frac{1}{n(x-1)}\] out of the log.
By using this inequality, sorry, we can got it, but may be we are entangled because of (x-1)

Answer 6

if x>2, then got it :D

Answer 7

what interval are you on?

Answer 8

oh nvm i see it, then you have a problem. it is clear what you have to do to prove uninform convergence or no? you have to show that this limit is independent of x, i.e. it depends only on n
it would be good to negate the definition of uniform convergence and see exactly what it says. then i believe you should be able to prove that this does not converge uniformly

Answer 9

do you know how to negate the definition?

Answer 10

in fact the proof that it is not uniform is pretty straightforward once you write what it is you have to show

Answer 11

@satellite73 yes it is. This sequence does not converge uniformly. Concemtrate your
thinking close to 1.

Answer 12

a hint would be to find \(x\) such that given any \(n\) you can make it larger than \(\log(2)\)

Answer 13

think @alfie went away though
the important part of the problem is not the proof, but rather understanding how to negate a definition.

Answer 14

If it is uniformly convergent, then there exits N, such that n>N one has
\[
\ln \left( 1 + \frac 1 {n(x-1)} \right) \le \frac 1 2
\]
for every x
Take
\[
x_n = 1 +\frac 1{n^2}
\]
in the above inequality and find a contradiction.

Answer 15

If we do that we get for every n >N
\[
\ln ( 1 + n) < 1/2
\]
Which cannot be.

Answer 16

okay our calculus professor didn't really explain it this way but sounds good to me. Thank you guys, if I could have given more best answers I'd have done that. Thanks ;)

Answer 17

I'm failing to understand why it is pointwise convergent; I've been struggling with this for a little in general, but I don't see explicitly how M is dependent on Epsilon, could somebody write it out or something?