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(1) Yes, your argument is correct. The fact that composition with $T^{-1}$ preserves Sobolev classes also needs to be proved, but the proof is immediate from consideration of what this composition does to Cauchy sequences (wrt $W^{1,p}$ norm) of smooth functions. (2) Yes, and this generalization is one of fundamental results for the theory of Sobolev ...

3

By the fundamental theorem of calculus, we have that $$u_\delta(x)=u_\delta (c)+\int_c^xu_\delta'(t)dt,$$ Can you conclude now?

3

No, not on $\mathbb R^n$. Take a smooth function $\phi$ with support contained in the unit ball. Consider the sequence $f_n(x) = \phi(x-3ne_1)$ where $e_1$ is a basis vector. This is a sequence of bumps going into infinity. Translation by $3ne_1$ does not change either $L^2$ norm or $W^{-1,2}$ norm (I write $W^{-1,2}$ for $(W_0^{1,2})^*$ here). Since the ...

2

Consider the interplay between Sobolev functions and absolutely continuous functions. Assume without loss of generality that $u,v\in W^{1,1}(\Omega)$, $uv\in L^1(\Omega)$ and $uDv+vDu\in L^1(\Omega)$. Moreover, let's assume that $u,v$ are absolutely continuous for a.e. segment of line parallel to the coordinate axes and contained in $\Omega$. Therefore, ...

2

Near $e_2$, you have the function approaching $1$ along the segment $\ell$; yet it's zero on the boundary of the disk. This will make $|\nabla u|$ blow up like $|x-e_2|^{-1}$ in between, guaranteeing $u\notin W^{1,p}$ for $p\ge 2$. I think $u\in W^{1,p}$ for $1\le p<2$. A more transparent description of (some version of) your function could be: Let ...

2

The only difference between $H^s_{\rm loc}(\mathbb R^n)$ and $H^s(\mathbb R^n)$ is the decay at infinity (of the function itself, and appropriate derivatives). So, if you want functions to not belong to $H^s(\mathbb R^n)$, make them not decay at infinity. Adding a nonzero constant will do the trick. Or add a polynomial, etc. Yes: the space of smooth ...

2

This answer has a couple of parts. Part I: A Computation Consider the simplest cast in $\mathbb{R}$. Let $\chi$ be the indicator function of $[-1,1]$. You have that $\|\Phi_n * \chi\|_{L^2} \approx 2^{- \frac{n}2}$. So in which Besov spaces does $\chi$ belong? First clearly we need $\alpha \leq \frac12$. When $\alpha = \frac12$ you have that $q = \infty$ ...

1

To prove that $u_{|N}\in W^{1,p}_0(N)$, it is sufficient to prove that $u^+_{|N}\in W_0^{1,p}(N)$. Your proof can be divided in three steps: I - Consider the sequence $\psi_n=\max\{u^+-1/n,0\}$. Once $u$ is continuous, for each point $p\in \partial N\cap \Omega$, there is a neighbourhood $U_{n,p}$ such that $\psi_n(x)=0$ for $x\in U_{n,p}$, however, this ...

1

If the domain of $A=-\Delta$ is chosen so that it is positive and selfadjoint, then the spectral theorem for $A$ is $$Au = \int_{0}^{\infty} \lambda dE(\lambda)u$$ with $u \in \mathcal{D}(A)$ iff $\int_{0}^{\infty}\lambda^{2}d\|E(\lambda)u\|^{2} < \infty$. More generally, $u \in \mathcal{D}(A^{s})$ for some $0 < s < \infty$ iff ...

1

The problem is that $C_0^\infty(\mathbb{R}^n)$ is not dense in $W^{1,\infty}(\mathbb{R}^n)$. Particularly, a function in $W^{1,p}(\mathbb{R}^n)$ for $1\leq p<\infty$ must decay "far outside". As an example, choose $v=1$ which is clearly in $W^{1,\infty}(\mathbb{R}^n)$, but does not have a zero trace or is in the completion of $C_c^\infty(\mathbb{R}^n)$ ...

1

The space $W_0^{1,\infty}(\Omega)$ consists of functions that tend to $0$ at the boundary of $\Omega$; the limit is understood in the classical way because $W^{1,\infty}$ functions have a continuous representative. See the discussion here. When $\Omega=\mathbb R^n$, the role of boundary is played by the point at infinity. Therefore, the zero extension of ...

1

For (b). Since $\lambda_n \to +\infty$, we have $\frac{1}{\lambda_n} \to 0$. Hence, from (A.4) and due to the fact that $$\frac{2}{|\Omega|}\int_{\Omega}|w_n|^{2/\alpha_n} \geq 0,$$ we get $$0 \leq \int_{\Omega}\left|\nabla w_n\right|^2 \leq \int_{\Omega}\left|\nabla ... 1 This is not true. The H^r norm controls the supremum of f if and only if 2r>n where n is the dimension of the space (n=1 in your case), by the Sobolev-Morrey embedding. In the borderline case r=1/(2n) the embedding fails. This is typical of embeddings. An example is easier to give in the periodic setting, on \mathbb R/(2\pi \mathbb Z). Let ... 1 Such a function could also be constructed the lines of Boundedness of functions in W_0^{1,p}(\Omega) if one chooses the functions f_i to be continuous which is clearly possible. For any x \in B, there is a neighborhood which intersects the support of finitely many of the f_i, hence the limit f is continuous at x. And since f is not bounded on ... 1 As far as I know, in general case the answer is negative. Even for the problem$$ u'_t = \mbox{div}(A(x,t) \nabla u) + f(x,t)  (with suitable initial-boundary conditions) the Strong maximum principle is unknown if $A(x,t)$ is only continuous (not smooth). (In this case you cannot use the classical technique of comparison with some "good" function, since ...

1

Consider $\Omega_1 = (0,2)$ and $\Omega_2 = (1,3)$. Then the constant functions $x\mapsto 1$ (one the respective intervals) are in $H^1(\Omega_i)$, but their sum (if extended by 0, otherwise, what do you mean by the sum?) is not continuous (has no continuous representative), hence is not in $H^1(\Omega_1 \cup \Omega_2)$.

1

If $\phi$ is a $C^\infty$ function with compact support then the convolution $u*\phi$ is also smooth, and $D(u*\phi)=(Du)*\phi=0$. So $u*\phi$ is locally constant by the usual calculus argument. Now let $\phi$ be a standard mollifier and let it converge to a delta, so that $u*\phi$ converges to $u$ a.e. (or in $L^1_{\text{loc}}$ if you prefer). So $u$ is ...

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Hint: You showed that if the problem is solvable then $\int_{\Omega}fdx=0$, suppose that $\int_{\Omega}fdx=0$ and show that the problem is solvable you need to proof a existence theorem, or use one if you know some. Edit: try to use Fredholm alternative for every $f ∈ L^2(Ω)$the boundary value problem (∗)\$ \triangle u = f \,\,in\,\, Ω,\,\,\, ...

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