I am interested in the optimization problem known as "analysis regularization": 


$$ {\rm argmin}\_{x \in \mathbb{R}^{p}}\frac{1}{2}\|y - Ax\|\_2^2 + \lambda \|D^T x\|\_1,$$
where $y \in \mathbb{R}^n$, $A \in \mathbb{R}^{n \times p}$, and $D \in \mathbb{R}^{p \times d}$. [In this paper](https://arxiv.org/abs/1109.6222), they state that the set of (global) minimizers to the above is nonempty and compact if and only if 
$$ {\rm kern}(A) \cap {\rm kern}(D^T) = 0.$$


How would one be able to prove this? They claim it to be a classic result, but I am not able to show it myself after a few attempts. I am unfamiliar with this area, so I may be overlooking a simpler argument. 


Any references or pointers to relevant material would be helpful. 



Existence of analysis regularization solution
Nothing is stated concerning $\lambda$. I will assume that (as always) the regularization parameter $\lambda >0$. Otherwise the whole thing does not really make sense.


The expression to be minimized is a convex function in $x$, so the set of minimizers is convex. Also every local minimizer is a global minimizer. 


If the intersection of the kernels is empty, then it is not too hard to show that
$$ \lim\_{r\to\infty} \left( \min\_{\|x\|\_2 = r} \left\{\frac{1}{2}\|y-Ax\|\_2 + \lambda\|D^Tx\|\_1\right\}\right) = +\infty . \tag{\*}$$
Assume this is not the case. Then choose a sequence $x\_k\to \infty$, so that the function stays bounded on that sequence. Then $x\_k/\|x\_k\|$ may be assumed to converge, say to $x^\*$. Then $D^T x^\*=0$. But then $x^\*$ is not in the kernel of $A$ and this gives a contradiction. Now (\*) implies that the set of minimizers is nonempty and compact.


If the kernels intersect, then we can first consider the problem on a complementary subspace to the intersection of the kernels. By the previous argument this restricted problem has a minimizer.
If you have a minimizer $\bar x$ and there are points $z$ such that $Az=D^Tz=0$, then of course $\bar x + \alpha z$ is a minimizer for any $\alpha \in \mathbb{R}$. So if the kernels have nontrivial intersection the set of minimizers is not compact. 


