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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.

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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
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.

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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
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.

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$u_{n}(x)$ 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 $x_0$.

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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.

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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

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