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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 $\mathbb{R} - \{x_n\}$

$\fbox{EDITED Proof Attempt}$

  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).

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The function $f$ is right-continuous as a uniform limit of right-continuous functions. –  Did Dec 19 '12 at 15:26
    
I just edited my proof attempt above incorporating your comment. Thanks for your help. –  user1770201 Dec 20 '12 at 20:09
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This question has an open bounty worth +100 reputation from user1770201 ending in 7 days.

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Verify the proof -- and in particular step (5) -- is correct.

1 Answer

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 intervals $[x_n, x_{n+1})$ so 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 $f_n \to f$.

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