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Knowing the definition is enough, here: the constraining condition is that the derivative of $f$ be in $L^4$, i.e. $x^{\alpha-1}\in L^4(0,1)$ or $4(\alpha-1)>-1$

(1) No, not an equality. Look at the characterization in terms of Fourier transforms (*). (2) It depends on $s$. For $n=1$ and $\frac12\le s<1$ the completion is a quotient of the semi-Hilbert space …

The obvious sufficient condition for $v\mapsto f(v)$ to map $H^1(\Omega)$ to itself ($\Omega$ bounded) is $f$ globally Lipschitz, i.e. $f'\in L^\infty(\mathbb R)$. Add $f(0)=0$ for general $\Omega$. …

I think the most natural assumption is that $D$ is a bilipschitzian image of a smooth domain $D_0$, since a change of variables $A:D_0\to D$ with $a|x-y|\le|A(x)-A(y)|\le b|x-y|$ ($a>0$) preserves $H^ …

Let $n=1$, $p=2$ for a partial answer: characterizing $1$-periodic functions $f\in L^1(0,1)$ such that $I(f):=\int_0^1 h^{-1}[\int_0^1(f(x+h)-f(x))^2\ dx]\ dh<\infty$. Let $\hat f(n):=\int_0^1 e^{2\p …

There is an explicit operator that maps $L^2$ isometrically onto $H^{s,\mu}_2$ :$$I_{s,\mu}u(x)=(1+|x|^2)^{-\mu/2}(I-\Delta)^{-s/2}u(x)$$The (inverse) Fourier transform $k_s(x)$ of $(1+|\omega|^2)^{-s …

Let $\Omega=[-1,1]$ and $u(x)=u(-x)=0$ for $x\in[2^{-2k+1},2^{-2k+2})$ and $u(x)=u(-x)=2^{-k}$ for $x\in[2^{-2k},2^{-2k+1})$, $k=1,\ldots,n,\ldots$. Doesn't this $u$ belong to your space $SBV\cap L^\ …

If you insist on "for all $t$ " as distinct from "for almost every $t$ ", one possible space is $C^1([0,\infty);H^{-1/2})\cap C^0([0,\infty);H^{1/2})$. A possible extension is the unique harmonic func …

Let $\frac12<s<1$, let $u\in H^s(\mathbb R)$, then also $u\in C_0$, and let $f$ be Lipschitz on $[-||u||_\infty \ ,+||u||_\infty]$ with $|f(x)-f(y)|\leq M|x-y|$. Then $$\int\int \frac{|f(u(t))-f(u(t') …

Hint at a partial answer to the revised question, extended to $p\in (0,\infty)$: for $\frac12<s<1$, $p\geq 1$ is necessary and sufficient. The "if" part is straightforward using the double integral …

"The" completion is not always a space of functions, for $N=1$ or $2$ for example it is a quotient $D^{-2}L^2/P_1$ (equivalence classes of functions $u\in H^2_{loc}$ with $\partial_i \partial_j u\in L …

No. Take $n=2$, $x=0$ (and $\alpha=1$, wlog). Define $f$ in polar coordinates as $f(r,\theta)=r^\beta g_r(\theta)$ where $\int_0^{2\pi} g_r(\theta)\ d\theta=1$ and $\lim_{r\to 0}g_r(0)=\infty$, which …

If it is true that $\int e^u<\infty$ whenever $u\in W^{1,3}$, for $d=3$, and $W^{1,3}$ is not a subset of $L^\infty$, then, since $W_0^{1,3}\subset H_0^1$, there certainly exist essentially unbounded …