9

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?

Marter Js
  • 515

4 Answers4

7

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.

N8tron
  • 2,507
1

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.

amWhy
  • 209,954
Platonix
  • 701
  • 1
  • 4
  • 21
  • 2
    It seems this objection has been addressed in the problem statement by defining sign$(0)$ as $1$. – David K May 22 '14 at 02:48
0

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 $0$.

Mansi Tyagi
  • 175
0

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.

David K
  • 98,388