# Uniform Convergence

• Nov 15th 2009, 01:38 PM
nmatthies1
Uniform Convergence
Consider $f(x)=\sum_{n=1}^{\infty} \frac{1}{n(1+nx^2)}$.

(a) For what values of x does the series converge?

By comparison test, $f(x)=\sum_{n=1}^{\infty} \frac{1}{n(1+nx^2)}$ converges $\forall x \in \mathbb{R}$\{ $0$}.

(b) On what intervals of x in the form $(a,b)$ does the series converge uniformly?

Now since $\left |\frac{1}{n(1+nx^2)} \right|=\frac{1}{n+n^2x^2}\leq \frac{1}{n^2x^2}$ $\forall x \in \mathbb{R}$\{ $0$}, by M-Weierstrass Test the series converges uniformly on $(- \infty, 0)$ and $(0, \infty)$.

(c) On what intervals of x in the form $(a,b)$ does the series fail to converge uniformly?

0?

(d) Is f continuous at all points where the series converges?

No idea how to do this.

Could someone please check the answers that I have and help me for (d)?
• Nov 15th 2009, 04:44 PM
Enrique2
Quote:

Originally Posted by nmatthies1
Consider $f(x)=\sum_{n=1}^{\infty} \frac{1}{n(1+nx^2)}$.

(b) On what intervals of x in the form $(a,b)$ does the series converge uniformly?

Now since $\left |\frac{1}{n(1+nx^2)} \right|=\frac{1}{n+n^2x^2}\leq \frac{1}{n^2x^2}$ $\forall x \in \mathbb{R}$\{ $0$}, by M-Weierstrass Test the series converges uniformly on $(- \infty, 0)$ and $(0, \infty)$.

Be careful, you need a uniform bound for $1/x^2$ for applying M-test. Your argument is correct for the compact subsets of $(-\infty,0)\cup (0,\infty)$, when such a bound always exists.

Just observe that this gives you an answer for (d), for each
$x\neq 0$ you have that there exists $x\in[a,b]\subseteq \mathbb{R}\setminus\{0\}$ such that the series converges uniformly on $[a,b]$.
Now, recall that the uniform convergence of continuous functions preserves continuity. Actually M-test gives continuity whenever the terms of the series are.
• Nov 16th 2009, 01:26 PM
nmatthies1
Quote:

Originally Posted by Enrique2
Be careful, you need a uniform bound for $1/x^2$ for applying M-test. Your argument is correct for the compact subsets of $(-\infty,0)\cup (0,\infty)$, when such a bound always exists.

Just observe that this gives you an answer for (d), for each
$x\neq 0$ you have that there exists $x\in[a,b]\subseteq \mathbb{R}\setminus\{0\}$ such that the series converges uniformly on $[a,b]$.
Now, recall that the uniform convergence of continuous functions preserves continuity. Actually M-test gives continuity whenever the terms of the series are.

What do you mean by uniform bound?
Also, I can't quite seem to get my head around (d), why is it that uniform convergence guaratees preservation of continuity?
• Nov 16th 2009, 01:52 PM
Enrique2
An uniform bound on [a,b] (or (a,b) is a bound for every $x\in [a,b]$. It is a necessary condition in M test that $f_n(x)\leq M_n$ for each $x\in [a,b]$ and $\sum_nM_n$ being convergent. Now look that in an interval (0,b) you have
that the bound $\frac{1}{xn^2}$ depends on $x$ and goes to infinity as $x$ goes to 0. This is not allowed in the M-test, you need a bound independent of $x$, i.e. valid for all the interval.

The fact of that uniform convergence of a sequence of continuous functions is again continuous is a well known theorem, I don't remember the name of the author. That is because C(K) endowed with the uniform norm is a Banach space if K is compact. Sure that you can check it in a book of Analysis, I suppose in Apostol for example.