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Let $C(X)$:space of continuous functions on a compact space.

Consider $f$ and $g :C(X)\rightarrow \mathbb{R}$ are upper semi continuous.

suppose for every $T\in C(X)$ set of $f(T)$ and set $(f+g)(T)$ are closed interval(connected set).Can we say set of $g(T)$ is closed interval(connected set),as well?

If Not,under which condition we have it.

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  • $\begingroup$ What does "set of $f(T)$" mean? If $f$ and $T$ are both fixed then $f(T)$ is a number. $\endgroup$
    – Nik Weaver
    Sep 22, 2018 at 13:28
  • $\begingroup$ @NikWeaver Yes f(T) is number.I mean $\{ f(T): T\in C(X) \}$is closed interval.Is it clear? $\endgroup$
    – Adam
    Sep 22, 2018 at 16:45
  • $\begingroup$ Got it, thank you. $\endgroup$
    – Nik Weaver
    Sep 22, 2018 at 17:25
  • $\begingroup$ You're welcome, $\endgroup$
    – Adam
    Sep 22, 2018 at 17:56

1 Answer 1

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The answer is no. First ask this question for functions from $\mathbb{R}$ to $\mathbb{R}$. There are easy counterexamples: take $f(t) = t$ and let $g$ be the characteristic function of the interval $(-\infty,0)$. Then the ranges of $f$ and $f + g$ are both $\mathbb{R}$ but the range of $g$ is not connected.

You can turn that into a counterexample on $C(X)$ by fixing a point $x \in X$ and composing the $f$ and $g$ described above with the evaluation functional $\phi \mapsto \phi(x)$ from $C(X)$ to $\mathbb{R}$ (or, really, any nonzero linear functional on $C(X)$).

I can't think of any good condition that would guarantee this. Just knowing that the range of a function is connected tells you very little about the function.

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  • $\begingroup$ Thanks for your answer.what about f=g? Again answer is no? Can't we use upper semi continues property? $\endgroup$
    – Adam
    Sep 22, 2018 at 17:55
  • $\begingroup$ E.g Inverse of closed of upper semi continuous function is closed. $\endgroup$
    – Adam
    Sep 22, 2018 at 17:58
  • $\begingroup$ If $f = g$ and you assume the range of $f$ is connected, then the range of $g$ will be connected. What is the question? $\endgroup$
    – Nik Weaver
    Sep 22, 2018 at 19:01

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