Let $H: \mathbb{R} \rightarrow \mathbb{R}$ s.t. $H(x) = 0$ whenever $x < 0$ and $H(x) = 1$ whenever $x \ge 0$. Let $\sum_{n=1}^\infty a_n$ be a series of positive terms which converges. Let $\{x_n\} \subset \mathbb{R}$ be a countable set of real numbers.

Define $f: \mathbb{R} \rightarrow \mathbb{R}$ s.t. $f(x) = \sum_{n=1}^\infty [a_n H(x - x_n)]$

From this definition we can see that $f$ is non-negative, non-decreasing, and bounded above by $\sum a_n$.

Now I want to show independently that

(i) $f$ is right continuous on all $x \in \mathbb{R}$

(ii) $f$ is not continuous on any $x_n$

(iii) $f$ is continuous on all $\mathbb{R} - \{x_n\}$

$\fbox{First Proof Attempt}$

  1. For all $n \in \mathbb{N}$, let $f_n = a_n H(x-x_n)$ so that if $x < x_n$, $f_n(x) = 0$ and if $x \ge x_n$, $f_n(x) = a_n$.

  2. Consider that for all $n \in \mathbb{N}$, we have that $|f_n| \le a_n$ with $\sum a_n \le M$ for some $0 < M \in \mathbb{R}$ (since this series converges by hypothesis). This implies $\sum f_n \rightarrow f$ uniformly (a fact from Real Analysis). For notational simplicity, let $\sum_{i=1}^n f_n = g_n$ so that we have $g_n \rightarrow f$ uniformly.

  3. The uniform convergence of the $g_n$ now implies that $\forall x \in \mathbb{R}$, $\underset{t \rightarrow x+}{lim} \underset{n \rightarrow \infty}{lim} g_n(t) = \underset{n \rightarrow \infty}{lim} \underset{t \rightarrow x+}{lim} g_n(t)$ (another fact from Real Analysis) so that since each $f_n$ is right continuous (and hence $g_n$, a finite sum of right continuous functions), we have that $\underset{t \rightarrow x+}{lim} f(t) = f(x)$ implying $f$ is right continuous.

  4. Similarly, if we let $b \in \mathbb{R} - \{a_n\}$, then it is easy to see that each $f_n$ is continuous at $b$ since each $f_n$ is locally constant on $b$. Then the finite sums $g_n$ will also be locally constant on $b$ so that since the uniform continuity of the $g_n \rightarrow f$ implies $\underset{t \rightarrow b}{lim} \underset{n \rightarrow \infty}{lim} g_n(t) = \underset{n \rightarrow \infty}{lim} \underset{t \rightarrow b}{lim} g_n(t)$, we have that $\underset{t \rightarrow b}{lim} f(t) = f(x)$ so that $f$ is continuous on $b$ as desired.

  5. Finally, if we consider an arbitrary $x_k \in \{x_n\}$, we can observe that $f_k$ is not continuous on $x_k$ (since, in particular, $f_k$ is not left continuous at $x_k$). This then implies any finite sum $g_N$ s.t. $N \ge k$ is discontinuous at $x_k$ so that since $g_n \rightarrow f$ uniformly, we have $\underset{t \rightarrow x_k}{lim} \underset{n \rightarrow \infty}{lim} g_n(t) = \underset{n \rightarrow \infty}{lim} \underset{t \rightarrow x_k}{lim} g_n(t)$ implies $\underset{t \rightarrow x_k}{lim} f(t) = \underset{n \rightarrow \infty}{lim} \underset{t \rightarrow x_k}{lim} g_n(t) = undefined$ since each $\underset{t \rightarrow x_k}{lim} g_n(t)$ is undefined (i.e., such limit doesn't exist).

In particular, I'm worried about step (5).

$\fbox{John Mangual's Proof}$

  1. $\sum f_n \rightarrow f$ uniformly via the Weiestrauss $M$-test.
  2. Let $\{x_{n_{k}}\}$ form an ordering of the countable $\{x_n\}$.
  3. Then $f(x)$ is constant on each $[x_{n_k},x_{n_{k+1}})$.
  4. Then $f(x)$ is right continuous on all real $x$ from (3).
  5. Furthermore, if $x \notin \{x_n\}$, then we have that $x \in [x_{n_k},x_{n_{k+1}})$ for some $k$. Hence $f$ is continuous at such $x$ since $f$ is constant in small enough neighborhoods of $x$ (i.e., neighborhoods inside $[x_{n_k},x_{n_{k+1}})$).
  6. Finally, at each $x_n$, we have that $f$ is discontinuous since

    $$ f(x_n - \delta) = \sum_{i \in S} a_i \ne a_n + \sum_{i \in S} a_i = f(x_n + \delta) $$

    where $S = \{i : x_i < x_n - \delta\}$.

    This completes the proof.

  • 2
    $\begingroup$ The function $f$ is right-continuous as a uniform limit of right-continuous functions. $\endgroup$ – Did Dec 19 '12 at 15:26
  • $\begingroup$ I just edited my proof attempt above incorporating your comment. Thanks for your help. $\endgroup$ – user1770201 Dec 20 '12 at 20:09

I think you lack intuition for this type of problem. The infinite sum of heaveside functions is a step function or even a renewal process if you let the $a_n$ and $x_n$ be random numbers.

Why not use mathematica and let $x_n = \frac{k}{M}$ for $M >> 1$ and $a_n = f(x_n)$, for example?


Your function is constant on the countably many intervals $\color{#EEEEEE}{[x_n, x_{n+1})}$ determined by how the $x_n$ are ordered on the real line. Even if the $x_i$ accumulate somewhere:

  • $f$ is right-continuous on $\mathbb{R}$
  • $f$ is continuous away from the jumping points $\{ x_n \}$.

For $\epsilon > 0$ sufficiently small, let $S =\{ i: x_i < x_n - \epsilon\}$. At the points of discontinuity, we show $f$ has different values on the left and on the right.

  • $f(x_n - \epsilon) = \sum_{i \in S } a_i$
  • $f(x_n + \epsilon) = a_n + \sum_{ i \in S } a_i$

Uniform convergence ensures that when we prove these statements for the finite sums $f_n$, they pass to the limit $f_n \to f$.

Example: $ \displaystyle \sum_{n=1}^\infty \frac{1}{n^2} H\left(x - \frac{1}{n}\right) $

enter image description here

  • $\begingroup$ I appreciate your answer. Is there anything written in my above attempt that is explicitly wrong? $\endgroup$ – user1770201 Apr 24 '14 at 13:56
  • $\begingroup$ @user1770201 You have not ruled out that for very large $N$, the limit might exist after all. I think you should demonstrate that $\displaystyle \lim_{n\to\infty} g_n(x_n - \epsilon) \neq \lim_{n\to\infty} g_n(x_n+\epsilon)$ and explain clearly what $\epsilon(n)$, the left and right sides should be. $\endgroup$ – cactus314 Apr 24 '14 at 15:11
  • $\begingroup$ (1) In your explanation, is it not the case that you have the values for $f(x_n- \epsilon)$ and $f(x_n + \epsilon)$ switched? (2) It doesn't seem quite right to say $f(x)$ is constant on $[x_n, x_{n+1})$ since there could be another $x_k \in [x_n, x_{n+1})$. Is this not the case? $\endgroup$ – user1770201 Apr 24 '14 at 15:45
  • $\begingroup$ @user1770201 Does that look better? My biggest concern is that some subsequence $x_{n'}$ accumulates near some $x_m$ and $f$ might be continuous there. The Cantor function or Devil's Staircase is such an example. $\endgroup$ – cactus314 Apr 26 '14 at 2:45
  • $\begingroup$ I wrote up my interpretation of your proof above. I don't think it has any holes. If you verify that my write up is correct, I think that will wrap up this discussion. $\endgroup$ – user1770201 Apr 26 '14 at 16:05

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