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2

The proof looks correct to me, well done! A couple of nitpicks: in the definition of $V \subset \subset U$ you forgot to include that $\overline{V}$ needs to be compact; the very last $\delta$ should be $\delta^p$. The only thing that I would like to add is an explicit definition for $V_k$: $$V_k := \Big\{x \in U : \text{dist}(x, \partial U) > ...


1

The right hand inequality is easy. For the left hand inequality: $$ I_p(h_1 + h_2) \le I_p(h_1) + I_p(h_2) $$ So $$ I_p(h)^p \le (I_p(\tfrac12h + k) + I_p(\tfrac12h - k))^p \le 2^{p-1}(I_p^p(\tfrac12h + k) + I_p^p(\tfrac12h - k)) $$ So if $|h| \le \delta$, $$ I_p(h)^p \le 2^{p-1} v_n^{-1} (\delta/2)^{-n} \int_{|k| \le \delta/2}(I_p^p(\tfrac12h + k) + ...


0

Since any function in $W^{1,p}(I)$ equals the integral of its weak derivative, which is in $L^p(I)$ and hence $L^1(I)$, it is absolutely continuous. Hence any function which is bounded and continuous but not absolutely continuous will be a counterexample. For instance, the Cantor staircase function.


1

I think that a direct proof is possible: Let $(u_k)$ be a Cauchy sequence in $\mathcal{B}$. Then, $(u_k)$ is Cauchy in $L^2(0,T; H^1_0)$ and $(\partial_tu_k)$ is Cauchy in $L^2(0,T; H^{-1})$. Since these are Banach spaces, we conclude that there exists $u\in L^2(0,T; H^1_0)$ and $w\in L^2(0,T; H^{-1})$ such that $$\left\{\begin{align}u_k\to ...


1

Is the regularity of u necessary to deduce this result? No. This proof does not seem to require convergence in a particular space, as in it seems to me that it would work, so long as we had convergence into two Banach spaces [...] I agree with you provided that the Banach space $Y$ is separable reflexive and the Banach space $X$ is continuously ...


0

Here is a possible explicit calculation: Let $A$ be the area of the unit sphere $S(0,1)=\{x\in\mathbb{R}^n;\ |x|=1\}$. Employing polar coordinates (see Theorem 2.49 in the Folland's book) we conclude that $$\int_{B(0,\varepsilon)}1\ dx=\int_0^\varepsilon\int_{\partial B(0,1)} r^{n-1}\ dS\ dr=A\int_0^\varepsilon r^{n-1}\ dr=A\frac{\varepsilon^n}{n}.$$ ...


1

Rearrange the weak formulation and you have $$\int_\Omega\nabla u\cdot\nabla v=\int_\Omega (f-u_t)v,$$ with the necessary assumption that $f\in L^2(0,T;L^2(\Omega))$, $u(t)$ can be considered to be the weak solution of Laplace's equation with right had side in $L^2(\Omega)$, for a.e. $t\in(0,T)$. Elliptic regularity indeed justifies that either $u(t)\in ...


0

Actually the log-Sobolev inequality for Lebesgue measure $\mathscr{L}^n$ is true, up to losing a multiplicative constant. In order to show it we recall that the original log-Sobolev inequality holds for the standard Gaussian measure $\mu=\frac 1{(2 \pi)^{n/2}} e^{-|x|^2/2}$ in the form (slightly different than what stated in the problem): ...


1

The operator $$ Tf = f'-\frac{x}{\sqrt{1+x^{2}}}f $$ is a bounded operator from $X$ to $Y$ because $$ \begin{align} |Tf| & \le |f'|+|f|,\\ |Tf|^{2} & \le |f'|^{2}+|f|^{2}+2|f'||f| \\ & \le 2|f'|^{2}+2|f|^{2} \\ \|Tf\|_{Y}^{2} & \le 2\|f\|_{X}^{2}. \end{align} $$ To ...


0

No, it will not be in $H^2(\Omega)$ in general. A one-dimensional example: let $\omega=(-1,1)$, $\Omega=(-2,2)$, and $f(x) = x^2-1$. Then $f\in H_0^1(\omega)\cap H^2(\omega)$, but its zero extension is not in $H^2(\Omega)$ (for one thing, $H^2\subset C^1$ in this case). A similar example works in higher dimensions: take two concentric balls, with ...


1

Since your region has a simple geometry this should be a consequence of the "absolutely continuous on lines" characterization of $H^1$, e.g. section 4.9.2 in Evans and Gariepy. By the way, the zero extension of a function in $H^1_0(\Omega)$ will belong to $H^1(\mathbb R^n)$ with no restriction whatsoever on the boundary of $\Omega$.


0

See http://home.iitk.ac.in/~tmk/courses/mth656/main.pdf, Theorem 2.4.1 for a complete proof and the statement.


1

Let $A=-\frac{d^{2}}{dx^{2}}$ on the domain $\mathcal{D}(A)=W^{2}_{2}(\mathbb{R})$. The restriction $A_{0}$ of $A$ to $\mathcal{C}_{0}^{\infty}(\mathbb{R})$ has a closure $\overline{A_{0}}=A$. In other words, the closure of the graph of $A_{0}$ is equal to the graph of $A$. If $V$ is any bounded measurable function on $\mathbb{R}$, then $V$ defines a ...


1

If we are on $H^2_0$, then both $u$ and it's first order derivatives are zero on the boundary (let's do this in $2$D, since it generalises easily), and so we see that $$\|\Delta u\|_2^2=\int_\Omega u_{xx}^2+2u_{xy}^2+u_{yy}^2=2\int_\Omega u_{xx}^2+u_{yy}^2=2|u|_{H^2}^2,$$ note that the latter equality is deduced by integrating by parts, possibly arguing by ...


2

Let $f$ be the cantor function on $[0,1]$. Let $g(x) = (1-x) f(x)$ on $[0,1]$. Then, $g$ is continuous and $g(0) = g(1) = 0$. For almost every $x\in[0,1]$ we have $$ g'(x) = \underbrace{(1-x) f'(x)}_{=0} - f(x) = -f(x) < 0.$$ If $g$ is (locally) absolutely continuous, then $g$ would be strictly decreasing, which contradicts $g(0) = g(1)$. Thus, $g$ is ...


2

First, from (i) we get $$ \|u(t)-u(s)\|\le\int_s^t \|u'(\tau)\|d\tau, $$ which proves that $u\in C([0,T],X)$. It remains to show the norm bound. Second, integrating $$ u(t)-u(s) = \int_s^t u'(\tau)d\tau. $$ with respect to $s$ from $0$ to $T$ yields $$ \int_0^T u(t)-u(s)ds = \int_0^T\int_s^t u'(\tau)d\tau\, ds\\ = \int_0^t\int_s^t u'(\tau)d\tau\, ds - ...


0

No, this intuition is not correct, because of the issue of domains with cracks. For an example, consider the unit disk $D\subset\mathbb{R}^2$, and the domain $\Omega$ which is obtained from $D$ by removing several straight lines. The domain $\Omega$ is not connected, even though $\bar{\Omega} = D$. This means that there can be piecewise constant functions ...


0

The answer posted by Jose27 is correct. For more details and related things, see the reference here: http://www.iadm.uni-stuttgart.de/LstAnaMPhy/Weidl/fa-ws04/Suslina_Sobolevraeume.pdf, especially Theorem 5.


1

1) Partitions of unity and local trivializations. Do the standard Sobolev norms and add them. This will depend on the choice of trivializations; up to equivariance it will not. 2) With he above norm, you can define Sobolev completions of the appropriate spaces of sections. Because locally your operator is just a differential operator in the standard sense, ...


3

A simple counterexample is $A=(0,1)$, $B=(-1,1)$. Set $v=1$ on $A$. Then $v\in H^1(A)$, but the extension by zero is not in $H^1(B)$: $v$ is discontinuous on $B$, but by Sobolev embeddings $H^1(B)$ is continuously imbedded into $C(\bar B)$. For regular domains, $H^1$ functions that can be extended by zero and still are $H^1$ are precisely the funcitons in ...


1

Once you convolve $u, u'$ with say a gaussian, they become smooth $\forall t > 0$ (with compact support). Then this equality comes down to "moving the derivative inside the integral": \begin{eqnarray*} \frac{d}{dt} \|u^{\epsilon}(t)- u^{\delta}(t) \|^2_{L^2(U)} &=& \frac{d}{dt} \int_U |u^{\epsilon}(t)(x) - u^{\delta}(t)(x) |^2 dx \\ &=& ...


1

Guessing First I would make a quick guess based on the chart of function space. It groups together Sobolev spaces $W^{s,p}$ with the same value of $\frac n p -s$, because these are related by the embedding theorem. While the inclusion provided by this theorem is strict, the sharpness of the theorem still makes "the spaces with equal $\frac n p -s$ are ...


0

You should consider the space often denoted $H^1_E(\Omega)=\{v\in H^1(\Omega):v|_{\Gamma^c}=0\}$.


2

No, in general this is false. Consider as an elementary example $s=0$, $q=2$ (so that we don't have to deal with fractional derivatives or anything like this). We are inspired by the following observation: if $f(x)=C$ is a nonzero constant function, then $\|f\|_{\dot{H}^2} = \|-\Delta f\|_{L^2} = 0$, while naively "$\|f\|_{L^2} = +\infty$". There are scare ...


1

The embedding does not have a standard name. For example, no particular name is used in the article Another Note on the Inclusion $L^p(μ)⊂L^q(μ)$ which is entirely about this fact. Some of the ways to refer to it: since the Lebesgue spaces are nested... by the nested property of the Lebesgue spaces... the inclusion between Lebesgue spaces... by Jensen's ...


3

I think the authors are somewhat inconsistent in what they mean by the domain of an operator. Let's ignore the boundary conditions for now and focus on the order of smoothness. The domain of $(-\Delta)^s$ as an operator into $L^2$ is a Sobolev space of order $2s$. Indeed, $s$ fraction of the Laplacian is like $2s$ derivatives. The domain of ...



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