Show that there is no $C\gt 0$ such that $\|u\|_{L^1}\le C\|Du\|_{L^1}$ for all $u\in C_c^\infty(\mathbb{R}^2)$.
Earlier I was able to prove that it is true, if you replace the norm on the left hand side by the $L^2$ norm. I know its a special case of the Gagliardo–Nirenberg–Sobolev inequality, but I can't find a counterexample to this one.
Let $u$ be a smooth cutoff, supported on $B(0,1)$, and set $$u_n(x)=\frac{1}{n^2}u\left(\frac{x}{n}\right).$$ Then, the change of variables $x=yn$ shows that $$\int_{\mathbb R^2}|u_n(x)|\,dx=\int_{\mathbb R^2}\frac{1}{n^2}\left|u\left(\frac{x}{n}\right)\right|\,dx=\int_{\mathbb R^2}\frac{1}{n^2}|u(y)|n^2\,dy=\|u\|_1,$$ while $$\int_{\mathbb R^2}|Du_n(x)|\,dx=\int_{\mathbb R^2}\frac{1}{n^2}\left|\frac{1}{n}Du\left(\frac{x}{n}\right)\right|\,dx=\int_{\mathbb R^2}\frac{1}{n^3}|Du(x)|n^2\,dx=\frac{1}{n}\|Du\|_1.$$ So, if such a $C$ exists, we should have that $$\|u\|_1=\|u_n\|_1\leq C\|Du_n\|_1=\frac{C}{n}\|Du\|_1,$$ and letting $n\to\infty$ leads to a contradiction.