<s>[INST] $f(x)$=$\left\lfloor { x }^{ 2 } \right\rfloor -\left\lfloor x \right\rfloor ^{ 2 }$ is discontinuous for all integer values of x except only at x=1 How to prove that $f(x)$=$\left\lfloor { x }^{ 2 } \right\rfloor -\left\lfloor x \right\rfloor ^{ 2 }$ is discontinuous for all integer values of x except only at $x=1$ ?
Ya,even I used intuition at the first go taking some trial values and checking.Then I plotted the graph on wolfram alpha.But if you can think of some other rigorous proof for this one,let me know.Thanks!
 [/INST] The approach here is to find formula that are equivalent to $f$ locally. The idea being that it is simpler to spot the discontinuity from the locally
equivalent formula.
Note that $\lfloor x \rfloor = 1_{[0,\infty)}(x)-1$ for $|x|<1$. In particular,
for small positive $x$, we have $\lfloor x \rfloor =0$. Also note that
$\lfloor x \rfloor^2 = - \lfloor x \rfloor$ for small $x$.
Consider $f(n+\epsilon)$ for integer $n$ and small $\epsilon$.
If $n=0$, we have $\lfloor \epsilon^2 \rfloor = 0$.
Hence
$f(\epsilon) = -\lfloor \epsilon \rfloor^2 = \lfloor \epsilon \rfloor$ for small $\epsilon$, and so $f$ is discontinuous at $x=0$.
For $n \neq 0$ and sufficiently small $\epsilon$,
we have $\lfloor (n+\epsilon)^2 \rfloor = n^2 +\lfloor n \epsilon + \epsilon^2 \rfloor= n^2 + \lfloor n \epsilon\rfloor$, and
$\lfloor n+\epsilon\rfloor = n + \lfloor \epsilon\rfloor$. Hence
$\lfloor n+\epsilon\rfloor^2 = n^2 + 2n \lfloor \epsilon\rfloor + \lfloor \epsilon\rfloor^2$,
and so
$f(n+\epsilon) = \lfloor n \epsilon\rfloor -2n \lfloor \epsilon\rfloor - \lfloor \epsilon\rfloor^2 = \lfloor n \epsilon\rfloor -(2n-1) \lfloor \epsilon\rfloor$.
If $n<0$ this gives
$f(n+\epsilon) =\lfloor- \epsilon\rfloor +(1-2n) \lfloor \epsilon\rfloor$, from which we see that $f(n) = 0$ and $f(n+\epsilon) =-1$ for small positive $\epsilon$.
If $n>0$, we have $f(n+\epsilon) =\lfloor \epsilon\rfloor +(1-2n) \lfloor \epsilon\rfloor = 2(1-n) \lfloor \epsilon\rfloor$, hence $f$ is continuous
for $n = 1$, otherwise $f(n) = 0$ and $f(n+\epsilon) = -2(n-1)$ for small
negative $\epsilon$.
</s>