I am looking to establish a result of the following form, or to find counterexample where this fails.

If $1 \leq p < \infty$ and $u \in L^p(\mathbb R^d)$ has weak derivatives up to second order with $D^2u \in L^p(\mathbb R^d),$ then $u \in W^{2,p}(\mathbb R^d).$

This result is true for $p \geq 2,$ which is a standard result. The solution is presented in this question, by proving that there is a constant $C$ depending on $p,d$ such that,

$$ \lVert Du \rVert_{L^p(\mathbb R^d)} \leq C\left(\lVert u \rVert_{L^p(\mathbb R^d)}\right)^{1/2}\left(\lVert D^2u \rVert_{L^p(\mathbb R^d)}\right)^{1/2}, $$

For all $u \in C^{\infty}_c(\mathbb R^d)$ and extending by mollification. General case $1 \leq p < \infty$ also holds on bounded domains by appealing to compactness theorems, but that obviously doesn't generalise to this case.

My progress so far: I first try prove it for $p=1,$ in hope that some kind of interpolation result would give $1<p<2.$ To do this it suffices to prove an inequality of the form,

$$ \int_{\mathbb R} |\frac{\partial f}{\partial x_i}(x_1,\dots,\tilde x_i,\dots,x_d)| d\tilde x_i \leq C\left( \int_{\mathbb R} |f(x_1,\dots,\tilde x_i,\dots,x_d)| d\tilde x_i + \int_{\mathbb R} |\frac{\partial^2 f}{\partial x_i^2}(x_1,\dots,\tilde x_i,\dots,x_d)| d\tilde x_i \right), $$

for all $f \in C^{\infty}_c(\mathbb R^2),$ $1 \leq i \leq d$ with $C = C(d)$ some constant. Indeed once we have such a result, we can integrate in the remaining variables to deduce the result. So without loss of generality, we can restrict to the case $d=1$ and seek an inequality of the form,

$$ \lVert f' \rVert_{L^1} \leq C \left( \lVert f \rVert_{L^1} + \lVert f'' \rVert_{L^1}\right). $$

Moreover since we have assumed $f'$ is continous, we can partition $\mathbb R$ into a disjoint union of intervals $[a_i,b_i)$ covering $\mathrm{supp}(f)$ such that $f$ is monotone on $[a_i,b_i).$ On these intervals it suffices to find a constant $C$ independent of $a_i,b_i$ such that the above inequality holds on $[a_i,b_i],$ with a constant $C$ independent of $a_i,b_i.$

This is where I'm stuck, I'm not sure if this can be proven.


It turns out this result is a special case of the Gagliardo–Nirenberg interpolation inequality, which states the following.

Let $d \in \mathbb N,$ $1 \leq p, q, r < \infty$ and $m<j \in \mathbb N$ such that, $$ \frac1p = \frac jd + \left(\frac1r - \frac md\right)\alpha + \frac{1-\alpha}q, $$ for some $\alpha \in (j/m,1).$ Then if $u \in L^q(\mathbb R^d)$ such that $D^mu$ exists (weakly) and is in $\mathbb L^r(\mathbb R^d),$ then $D^ju \in L^p(\mathbb R^d.$ Morevoer there is a constant $C = C(d,p,q,r,m,\alpha)$ such that, $$ \lVert D^j \rVert_{L^p(\mathbb R^d)} \leq C \lVert D^mu \lVert_{L^r(\mathbb R^d)}^{\alpha} \lVert u \rVert_{L^q(\mathbb R^d)}^{1-\alpha}.$$

Indeed taking $p=q=r,$ $j = 1$ and $m=2$ and $\alpha = 1/2$ gives the desired result. A relatively clean proof of this result is given in chapter 13.3 of "Partial Differential Equations III: Nonlinear Equations" by Michael Taylor.


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