Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know that for a series of non-negative, continuous functions $u_{n}(x)$, a sufficient condition for uniform convergence of $\sum u_{n}(x)$ to $u(x)$ is for $u(x)$ to be continuous in $I\subset \mathbb{R}$.

But I can't think of an example where $\sum f_{n}(x)\to f(x)$ uniformly in $I\subset \mathbb{R}$ but $f(x)$ is discontinuous in $I\subset \mathbb{R}$.

One of $(f_{n}(x))$ must obviously be discontinuous in the interval because (as others have written below) a uniform limit of continuous functions is continuous.

share|improve this question
    
Technically speaking, your first sentence is not always true. However, by Dini's Theorem (see Jonas' answer below), it can be fixed if you assume that your interval $I$ is compact. –  Jesse Madnick Jan 31 '11 at 22:16
    
@Jesse: Sorry, I wasn't reading properly. Thank you. I'll delete my incorrect comment to avoid confusion. –  Jonas Meyer Jan 31 '11 at 22:25
    
Sorry, now I'm misreading. I deleted my previous post. (My post at the top is still right, though.) –  Jesse Madnick Jan 31 '11 at 22:37

4 Answers 4

up vote 3 down vote accepted

One way to get an example for the clarified question is to let $u_1$ be any discontinuous function on $[0,1]$, like $u_1(x)=0$ if $0\leq x\leq \frac{1}{2}$ and $u_1(x)=1$ if $\frac{1}{2}\lt x\leq 1$, and let $u_n(x)=0$ for $n\geq 2$. Then the series $\sum u_n(x)$ converges uniformly to $u_1(x)$.

If the $u_n$'s are continuous, you won't be able to find such an example, because a uniform limit of continuous functions is continuous.


Regarding your first statement, it isn't necessarily true if the interval is not compact. For an example of a continuous function that is the nonuniform limit of an increasing sequence of nonnegative continuous functions on an interval, consider $f(x)=1$ on $(0,1)$, and $f_n(x)=1-x^n$. To phrase this in terms of a series of nonnegative functions, take $u_1=f_1$ and $u_n=f_n-f_{n-1}$ for $n\geq2$.

However, if the interval is compact, then it is true that if a monotonically increasing sequence of continuous functions converges pointwise to a continuous function, then the convergence is uniform. This is called Dini's theorem.

share|improve this answer
    
please see my edit, I wasn't clear before –  daniel.jackson Jan 31 '11 at 22:23
    
@daniel.jackson: I think I've added an example that addresses the update. –  Jonas Meyer Jan 31 '11 at 22:32
    
@daniel.jackson: If you want all of the functions to be discontinuous, see milcak's and Jesse Madnick's answers. In my example, only $u_1$ is discontinuous. –  Jonas Meyer Jan 31 '11 at 22:45
    
thanks, sorry for the confusion. –  daniel.jackson Jan 31 '11 at 22:51
1  
Another important result is the Baire-Osgood theorem which states (if I recall correctly) that if $(f_n)$ is a sequence in $C(\mathbb{R})$ that converges pointwise to a function $f: \mathbb{R} \to \mathbb{R}$, then $f$ has a point of continuity. That is to say that the limit of a sequence of continuous functions over a non-compact space does have some nice properties (I am fairly certain that the key ingredient here is that $\mathbb{R}$ is completely metrizable, but don't quote me on that) –  kahen Jan 31 '11 at 23:02

Suppose your interval $I = [-1,1]$.

Let $u_n(x) = 2^{-n}$ for $x \in [-1,0)$ and $u_n(x) = 2^{-n} + 1$ for $x \in [0,1]$. Then $\sum u_n(x) \to u(x)$ uniformly, where $u(x) = 1$ on $[-1,0)$ and $u(x) = 2$ on $[0,1]$.

For general intervals $I$... well, just change the bounds.

share|improve this answer
    
$u_{n}(x)$ in $[0,1]$ doesn't go to $0$ here so it diverges but it can be fixed by taking $u_n(x) = \frac{1}{3^{n}}$ for example. –  daniel.jackson Feb 1 '11 at 8:35

EDIT: If you want discontinuous functions whose sum converges uniformly to a discontinuous function, consider the characteristic function of the rationals:

$$ \lambda(x) = \left\{ \begin{array}{lr} 1 & : x \in \mathbb{Q}\\ 0 & : x \notin \mathbb{Q} \end{array} \right.$$

Then let $\lambda_n = \frac{1}{2^{n+1}} \cdot \lambda$. The sum will indeed converge uniformly to $\lambda$ but will clearly be discontinuous. In fact if $\lambda$ is any discontinuous function, this construction will give you your desired result.


However, if the $u_n$'s are all continuous, this cannot happen. Let $$g_n = \sum_{k=1}^{n} u_k$$ Then each $g_i$ is countinuous being the sum of continuous functions. But since $g_n \rightarrow u$ uniformly, it must be the case that $u$ is continuous as well. Here's a quick proof:

Suppose we want to show that $u$ is continuous at some $x_0 \in I$. Fix some $\epsilon > 0$. Find $N$ large enough such that $|g_n (x) - u(x) | < \frac{\epsilon}{3}$ for all $x \in I$ and $n \ge N$. Choose some $\delta$ such that $| g_N (x) - g_N (x_0) | < \frac{\epsilon}{3}$ whenever $|x - x_0 | < \delta$, which we can do as $g_N$ is continuous.

Now we have that:

$$|u(x) - u(x_0)| \le |u(x) - g_N(x)| + |g_N (x) - g_N (x_0) | + |g_N(x_0) - u(x_0)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon $$

by applying the triangular inequality. Hence $u$ is continuous at $x_0$.

share|improve this answer

This follows from the following:

Theorem. If the limit function $f(x)$ of a pointwise convergent sequence of continuous functions $\{f_{n}(x) \}$ is discontinuous, then the convergence of the sequence $\{f_{n}(x) \}$ is nonuniform.

So the contrapositive is: $\{f_{n}(x) \} \to f \ \text{uniformly} \Longrightarrow f(x) \ \text{is continuous}$.

Hence this is impossible.

share|improve this answer
    
By "This follows", do you mean "This is impossible"? –  Jonas Meyer Jan 31 '11 at 22:15
    
@Jonas Meyer: Yes –  PEV Jan 31 '11 at 22:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.