Unbounded function of bounded variation (in $\mathbb R^d$, $d>1$) I have seen this stuff tons of times, but every time I see it I got stuck. Today is the right day to clean my ideas once and for all. 
It is very well known that if $f \colon [a,b] \to \mathbb R$ is a function of (pointwise) bounded variation, then it is bounded (there are a lot of questions about this point, see for instance Does bounded variation imply boundedness ).  

Question. Is the above true if we consider BV functions $f \colon \Omega \to \mathbb R$, being $\Omega \subset \mathbb R^d$ an open set? 

Here BV means in the distributional sense, i.e. the gradient $Du$ is a finite Radon measure. 
The answer - I am pretty sure - is "No". Could you please recall me how one can build a counterexample? 
Thanks.
 A: Recall that the Sobolev space $W^{1,1}(\Omega)$ is a (proper) subspace of $BV(\Omega)$. If $x_0 \in \Omega$, then $|x - x_0|^{-\alpha}$ is unbounded and belongs to $W^{1,1}(\Omega)$ provided $\alpha < d - 1$. (Tinker with this function if $\Omega$ is unbounded)
A: (i) How I figured this out. (ii) An example. (iii) A simpler example. (iv) Then a minute later I realize it should have been obvious, even without looking at a single explicit example.
i. A minute ago I had no idea. I say let's think about the plane. If there exists $c$ so that $$\sup_{x,y}|f(x)-f(y)|\le c||Df||_1\quad(*)$$then yes, if not then we can build a counterexample.
So I start with the simplest example I can think of: In the plane, $$f_r(x)=(r-|x|)^+.$$A little conical pyramid thingie. On the one hand $f_r(0)-f_r(x)=r$ if $|x|=r$. On the other hand $|Df|=1$ in the disk of radius $r$ and vanishes elsewhere, so $||Df_r||_1=\pi r^2$.
So (*) is violated if $r\to0$, so there should be a counterexample.
ii. And then sure enough, just let $f=\sum_{n=1}^\infty f_{1/n}$. This blows up at the origin but has $||Df||_1<\infty$.
iii. Then looking at how that example works it's clear there should be something simpler. For example $$f(x)=(\log(1/|x|)^+,$$still in the plane. Blows up at the origin. But $|Df(x)|=1/|x|$ for $|x|<1$, vanishes elsewhere, so $||Df||_1<\infty$.
iv. Smack forehead. It's obvious that (*) cannot hold, by homogeneity! Say $t>0$ and $g(x)=f(tx)$. The left side of (*) is not changed, but it's clear that the right side is multiplied by $t^\alpha$ for some $\alpha\ne0$. Even if we have no idea what $\alpha$ is this makes it clear (*) cannot hold.
Comment This homogeneity thing can be very useful, both for seeing that some inequality cannot hold and for improving an inequality to the "right" inequality. At one time I thought the second point was sort of "mine", then I nnoticed Terence Tao making a big deal of it - see his blog for various examples.
My favorite example deals with Fourier transforms, let's say in one variable. Suppose you've shown that $$||\hat f||_1\le c(||f||_2+||f'||_2).$$By considering $g(x)=f(tx)$, show that in fact $$||\hat f||_1\le c\left(||f||_2\,||f'||_2\right)^{1/2}.$$
