Suppose I have sequence of functions $\; f_n : [a, b] → \mathbb R$ that converges uniformly, i.e. $f_n \rightrightarrows f$ as $n \to \infty.$

Consider two properties of $\;f_n$:

  1. $f_n$ has countably many jump discontinuities, $\forall\; n \in \mathbb N.$
  2. $f_n$ has no jump discontinuities, $\forall\; n \in \mathbb N.$

My question is which of these properties necessarily carries over to $f.$ How can I can this be (dis)proven?

  • $\begingroup$ What is there in (1). If $f_n$ may or may not have jump discontinuties, then is it necessary that $f$ also may or may not have jump discontinuties? Isn't this trivial? aside another note : number of jump discontinuties of a function in $\mathbb{R}$ is always countable. $\endgroup$ – Rajesh Dachiraju Feb 22 '16 at 16:24
  • $\begingroup$ This sounds like a home-work problem to me! $\endgroup$ – Rajesh Dachiraju Feb 29 '16 at 16:10

Looking at only jump discontinuities is too restrictive.

Property 1. holds automatically, as proved e.g. here, since a function defined on an interval in $\mathbb{R}$ always has only countably many (possibly none) jump discontinuities.

But if we're looking at jump discontinuities only, property 2. doesn't carry over to the uniform limit, since an essential discontinuity of the $f_n$ can become a jump discontinuity of the limit function. Consider for an example

$$f_n(x) = \begin{cases} \qquad 0 &, x \leqslant 0 \\ 1 + \frac{1}{n}\sin \frac{1}{x} &, x > 0. \end{cases}$$

Then each $f_n$ has an essential discontinuity at $0$, but the amplitude of the oscillation decreases to $0$, and the uniform limit of the $f_n$ has a jump discontinuity at $0$.

We have the analogue of 2. if instead of only jump discontinuities we look at a more general type of discontinuity. Let's say that a function $f$ has a jump-type discontinuity (or a generalised jump discontinuity) at $x$ if either

$$\limsup_{y \nearrow x} f(y) < \liminf_{z \searrow x} f(z),\quad\text{or}\quad \liminf_{y\nearrow x} f(y) > \limsup_{z \searrow x} f(z).$$

Informally, a jump-type discontinuity is a superposition of a jump discontinuity and possibly an essential discontinuity that is not too wild.

Looking at jump-type discontinuities instead of jump discontinuities, both properties carry over to the uniform limit. Property 1. again carries over because a function has only countably many jump-type discontinuities - the proof here doesn't in fact use the existence of the one-sided limits, it works with small modifications in the wording for our jump-type discontinuities. The carrying-over of property 2. is not automatic without looking at the functions themselves. Let us formulate it as

Suppose $f_n \colon (a,b) \to \mathbb{R}$ is a sequence converging uniformly to $f$. If $f$ has a jump-type discontinuity at $x \in (a,b)$, then there is an $N \in \mathbb{N}$ such that $f_n$ has a jump-type discontinuity at $x$ for all $n \geqslant N$.

A proof is not difficult, suppose without loss of generality that

$$L := \limsup_{y \nearrow x} f(y) < M := \liminf_{z \searrow x} f(z).$$

Let $\varepsilon = \frac{1}{3}(M-L)$. By uniform convergence, there is an $N\in \mathbb{N}$ such that $\lvert f_n(t) - f(t)\rvert \leqslant \varepsilon$ for all $n\geqslant N$ and all $t\in (a,b)$. Then we have

$$\limsup_{y \nearrow x} f_n(y) \leqslant L + \varepsilon < M - \varepsilon \leqslant \liminf_{z \searrow x} f_n(z),$$

so $f_n$ has a jump-type discontinuity at $x$ for $n \geqslant N$.

The essence is that if $f_n$ converges uniformly to $f$, then $f$ cannot behave worse at $x$ than almost all of the $f_n$ behave at $x$. But it can behave better, and thus essential singularities of the $f_n$ might become jump discontinuities of $f$.

  • $\begingroup$ In your example, how would the uniform limit of $f_n$ be defined explicitly? $\endgroup$ – Rebecca Bonham Nov 12 '19 at 2:42
  • 1
    $\begingroup$ The limit function is $$f(x) = \begin{cases} 0 &\text{if } x \leqslant 0 \\ 1 &\text{if } x > 0 \end{cases}$$ @RebeccaBonham. We have $\lvert f_n(x) - f(x)\rvert \leqslant \frac{1}{n}$ for all $x$ and all $n$. $\endgroup$ – Daniel Fischer Nov 12 '19 at 12:09

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