$L^1$ Estimates involving bi-Laplacian The following inequality can be shown to be true in the cases $p>1$:

If $n\ge 5$, $\frac{1}{q} = \frac{1}{p} - \frac{4}{n}$, then there exists $C_{p,q,n}>0$ such that, for every $f \in C^{\infty}_c$,
$$ \|f\|_q \le C_{p,q,n} \|\Delta(\Delta f))\|_p$$

So, this one is easily done by writing explicitly who's the fundamental solution (in dimension $n\ge 5$) for the inhomogeneous biharmonic problem $\Delta^2 f = g$ and comparing with some Riesz potential.
The thing is: is it true for the critical case $p=1$?
Some helpful ideas:
1 - Use, in some smart way, the Gagliardo-Nirenberg-Sobolev inequality
$$ \|f\|_q \le C \|\nabla f\|_p $$
Which holds when $\frac{1}{q} = \frac{1}{p} - \frac{1}{n}$, and $\infty > p \ge 1$
2 - Try to use the $W^{2,p}$ boundedness for the Laplacian, i.e., that
$$ \|D^2 f \|_p \le C_p \|\Delta f\|_p$$
Which holds when $ 1 < p < \infty$
Any good ideas?
 A: No, this fails in the critical case. As you know, the fundamental solution is $\phi(x)=|x|^{4-n}$. This fails to be in $L^q$ when $q=n/(n-4)$. This is not yet an example because $\Delta^2 \phi$ is a point mass, not an $L^1$ function. But you shouldn't expect the convolution of $\phi$ with a general $L^1$ function to improve integrability in a quantitative way. Functions of class $L^1$ can come as close to being a point mass as we want. 
More concretely: in $n=5$ dimensions, take $f(x) = r^{-1}(\log r)^{-1/5}$ where $r=|x|$. Note that $f\notin L^5$. Compute the Laplacian in the usual way: for radial functions $g$ it is 
$r^{-4}(r^4 g_r )_r $. Sage tells me that 
$$
\Delta f(x)  = {-\frac{{\left(10 \, \log\left(r\right) - 3\right)} {\left(5 \, \log\left(r\right) + 2\right)}}{25 \, r^{3} \log\left(r\right)^{\frac{11}{5}}}}
$$
and 
$$
\Delta^2 f = 
{-\frac{6 \, {\left(125 \, \log\left(r\right)^{3} + 125 \, \log\left(r\right)^{2} - 110 \, \log\left(r\right) - 176\right)}}{625 \, r^{5} \log\left(r\right)^{\frac{21}{5}}}}
$$
The singularity at $0$ is $r^{-5}(\log r)^{-6/5}$, therefore $\Delta^2 f \in L^1$.
Of course this example is not $C^\infty $ smooth, but if the estimate was true for smooth functions, it would hold in general by approximation.
