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If you have a compact subset $K$ of some open set $\Omega$ I understand your question as : can I find a smooth set $K'$ such that $K\subset K' \subset\Omega$ ? The natural strategy would be to use the …

I am giving another answer which does use the Vitali-covering lemma. Your assumption can be seen as a way to ensure a uniform $\mathrm{L}^1$ approximation by a specific regularizing kernel ; by the wa …

Here's another way to get the same conclusion, maybe a tiny bit more elementary than the answers above. First, the question can be localized replacing $T$ with $T\theta$ for some arbitrary bump functi …

A few years ago I asked a reference about the Hardy maximal function on the flat torus. Mateusz Kwaśnicki kindly answered in a comment, and confirmed my conviction that basically everything which is k …

Thanks Daniele and Willie for these nice answers. Willie : I tried this doubling variable thing but got stuck : I was writing $|h(x+y)|$ as the product of two elements respectively in $L^\infty_y(L^{p …

I am searching for a reference for the (sharp) Hardy maximal function on the torus $\mathbb{T}^2:=\mathbb{R^2}/\mathbb{Z}^2$, for instance I would need result result of the following type : if $g\in H …

You can also use that for $g\in L^p(\mathbf{R})$, \begin{align*} \|\tau_h g-g\|_p\lesssim |h|, \end{align*} characterizes elements $g$ of $W^{1,p}(\mathbf{R})$ for $p>1$ and elements of $BV(\mathbf{R} …

Withtout loss of generality, we can assume $p=0$ and $f(0)=0$. Also, the problem is purely local : we can assume that $f$ and all the functions $f_\epsilon$ are compactly supported in the unit ball …

For $1\leq p,q \leq \infty$ such that $\frac1p +\frac1q\geq 1$, Young's inequality states $\|f\star g\|_r\leq \|f\|_p\|g\|_q$ (we work on $\mathbf{R}^d$ here), where $1+\frac1r = \frac1p+\frac1q$. Equ …

For $k=1$, the proof works the same on $\mathbb{R}$ ; you only need to check that compactly supported functions (no smoothness here) are dense in $H^1(\mathbb{R},\rho(x)dx)$ and this can be done using …

Aleksei already answered, but here is another way to see it. Since $\Omega$ is bounded, any bound in the $L^q(\Omega)$ norm leads tightness in the $L^p(\Omega)$ norms with $p<q$ : this is a consequenc …

I think in dimension 1 you won't be able to produce a counterexample, see §4 of V. V. Zhikov, "Weighted Sobolev spaces", Mat. Sb., 189:8 (1998), 27–58; Sb. Math., 189:8 (1998), 1139–1170 For some co …

No, this is is a classical counterexample: $f:t\mapsto t^{-\ln(t)}\sin(2\pi\ln(t))$, it is continuous and has vanishing moments. Indeed, for any natural integer $n$, the change of variable $u=\ln(t)$ …

The unit cube has lipschitz regularity so yes, you have an extension operator. Note that your extra condition on the compactness of the support can be ensured easily once you have an extension (just m …