$f(x)=x^2-[x^2]$ is continuous?

Question

$f(x)=x^2-[x^2]$ is continuous? If not than say it has first order or second order breakpoints.

Can we say because $x^2$ is continuous $[x^2]$ is not than $f(x)$ is not continuous?

$[x]$ is $floor$ function.

Answer 1

$f(x)=x^2-\lfloor x^2 \rfloor$ of course is not continuous - consider, for example $x=\pm 1$. In general, $\pm \sqrt{n}$ are "jump" breakpoints.

When $f$ is continuous and $g$ not, then $\phi=f\pm g$ cannot be continuous, because if it is, then also become continuous $g=\phi -f$, which contradicts assumption.

Answer 2

$f(x)=x^2-[x^2]=\{x^2\}$, where $\{\}$ is fractional part function.
$f$ is not continuous at any points of the type $\pm\sqrt n$, where $n\in \mathbb N$. Although $f$ is continuous at $x=0$.
At points $x=\pm \sqrt n, g(x)=[x^2]$ is discontinuous but $h(x)=x^2$ is continuous at $x=\pm \sqrt n$ so if $f(x)$ is continuous then so is $g(x)=h(x)-f(x)$, which is contradiction.

Answer 3

Well, play around when $x$ is just under and just over an integer.

$x^2 -[x^2]\approx 0$ if $x =n + \delta$ for a very small $\delta > 0$.

That is to says if $x = n + \delta$ then $x^2 - [x^2] = (n+\delta)^2 - [(n+\delta)^2] =n^2 + 2\delta n + \delta^2 -[n^2 + 2\delta n + \delta^2] = (n^2+2\delta n + \delta^2) - n^2 = 2\delta n + \delta^2\approx 0$

But if $x = n -\delta$ for a very small $\delta > 0$ then

$x^2 = n^2 - 2n\delta +\delta^2 < n^2$ (assuming $\delta$ is small enough and $x \ne 0$ so $[x^2] = [n^2 - 2n\delta + \delta^2] = n^2 -1$.

So $x^2 -[x^2] = -2\delta + \delta^2 + 1 \approx 0$.

Or more formally $\lim_{x\to n+\in \mathbb Z; n\ne 0} x^2 -[x^2] = 0$ but $\lim_{x\to n-\in \mathbb Z; n\ne 0} x^2 -[x^2] = 1$.

We can see we will also have the same if $x$ is near a square root of an integer (other than $0$).

Answer 4

$$f(x)=x^2-[x^2] \implies f(n)=0$$ $$\implies f(n^{+})=\lim_{h \to 0} f(n+h)=\lim_{h \to 0}(n+h)^2-[(n+h)^2]$$ $$=\lim_{h \to 0} (n^2+h^2+2nh-[n^2+2nh+h^2])=\lim_{h \to 0}(n^2+h^2+2nh-n^2)=0$$ Next, $$f(n^-)=\lim_{h \to 0} (n-h)^2-[(n-h)^2]=\lim_{h\to 0} n^2-2nh+h^2-[n^2-2nh+h^2]=\lim_{h \to 0} n^2-2nh+h^2-n^2+1=1$$ As $f(n^+) \ne f(n^-)$, $f(x)$ is discontinuous at every integer.