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If $f$ any $g$ be two functions defined from $[0,1]$ to $[0,1]$ with $f$ strictly increasing. Then

  1. if $g$ is continuous, is $f\circ g$ continuous?

  2. if $f$ is continuous, is $f\circ g$ continuous?

  3. if $f$ and $f\circ g$ are continuous, is $g$ continuous?

Here, $f\circ g$ implies composition of $f$ and $g$. I think the answer to the third is yes by using the fact that preimage of an open set under a continuous map is open? Any idea .Thanks.

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vidyarthi
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    Possible duplicate of this: http://math.stackexchange.com/questions/318614/composition-of-continuous-functions – Zaros Dec 09 '16 at 04:39
  • @lordoftheshadows no, this is not duplicate, here f and g are any functions with f increasing – vidyarthi Dec 09 '16 at 04:44
  • One can prove that a strictly increasing function has at most finitely many discontinuities. This may prove useful to you. – A. Thomas Yerger Dec 09 '16 at 04:45
  • @AlftedYerger but how to show that the preimage of any open set is open under the composition map? – vidyarthi Dec 09 '16 at 04:51
  • Also, it is not true in general that the image of an open set under a continuous map is open. Consider $f(x) = x^2$ defined on $(-1,1)$. This maps to $[0,1)$ which is not open, although sure $x^2$ is continuous. What you mean is that the PREimage of an open set is open. Also, my comment is incorrect. There can be countably many discontinuities. Small but important goof. – A. Thomas Yerger Dec 09 '16 at 04:51
  • @AlfredYerger spelling mistake.modified the question – vidyarthi Dec 09 '16 at 04:53
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    For the third, do you mean "if $g$ and $f \circ g$ are continuous, then $f$ is continuous"? – Ethan Dec 09 '16 at 04:56
  • @Ethan thanks for pointing out. Edited the question. – vidyarthi Dec 09 '16 at 04:59
  • Ah, cool. In that case, yes, your argument using open sets works. – Ethan Dec 09 '16 at 05:00
  • For $(2)$ the answer is no. To see this take $f(x)=x$ and $$g(x)=\displaystyle\begin{cases}0&\text{if}\ x\in[0,1]\cap \mathbb{Q}\1&\text{else}\end{cases}$$ Now observe that $f$ is strictly increasing and continuous but, $$(f\circ g)(x)=\displaystyle\begin{cases}0&\text{if}\ x\in[0,1]\cap \mathbb{Q}\1&\text{else}\end{cases}$$ –  Dec 09 '16 at 05:16
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    Note that since the identity function $\mathrm{id}$ on $[0,1]$ is continuous and strictly increasing, and furthermore has the property that for any $f:[0,1]\to [0,1]$ we have $\mathrm{id}\circ f = f \circ \mathrm{id} = f$, each of (1) and (2) would imply that all functions are continuous. – celtschk Dec 09 '16 at 10:01

4 Answers4

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$(3)$:Since $f$ is increasing it is injective and hence has a left inverse .

Moreover $f$ is given to be continuous so the left inverse is also so and hence $f^{-1}\circ (f\circ g)=g$ is continuous which proves $3$.

$(1)$: $f(x)=$\begin{cases} x & x\in [0,\frac{1}{2}]\\\frac{x}{2}&x\in (\frac{1}{2},1]\end{cases}

and $g(x)=x$,then $f\circ g$ is discontinuous at $\frac{1}{2}$.

$(2)$: $f(x)=x^2$;

$g(x)=$\begin{cases} 1 & x\in \Bbb Q\cap[0,1]\\0 &x\in \Bbb Q^c\cap [0,1]\end{cases};then $f\circ g$ is discontinuous

Learnmore
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In plain English:

  1. if $g$ is continuous, is $f\circ g$ continuous?

Not necessarily. We know that $f$ is strictly increasing, but that does not imply that it is continuous.

Counter-example: Define $f$ as any strictly increasing, non-continuous function.

In other words, stating that the input to $f$ "changes smoothly" (i.e. $g$ is continuous) states nothing whatever about the output of $f$.

  1. if $f$ is continuous, is $f\circ g$ continuous?

Not necessarily. $g$ could be any arbitrary function; it may not be continuous.

Describing a function as "continuous" states that if the input changes smoothly, the output changes smoothly. If the input (in other words, the output of $g$) jumps around arbitrarily (discontinuous), the output of $f$ may not change smoothly.

  1. if $f$ and $f\circ g$ are continuous, is $g$ continuous?

Yes, but note that the information provided at the beginning of the question, that $f$ is strictly increasing, is necessary to prove this point.

If this restriction is omitted, the following would be a counter-example:

$g(x) = \begin{cases} x \le 0.5 && 0.2 \\ x \gt 0.5 && 0.8 \end{cases}$

$f(x) = 4x^2 - 4x +1$

Note the following attributes of the above function definitions:

  • $f$ is continuous.

  • $f \circ g$ is continous.

  • $g$ is not continuous.

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Wildcard
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    OP edited the third question. It's asking if $g$ is continuous. – Ethan Dec 09 '16 at 05:03
  • To say that a proposition doesn't necessarily hold you need to produce a counterexample satisfying all the assumptions of the proposition, yet giving just the negation of the conclusion of the proposition. For $(1)$ and $(2)$, where have you done that? –  Dec 09 '16 at 05:08
  • What counterexample can be given in first and second cases? – vidyarthi Dec 09 '16 at 05:09
  • @Ethan, thanks; I saw the edit and modified this answer accordingly. – Wildcard Dec 09 '16 at 05:22
  • @user170039, these conclusions proceed directly from the definition of "continuous." I've updated my answer to illustrate this better. – Wildcard Dec 09 '16 at 05:24
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So the third case is shown to be true properly by learnmore. For the counterexample to case 2), maybe we can take $f=x$ for case 2) and $g=\begin{cases} \frac{x}{2} && x\in[0,0.5)\\ \frac{x+1}{2} && x\in[0.5,1]\end{cases}$. Similar type could be given for case 1)

vidyarthi
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I think, for this kind of question, it's best to go with your gut and see if you can't produce a few examples that corroborate or refute it before trying to prove it. Can you think of any examples or counterexamples to either (1) or (2)? Your proposed answer to (3) is indeed correct, so let's focus on (1) and (2).

For (2), the only things we're given are that $f$ is continuous and strictly increasing. But $g: [0,1] \to [0,1]$ could be pretty much anything! So, your first hunch should probably be "no, $f \circ g$ need not be continuous." Can you find a counterexample?

For (1), try to use the same type of reasoning.

Ethan
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    I'm a new user: am I supposed to just give away the actual answer rather than supplying hints? I honestly don't know, what's the protocol for this? – Ethan Dec 09 '16 at 05:11
  • I mean the main idea is to supply hints, and just let the OP know if he/she is on the right track or not. – Felicio Grande Dec 09 '16 at 05:33