Function Composition and Jump Discontinuities

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.

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.

• nice result. +1 – JEM Nov 3 '17 at 1:10

Let $f$ be a constant function.

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

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

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