6
$\begingroup$

I have some exercise that asks me this:

Find $f$ and $g$ discontinuos such that $f+g$ is continuous. This is what I tought:

$$f = \mbox{sign}(x)$$ $$g = -\mbox{sign}(x)$$

Where $\mbox{sign}(x)$ is the function that maps to $1$ if $x\ge0$ and $-1$ if $x<0$. Then, the two are discontinuous in $0$, but the sum, is continuous, because it's equal to $0$ in every point. First of all, is this example correct?

The exercise also asks me to find examples such that:

$f,g$ discontinuous, but $f$ composed with $g$ is continuous. (need help)

$f$ continuous, $g$ discontinuous but the composite $f$ with $g$ is continuous.

(for the second one, if I take $g=\mbox{sign}(x)$ and $f=|x|$, then $fog = 1$. Is this right?

Could you guys give at least a hint, or a less poor example?

$\endgroup$
  • 1
    $\begingroup$ Your examples are simple, but not bad. Simple can be good. $\endgroup$ – David K May 22 '14 at 3:09
4
$\begingroup$

You can come up with some really interesting examples of compositions that end up being continuous where the components are not. Here's my favorite:

$f(x)=\begin{cases} 0 & \text{if }x \text{ is irrational}\\ 1 & \text{if }x \text{ is rational} \end{cases}$

Then $f\circ f$ is the constant function $1$ which is continuous.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Your examples are wonderful, with one correction: your sign function is actually not a function, because $\operatorname{sign}(0)$ maps to both $1$ and $-1$ (in high school algebra parlance, this fails the "vertical line test"). This is easy to correct, however. Just let $\operatorname{sign}(0)=0$. An example of discontinuous functions that compose to a continuous function runs thus: let $f(x)=0$ if $x=0$, and $f(x)=1$ otherwise; let $g(x)=1$ if $x=0$ or $x=1$, and $g(x)=2$ otherwise. $f(g(x))$ is thus continuous, while $f$ and $g$ are discontinuous.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ It seems this objection has been addressed in the problem statement by defining sign$(0)$ as $1$. $\endgroup$ – David K May 22 '14 at 2:48
0
$\begingroup$

If $f$ is discontinuous and has a discontinuous inverse, then let $g = f^{-1}$, so that $f \circ g$ is the identity function.

As a candidate for such an $f$, I offer $f(x) = 1 + 2 \lfloor x \rfloor - x$. But there are uncountably many such functions.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

You can also take $f(x)=sgnx$ and $g(x)=x(1-x^2)$. Here $f$ is discontinuous at $0$ and $g$ is continuous at $f(0)$ but the composite $gf$ is continuous at $0$.

| cite | improve this answer | |
$\endgroup$

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.