Let $\Omega \subset \mathbb{R}^n$ be a bounded Lipschitz set, and let $W^{k,p}(\Omega)$ denote the usual Sobolev space with $k \in \mathbb{N}$ being the order of the derivatives and $p \in [1, \infty)$ the rate of integrability. I know that there exists a result of this type: fix $\varepsilon >0$, then $$\Vert f\Vert_{W^{k-1,p}(\Omega)} \leq \varepsilon \Vert f\Vert_{W^{k,p}(\Omega)} + c(\varepsilon) \Vert f\Vert_{L^1(\Omega)}$$ for any function $f \in W^{k-1,p}(\Omega)$, where the constant $c(\varepsilon)>0$ depends on $\varepsilon$ but not on $f \in W^{k-1,p}(\Omega)$.

Do you know how to prove it, or have any reference to a book/paper where I can find the proof of this result?


1 Answer 1


This can be proved by contradiction. Let $\epsilon > 0$ be given, and $(f_j \mid j \in \mathbf{N})$ be a sequence of functions in $W^{k,p}(\Omega)$ with $\lvert f_j \rvert_{k-1,p} \geq \epsilon \lvert f_j \rvert_{k,p} + j \lvert f_j \rvert_{0,1}$. Rescale these functions to have $\lvert f_j \rvert_{k-1,p} = 1$ for all $j$. Then $\lvert f_j \rvert_{k,p} \leq \epsilon^{-1}$ and one can extract a subsequence that converges weakly in $W^{k,p}(\Omega)$ and strongly in $W^{k-1,p}(\Omega)$. Let $f \in W^{k,p}(\Omega)$ be their limit. At the same time $\lvert f_j \rvert_{0,1} \leq j^{-1}$, so that $f_j \to 0$ in $L^1(\Omega)$. Therefore $f = 0$, but this is absurd because the convergence in $W^{k-1,p}(\Omega)$ means that $\lvert f \rvert_{k-1,p} = 1$.

  • 1
    $\begingroup$ Just a remark. You are supposed to contradict : $\forall \epsilon >0, \exists C$ such that blah blah. And you start with $\forall \epsilon>0$ : ) (it is fine since you are proving more, of course). $\endgroup$
    – username
    Commented Apr 30, 2021 at 11:30
  • $\begingroup$ @Leo Moos thanks! It was easier than expected. $\endgroup$
    – vampip
    Commented Apr 30, 2021 at 13:20

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