Which of the following is true about $f(x)$?

Question

If $f(x)=x+\sin x$, then which of the following is true about $f(x)$?

$1.f(x)$ is uniformly continuous on $\mathbb{R}$.

$2.f(x)$ has bounded variation on $\mathbb{R}$.

$3.f(x)$ does not have bounded variation on bounded intervals of $\mathbb{R}$.

My attempt:Since $f'(x)=1+\cos x$ is bounded on $\mathbb{R}$ so $f(x)$ is uniformly continuous on $\mathbb{R}$

$3$rd is false because $$TV(f)_a^b=\int_a^b|f'(x)|dx\leq2(b-a)<\infty$$ Is this argument correct?

About $2$nd I have no idea.

Thanks

Answer 1

$\int_{\mathbb{R}}|1+cosx|\geq\sum_{k=1}^{\infty}\int_{-\pi+2k\pi}^{\pi+2k\pi}|1+cosx|dx\geq\sum_{k=1}^{\infty}\int_{-\pi+2k\pi}^{\pi+k2\pi}1dx=+\infty$

Answer 2

To show a function is not a Bounded variation, especially this function has a valid derivative defined, then you can just compute the $L^1$ norm of the derivative. Typically, we have $W^{1,1}(I)\subset BV(I)$ with $I\subset R$ is bounded or unbounded. Moreover, you could take the following definition, it is also in wiki that you suggest, $$\|u\|_{BV(R)}= \sup\left\{\int_R u\,\phi'\,dx,\,\,\phi\in C_c^1(R),\,\,|\phi\|\leq 1\right\}$$ Noticing that if your function $u$ is actually have a weak derivative, in your case you have derivative in classical sense is even better, you can integration by parts and have $$\int_R u\phi'\,dx=-\int_R u'\,\phi\,dx$$ and hence $$\|u\|_{BV(R)}= \sup\left\{\int_R u\,\phi'\,dx,\,\,\phi\in C_c^1(R),\,\,|\phi\|\leq 1\right\}=\sup\left\{-\int_R u'\,\phi\,dx,\,\,\phi\in C_c^1(R),\,\,|\phi\|\leq 1\right\}= \int_R |u'|dx$$ In your case, the last term is infinite and hence it is not $BV$.

Be careful that in general a function is not $W^{1,1}$ can be $BV$, just take $\chi_{[0,1]}$. It has bounded variation in $R$ but it is not $W^{1,1}(R)$