Not bounded function in $W^{1,1}(\mathbb{R})$

Question

I have seen in page 627 (Exercise 1) of Taheri's book: Function spaces and partial differential equations that if $n\geq2$, then $W^{1,n}(\mathbb{R}^n)$ does not embed into $L^\infty(\mathbb{R}^n)$. But, can we say when $n=1$? Is there any not bounded function in $W^{1,1}(\mathbb{R})$?

Answer 1

For $W^{1,1}(\mathbb R)$ the linked proof simplifies a little, for any $f\in C^1_c$, we have

$$|f(x)| = \left| \int_{-\infty}^x f'(t) dt \right| \le \int_{-\infty}^\infty |f'| \le \| f\|_{W^{1,1}}$$ then use the density of $C^1_c$ in $W^{1,1}$: if $f_n \in C^1_c$ converge to $f$ in $W^{1,1}$, the above inequality implies that $f_n$ is Cauchy and hence converges in $L^\infty$.

The result is true for all $W^{1,p}(\mathbb R)$, see the duplicate link.