1
$\begingroup$

We say that if $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $g(a)$ then $(f\circ g)(x)$ is continuous at $a$.

The converse: If $(f\circ g)(x)$ is continuous at $a$ then $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $g(a)$ is not necessarily true.

For example consider $g(x) = x+1 -2 H(x)$ where $H(x)$ is the Heaviside step function. $g(x)$ has a jump discontinuity at $x=0$. In fact,

$g(x) = \left\{\begin{array}{l}x+1\qquad x<0\\x-1\qquad x\geq0 \end{array}\right.$

However, if we choose$f(x)=x^2$ then

$(f\circ g)(x) = \left\{\begin{array}{l}(x+1)^2\qquad x<0\\(x-1)^2\qquad x\geq0 \end{array}\right.$

which is continuous for all real $x$.

Its seems plausible that if $g(x)$ has a jump discontinuity at $x=a$ then we will always be able to find some non-constant $f(x)$ that makes it continuous through composition. Can anyone state a counter example for this or maybe have seen a proof?

Extensions of this would be to a finite collection of discontinuities and then countable collections.

$\endgroup$
1
$\begingroup$

If $g$ has a jump discontinuity at $a$, then

$$f(x) = \left[x - \frac{g(a^+) + g(a^-)}{2}\right]^2$$

will have the property that $f(g(x))$ is continuous.

$\endgroup$
  • $\begingroup$ nice result. +1 $\endgroup$ – JEM Nov 3 '17 at 1:10
2
$\begingroup$

Let $f$ be a constant function.

$$\forall x \in \mathbb{R}, f(x) = 1$$

Hence $f(g(x))=1$ which is continuous.

$\endgroup$
  • $\begingroup$ I guess I should have said $f(x)$ not constant. Very nice Siong. $\endgroup$ – JEM Nov 3 '17 at 0:58

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.