1
$\begingroup$

Let $a,b>0\in\mathbb{R}$. Let $U$ be an domain in $\mathbb{C}^n$. Let $f:[a,b]\longrightarrow U$ be a piecewise continuous map. Then is $f[a,b]$ compact? If not compact, will it be bounded?

Ok. This is in the following context. I am given a piecewise smooth path $\gamma:[a,b]\longrightarrow U$. Where $\gamma(a)=z$ and $\gamma(b)=w$, for given $z,w\in U$. We are also given a function $\alpha:U\times\mathbb{C}^n\longrightarrow \mathbb{R}$, which is upper semicontinuous. Now it is said that $t\in[a,b]\longrightarrow \alpha(\gamma(t),\gamma’(t))$ is bounded and measurable. I wanted to know why the function is bounded. I know that $\gamma[a,b]$ is compact. And $\gamma$ being upper semicontinuous will attain its maximum on a compact set. But I am not sure about $\gamma’$.

$\endgroup$
7
  • 3
    $\begingroup$ What does "piecewise continuous" mean here? How many pieces? Finitely many? Are the pieces compact? Open? $\endgroup$ Aug 19, 2020 at 10:51
  • 1
    $\begingroup$ What have you tried? $\endgroup$
    – supinf
    Aug 19, 2020 at 10:53
  • $\begingroup$ The image of a compact space under continuous map is always compact. Regardless of what "piecewise" means and what $U$ is (as long as it is Hausdorff). $\endgroup$
    – freakish
    Aug 19, 2020 at 10:57
  • $\begingroup$ Please provide additional context, which ideally explains why the question is relevant to you and the community. Some forms of context include background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. $\endgroup$ Aug 19, 2020 at 11:18
  • $\begingroup$ @SahibaArora I have added the context. $\endgroup$
    – user531706
    Aug 19, 2020 at 13:25

1 Answer 1

0
$\begingroup$

Compact, not necessarily: On $[0,1]$ let $f(x) = x, 0\le x<1,$ $f(1)=2.$ Then $f([0,1]) = [0,1)\cup\{2\}.$

Bounded, yes: First, a lemma: If $f$ is continuous on $(a,b)$ and $f$ has finite limits at the end points, then $f(a,b)$ is bounded.

Proof: Suppose $\lim_{x\to a^+} f(x)=L,$ $\lim_{x\to b^-} f(x)=M.$ Let $\epsilon=1.$ Then there exists $\delta_a>0, \delta_a<(b-a)/3,$ such that $|f(x)-L|<1$ for $x\in (a,a+\delta_a).$ Thus for such $x,$

$$|f(x)| = |f(x)-L+L|\le |f(x)-L|+|L| <1+|L|.$$

Similarly, there exists $\delta_b>0,\delta_b<(b-a)/3,$ such that $|f(x)|<1+|M|$ for $x\in (b-\delta_b,b).$ It follows that $f$ is bounded on the set $(a,a+\delta_a)\cup (b-\delta_b,b).$

Since $f$ is continuous on the compact set $[a+\delta_a,b-\delta_b],$ $f([a+\delta_a,b-\delta_b])$ is compact, hence is bounded. It follows that $f(a,b)$ is bounded.

Now suppose $f$ is piecewise continuous on $[a,b].$ Then there exist points $a=x_0<x_1<\cdots <x_n=b$ such that $f$ is continuous on each $I_k=(x_{k-1},x_k)$ and has finite limits at the end points of $I_k.$ By the lemma, each $f(I_k)$ is bounded. The set $f(\{x_0,\dots x_n\})$ is also bounded. Therefore

$$f([a,b])=f(I_1)\cup \cdots \cup f(I_n)\cup f(\{x_0,\dots x_n\})$$

is bounded.

$\endgroup$

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.