Is it true that $f(x)=\Vert Ax-b\Vert_2^2+\lambda\Vert \Phi x\Vert_1$ always attains its infimum? Here, $A:\mathbb{R}^n\to\mathbb{R}^m$ ($m\leq n$) has maximum rank and $\Phi$ is any linear mapping (not necessarily of maximum rank). It seems reasonable that this is true. Denoting by $L=ker(A)\cap ker(\Phi)$, we can always assume $L=\varnothing$. Briefly, if $L\neq\varnothing$, then the function $\tilde f(x+L)=f(x)$ defined on the quotient $\mathbb{R}^n/L$ is well defined and does the job. So, it is enough to assume that $L=\varnothing$. In this case, I'd like to prove that $\lim_{|x|\to+\infty}f(x)=+\infty$. If $\lim_{|x|\to+\infty}f(x)$ is not infinity, then there exists $l\in\mathbb{R}_+$ and $\gamma:[0,+\infty)\to\mathbb{R}^n$ such that $\lim_{t\to+\infty}f(\gamma(t))=l$ and $\lim_{t\to+\infty}|\gamma(t)|=\infty$, where $|\cdot|$ denotes any norm on $\mathbb{R}^n$. Since $f$ is the sum of two non-negative quantities, both of its addenda have to be bounded at infinity, i.e. $\Vert A\gamma(t)-b\Vert_2^2\leq M_1$ and $\Vert\Phi \gamma(t)\Vert_1\leq M_2$, with $M_1,M_2\in[0,+\infty)$. Then, I'm trying to argue exploiting the fact that $\Vert Ax-b\Vert_2^2$ is a semidefinite positive hyper quadric and $\Vert\Phi x\Vert_1$ is basically a function that goes to infinity or remains constant.
Another approach is to show that $f$ attains its infimum along every direction, which is a straightforward computation, but this shall not be sufficient to deduce that $f$ attains its infimum. Unless somehow, the non-negativity of $f$ and its convexity help somehow.
Suppose $L=\{0\}$.
Assume there is a sequence $(x_k)$ and $M>0$ such that $\|x_k\|\to \infty$ and $f(x)\le M$. Define $d_k:=x_k / \|x_k\|$. Dividing the inequality $f(x)\le M$ by $\|x_k\|$ and $\|x_k\|^2$ shows $$ \lambda \|\Phi d_k\|_1 \to0 $$ and $$ \|Ad_k \|_2^2 \to 0. $$ Now $\|d_k\|=1$, so it has a converging subsequence with limit $d$, $\|d\|=1$. Then the above limits imply $d\in L$, hence $d=0$, a contradiction.