Let $f, g : [0, 1] \to \mathbb{R}$ be defined as follows: $f(x) = x^2 \sin (1/x)$ if $x = 0$, $f(0)=0$
$g(x) = \sqrt{x} \sin (1/x)$ if $x = 0, g(0) = 0$.
Which are functions of bounded variations?Every polynomial in a compact interval is of BV?
Could any one just tell me what is the main result to see whether a function is of BV? Derivative bounded?
Yes, bounded derivative implies BV. I explained this in your older question which condition says that $f$ is necessarily bounded variation. Since $f$ has bounded derivative, it is in BV.
A function with unbounded derivative could also be in BV, for example $\sqrt{x}$ on $[0,1]$ is BV because it's monotone. More generally, a function with finitely many maxima and minima on an interval is BV.
But $g$ has unbounded derivative and infinite number of maxima and minima on an interval. In such a situation you should look at the peaks and troughs of its graph and try to estimate the sum of differences $\sum |\Delta f_i|$ between them. It is not necessary to precisely locate the maxima and minima. The fact that $$g((\pi/2+2\pi n)^{-1})=(\pi/2+2\pi n)^{-1/2},\quad g((3\pi/2+2\pi n)^{-1})=-(3\pi/2+2\pi n)^{-1/2}$$ gives you enough information about $g$ to conclude it is not BV.
