1
$\begingroup$

Suppose $E:=(E,\|\!\cdot\!\|)$ is a Banach space and $I:=[\alpha,\beta]$ is a compact perfect interval. A function $f:I\to E$ is jump continuous if the one-sided limits $f(\alpha+),f(\beta-)$ and $f(x\pm)$ exist for all $x\in I^\circ$.

The following is Exercise VI.1.5 from Analysis II by Amann & Escher:

Prove that every jump continuous function has at most a countable number of discontinuities.

I know that monotone functions ($\mathbb{R\to R}$) have the same property, whose proof begins by choosing a rational number between $f(x-)$ and $f(x+)$. I tried to imitate, but I don't know what to do when $E$ is not totally ordered as $\mathbb{R}$ is.

The proof of this exercise for the case $E=\mathbb{R}$ can be found here, but I don't know how the proof can be modified for the general case. However, I think the proof here works for the general case, but it does not seem "natural" to me. I'm not familiar with the fact that a discrete subset of $\mathbb{R}$ is countable (which does not appear in Amann & Escher). Also, the exercise comes with no hint, so I think it should be easy in some sense. Maybe I'm missing something. Are there other proofs?

$\endgroup$
  • 1
    $\begingroup$ Since $I$ is compact, there can only be finitely many jumps of size $\geqslant 1/n$ by Bolzano-Weierstraß. $\endgroup$ – Daniel Fischer Jan 12 '18 at 16:16
3
$\begingroup$

Let $\varepsilon > 0$. Let $(t_n)_n$ be a sequence in $[\alpha , \beta]$ with $t_n \rightarrow t \in [\alpha , \beta]$ and $$\max \{||f(t_n+) - f(t_n)||,||f(t_n-) - f(t_n)||\} > \varepsilon$$ Without loss of generality assume $t_1<t_2<\ldots<t_n<\ldots$ Thus, for every $n$ there exists a $s_n> t_n$ or $s_n < t_n$ with $||f(s_n)- f(t_n)|| > \frac \varepsilon 2$, but with $s_n < s_{n+1} < t$.

So we have $\frac \varepsilon 2 < ||f(s_n)- f(t_n)|| \rightarrow ||f(t-)-f(t-)||=0$. Contradiction.

We conclude, that the set $\{t\in [\alpha , \beta ]: \max \{||f(t+) - f(t)||,||f(t-) - f(t)||\} > \varepsilon \}$ is finite. (Otherwise we find a convergent sequence with Bolzano-Weierstraß)

The set of discontinuities is

$$\bigcup\limits_{n=1}^\infty \{t\in [\alpha , \beta ] : \max \{||f(t+) - f(t)||,||f(t-) - f(t)||> \frac 1 n\}$$

hence countable.

$\endgroup$
1
$\begingroup$

For $x\in (\alpha, \beta)$ let $g(x)=\|f(x)-f(x^-)\| + \|f(x)-f(x^+)\|.$

For any $\epsilon >0$ let $y\in (\alpha,x)$ such that $$ (I)...\quad \forall z\in (y,x)\;(\;\|f(z)-f(x^-)\|<\epsilon).$$ For $z\in (y,x)$ take $z'\in (y,z)$ such that $\|f(z')-f(z^-)\|<\epsilon$ and take $z''\in (z,x)$ such that $\|f(z'')-f(z^+)\|<\epsilon.$

Then $$(II)...\quad \|f(z)-f(z^-)\|\leq \|f(z)-f(x^-)\|+\|f(x^-)-f(z')\|+\|f(z')-f(z^-)\|$$ and $$(III)...\quad \|f(z)-f(z^+)\|\leq \|f(z)-f(x^-)\|+\|f(x^-)-f(z'')\|+\|f(z'')-f(z^+)\|.$$

By $(I),$ each of the RHS terms in $(II)$ and $(III)$ is less than $\epsilon,$ so $$\forall z\in (y,x)\;(0\leq g(z)<6\epsilon).$$

Therefore $g(x^-)=0.$

Similarly, we show that $g(x^+)=0. $

From the case $E=\Bbb R$ we know that $g$ has only countably many discontinuities. So for all but countably many $x\in (\alpha,\beta)$ we have $g(x)=g(x^-)=g(x^+)=0.$ Now $g(x)=0$ implies $f(x)=f(x^-)=f(x^+)$ which implies that $f$ is continuous at $x.$

Addendum. Regarding a discrete $S\subset \Bbb R$ being necessarily countable: For $s\in S$ let $r_s>0$ such that $(s-r_s\;,s+r_s)\cap S=\{s\}$ and let $U(s)=(s-\frac {1}{2}r_s\;,s+\frac {1}{2}r_s).$

For unequal $s_1,s_2$ in S, suppose (by contradiction) that $t\in U(s_1)\cap U(s_2).$ We have $$(r_{s_1}\leq |s_1-s_2|)\; \land \; (r_{s_2}\leq |s_2-s_1|).$$ But we also have $$|s_1-s_2|\leq |s_1-t|+|t-s_2|<\frac {1}{2}r_{s_1}+\frac {1}{2}r_{s_2}\leq $$ $$\leq \frac {1}{2}|s_1-s_2|+\frac {1}{2}|s_2-s_1|=|s_1-s_2|,$$ implying $|s_1-s_2|<|s_1-s_2|,$ which is absurd. Therefore $U(s_1)\cap U(s_2)=\emptyset.$

Therefore $F=\{U(s):s\in S\}$ is a pair-wise disjoint family of non-empty open subsets of $\Bbb R.$ Each $U(s)$ contains a rational that does not belong to any other member of $F,$ so $F$ must be countable because $\Bbb Q$ is countable.

This result easily generalizes from $\Bbb R$ to any separable metric space.

$\endgroup$
  • $\begingroup$ An amusing example of a discontinuous $g:(0,1)\to \Bbb R$ for which $g(x^-)=g(x^+)=0$ for all $x\in (0,1)$ is: $g(x)=0$ when $x\in (0,1)\setminus \Bbb Q,$ and if $m,n$ are co-prime members of $\Bbb N$ with $m<n$ then $g(m/n)=1/n.$ Then $g$ is continuous at each $x\in (0,1)\setminus \Bbb Q$ and $g$ is discontinuous at each $x\in (0,1)\cap \Bbb Q.$ $\endgroup$ – DanielWainfleet Jan 13 '18 at 2:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.