discontinuity of $f(x) = \left \{ x \right \}* \left \{ -x \right \}$

Question

I have this function: $$f(x) = \left \{ x \right \}* \left \{ -x \right \}$$ in which $\left \{ x \right \}=x-[x]$

I have to establish if $f(x)$ is continuous or discontinuous somewhere and, in this last case, the kind of discontinuity. $f(x) = \left \{ x \right \}* \left \{ -x \right \}=(x-[x])*(-x+[x])=-(x-[x])^2=-x^2-[x]^2+2x[x]$

$\lim_{x\rightarrow n^{+}}(-x^2-[x]^2+2x[x]) =-n^2$

$\lim_{x\rightarrow n^{-}}(-x^2-[x]^2+2x[x]) =-(n-1)^2$

$s=-n^2+(n-1)^2=1-2n$

$\forall n \in Z$ $f(x)$ is discontinuous and it is of first kind. Is it right?

Answer 1

We just need to study the continuity at $x=n\in\Bbb Z$.

If $n<x <n+1,$

$$f (x)=(x-n)(-x+n+1) $$

thus $f (n^+)=0$.

If $n-1<x <n$

$$f (x)=(x-n+1)(-x+n) $$ thus $f (n^-)=0=f (n) $. $f $ is continuous at $\Bbb R $.

As $f $ is periodic $f (x+1)=f (x) $, you just need to study continuity at $x=0$.

Answer 2

Let me try.

The function is periodic with period 1 because $\{x+1\}=\{x\}$ and $\{-(x+1)\}=\{-x-1\}=\{-x\}$, so it sufficient to decide continuity in interval of length 1, e.g. $[0,1]$. The function $t\mapsto\{t\}$ continuous on $(0,1)$, so your function also, so it remains to check at 0 and 1.

$f(0)=f(1)=0$.

$x\to 0^+\implies \{\pm x\}\to 0^+\implies f\text{ continuous at 0}$,

$x\to 1^-\implies\{\pm x\}\to 0^+\implies f\text{ continuous at 1}$.

Function is continuous.

See the plot from Wolfram