Why is $\sum_{n=1}^{\infty} c_n I(x - x_n)$ not continuous when $x = x_n$?

Question

Let $I(x) = \begin{cases} \hfill 0 \hfill & x \leq 0 \\ \hfill 1 \hfill & x > 0 \end{cases}$.

If $\{x_n\}$ is a sequence of distinct points of $(a, b)$, and if $\sum\lvert c_n \rvert$ converges, the series $$f(x) = \sum\limits_{n=1}^{\infty} c_n I(x - x_n) \text{, where } (a \leq x \leq b)$$ converges uniformly, and $f$ is continuous for every $x \neq x_n$, but $f$ is not continuous when $x = x_n$.

If $x = x_n$, then every term is $0$, which is a continuous function. Why is the claim that it's not continuous?

Answer 1

Because at I(0) the limit from either side is different- I(x) is said to be discontinuous at this point - from the right hand side we have 1 , from the left hand side we have 0.

Answer 2

Fix $x \in [a,b]$, and let $S_x = \{ n \in \mathbb N \, | \, x > x_n \}$. Then the function $f$ can be described as $f(x) = \sum_{n \in S_x} c_n$. This is always a convergent series by assumption, but its value is not a continuous parameter of $x$ ; it admits a discontinuity of "jump type" ($\lim_{x \to c^+}$ and $\lim_{x \to c^-}$ exist but are not equal) and you can check that the difference between these two limits at $x_N$ is precisely $c_N$. This is essentially because if you go through the interval $[a,b]$ from left to right, you pick a jump at each $x_n$ of size $c_n$.

In particular, $f$ is continuous at $x_n$ if and only if $c_n = 0$.

Hope that helps,

Answer 3

Also, it is not true that every term is zero, only the terms with $x \le x_n$. If $x > x_n$, then the term is $c_n$.