If $ \Omega $ is a bounded domain, is a $ BV(\Omega) $ function also $ L^\infty(\Omega) $? Let $ \Omega $ be a (non-empty) bounded domain. The space of functions of bounded variation is defined by 
$$ BV(\Omega) = \{ u\in L^1(\Omega) \mid \|u\|_{TV} < \infty \} $$
where 
$$ |u|_{TV} = \sup\left\{ \int_\Omega u\operatorname{div}\phi\,dx \mid \phi \in C_c^1(\Omega), |\phi|\leq 1 \right\}. $$
Does $ u \in BV(\Omega) $ imply that $ u \in L^\infty(\Omega) $, i.e. is it the case that $ BV(\Omega) \subset L^\infty(\Omega) $? 
In case it isn't true, then $ \operatorname{ess}\sup_{x\in\Omega} |u(x)| = \infty $. $ u $ being infinite everywhere whould make the TV-norm infinite, which is a contradiction, so $ u $ must be finite somewhere, say at $ y $. I feel like this jump from finiteness at $ y $ to whereever the essential supremum is attained should make the TV-norm infinite again, thus contradicting my initial assumption. However, I have no clue so far about how to go about it technically. Maybe my gut feeling is just plain wrong?
 A: No, as if $n \geq 2$ this also fails in $W^{1,1}(\Omega).$ A standard counterexample is to take $\Omega = B(0,1) \subset \mathbb R^n$ and consider the function $f: \Omega \rightarrow \mathbb R$ defined by,
$$ f(x) = \left[ \log \log(1+ |x|^{-1}) \right]^{1/n}. $$
One can check that $f \in W^{1,1}(\Omega),$ but it is evidently unbounded.

Edit: Upon revisiting this question, it struck me that this counterexample is far more complicated than necessary (it gives an unbounded function in $W^{1,n}$). The essential observation necessary is that $W^{1,1}(\Omega) \subset BV(\Omega)$ contains unbounded functions, which can more simply be established with the function,
$$ f(x) = |x|^{-1/2}. $$
This lies in $W^{1,1}(B(0,1))$ for $n \geq 2.$
Also the result is true if $n=1.$ An easy way to show this is to establish an estimate of the form,
$$ \lVert u \rVert_{L^{\infty}(\Omega)} \leq C \lVert u \rVert_{BV(\Omega)} $$
for all $u \in C^1(\Omega) \cap BV(\Omega).$ This is straightforward if you use the alternative definition of BV in one dimension, which is equivalent under the $C^1$ assumption. Then the result follows by density.
