# Pointwise Limits and Uniform Convergence Help

1. Apr 23, 2010

### BlackTiger

Hi all
Came across this question which i have attempted to answer but am sure i have gone wrong somewhere, any help would be appreciated...

Suppose fn(x):R-->R, where fn(x)=x+(1/n) (n belongs to N). Find the pointwise limits of the sequences (fn), (1/n fn) and (fn2). In each case determine whether the convergence is uniform.

Now i managed to get all three as being uniformly convergent and am sure i went wrong somewhere. I am still struggling to understand uniform convergence completely. The pointwise limits i computed were x, 0, x2 respectively...sorry i havnt put my working up struggling to use latex.

2. Apr 23, 2010

### mrbohn1

Your limits are right. (Let $$f$$ be the limit of $$f_n$$). The first sequence is uniformly convergent: given $$\epsilon >0$$, choose N to be the lowest integer that is greater than:$$\frac{1}{\epsilon}$$.

Then, for all $$n>N$$:
$$|f-f_n|=|x-(x+\frac{1}{n})|=\frac{1}{n}<\frac{1}{N}<\epsilon$$.

For the second one, note that $$sup_x|f_n|$$ does not go to 0. This tells you it is not uniformly convergent. To prove this, suppose we are given $$\epsilon >0$$. For any choice of N, let $$x>\epsilon .N-\frac{1}{N}$$.
Then:

$$|f-f_N|=|0-\frac{x}{N}-\frac{1}{N^2}|=|\frac{x}{N}+\frac{1}{N^2}|>\frac{\epsilon .N-\frac{1}{N}}{N}+\frac{1}{N^2}=\epsilon$$.

The third one is very similar. I'm afraid all this might be a bit bewildering if you don't yet understand uniform convergence though! Keep thinking about it and eventually it will click.

PS - I may have made some mistakes here as I did it very quickly, so be sure to check everything!

Last edited: Apr 23, 2010