Prove that, the map $x\mapsto=[x]\cos^2{\pi x}$ is discontinuous at every integer points.

Question

I think that the map $x\mapsto=[x]\cos^2{\pi x}$ may have jump discontinuities at every integer point. However I cannot establish that fact in a rigorous manner.
Can anybody assist me to find the proper way to solve my question?
Thanks for assistance in advance.

Answer 1

Let, $c\in\Bbb{Z}$
Let, $g:\Bbb{R}\to\Bbb{R}$ is defined by $g(x)=[x]\cos^2{\pi x}\quad\forall x\in\Bbb{R}$
I'll prove that $g$ has jump discontinuity at $c$.
Note that, $g(c)=[c]\cos^2{\pi c}=c$
My claim is $\lim\limits_{x\to c-} g(x)=c-1$
Choose $\varepsilon>0$
As the map $x\mapsto\cos^2{\pi x}$ is continuous on the whone real line(Why?), we have $\lim\limits_{x\to c+} \cos^2{\pi x}=\lim\limits_{x\to c-} \cos^2{\pi x}=\cos^2{\pi c}=1$.
Now using $\lim\limits_{x\to c-} \cos^2{\pi x}=1$, for $\frac{\varepsilon}{|c-1|}>0$, we can have $\delta_\varepsilon>0$ such that
$|\cos^2{\pi x}-1|<\frac{\varepsilon}{|c-1|}\quad\forall x\in(c-\delta_\varepsilon, c)$
$\implies|(c-1)\cos^2{\pi x}-(c-1)|<\varepsilon\quad\forall x\in(c-\delta_\varepsilon, c)$ $\implies|[x]\cos^2{\pi x}-(c-1)|<\varepsilon\quad\forall x\in(c-\delta_\varepsilon, c)$ (because $x\in(c-\delta_\varepsilon, c)\implies[x]=c-1$)
$\implies|g(x)-(c-1)|<\varepsilon\quad\forall x\in(c-\delta_\varepsilon, c)$
Therefore, for any $\epsilon>0,\exists\delta_\varepsilon>0$ such that $|g(x)-(c-1)|<\varepsilon\quad\forall x\in(c-\delta_\varepsilon, c)$
$\implies\lim\limits_{x\to c-} g(x)=c-1\ne c=g(c)$
$\implies g$ has (left)jump discontinuities at every integer points on $\Bbb{R}$.