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1

For $H^1$- functions, this does not hold in general. It is basically the same argument that $L^2$- functions generally do not vanish at $\infty$, you just have to take such a function and integrate it, for example a bump function where the bumps get thinner when you go outside. For $H^2$-functions however, it does hold: Take $$\int_0^a ...


1

The argument is the following: Let $v_n$ be a sequence in $W^{1, p}_0(\Omega)$ for some $\Omega \subset \mathbb R^N$ and $p >N$. If $v_n \to v$ weakly in $W^{1,p}_0(\Omega)$, then $||v_n||_{1, p}$ is uniformly bounded. By the Sobolev Embedding (Theorem 7.17) and the fact that $$C^{0, \alpha}(\overline \Omega) \to C(\overline \Omega)$$ is compact, there ...


0

Yes, this is true. A convex function is always continuous and actually locally Lipschitz. @Yes already provided you a nice proof in 1-D but let me point out that this result also holds in high-dimensions. That is, given $f: \mathbb R^N\to \mathbb R$ is convex, we have $f$ is locally Lipschitz on $\mathbb R^N$. You can find proof at page 236 in this book, ...


0

This is true. In fact, it is locally in $W^{1,p}$ for any $p\in [1,\infty]$ because a convex function is locally Lipschitz.


2

In the case $q<p^*$ the Rellich–Kondrachov theorem applies, if your domain $\Omega$ has the required properties. If $q=p^*$, the $L^q$ norm is still controlled by the $W^{1,p}$ norm; and since a weakly convergent sequence is norm-bounded, $L^q$ norm is bounded. Moreover, you will have weak convergence in $L^q$, since linear maps preserve weak ...


0

1) If $U$ is not bounded, just write it as a countable union of open bounded sets, however, I don't see where you need to use the boundedness of $U$. 2) Yes, this is true and the proof can be found in a lot of places. For example, it can be found in Brezis book chapter 4.


1

Suppose that $f_n$ is a Cauchy sequence in $W^{1,2}$ and write $$f_n(b)-f_n(a)=\int_a^b f_n'(t)dt.\tag{1}$$ By one hand, there is $f\in L^2$ such that $f_n\to f$. On the other hand, there is $g\in L^2$, such that $f'_n\to g$. From the estimate $\|f_n\|_\infty\le K\|f_n\|_{1,2}$, we must conclude that $f_n(x)\to f(x)$ for all $x\in\mathbb{R}$ and by using ...


1

The following claim is not true in general. Let $X$ and $Y$ be Banach spaces and $X_1 \subset X$ a subspace. If $T: X_1 \to Y$ is a continuous bounded operator, then there exists a continuous linear extension to the whole space $X$, i.e. there exists a bounded linear operator $\hat T: X \to Y$ such that $ \hat{T}\restriction_{X_1} = T$. Take for ...


0

If $U$ is a bounded domain with regular boundary then, one approach is the following: Consider the problem $$ \left\{ \begin{array}{rl} -\Delta u=f &\mbox{ in}\ U, \\ u=0 &\mbox{on } \partial U. \end{array} \right. $$ Once $f\in C(\overline{U})$, we also have that $f\in L^p(U)$ for each $p\in [1,\infty)$. This implies in particular (see ...


0

It is not true that they are equal, instead the left hand side is bounded by the right hand side multiplied by some constant that is independant of $\epsilon$, it seems to be part of the the $C$ in the example. This fact arises because $\partial B(0,\epsilon)$ is an $n-1$ dimensional object, so $$\int_{\partial B(0,\epsilon)}\,dS=|\partial B(0,\epsilon)|\le ...


0

We know from the comments above for $h:= w_s \hat u \in L^2(\mathbb R^n)$ that it holds $$ \langle w_s \hat u, \varphi \rangle = \langle \hat u, w_s \varphi \rangle = \langle h, \varphi \rangle_{L^2(\mathbb R^n)} $$ for all $\varphi \in S(\mathbb R^n)$ and if we choose $\varphi := w_{-s} \psi \in \mathcal S (\mathbb R^n)$ with $\psi \in \mathcal S (\mathbb ...


0

HINT: One has $ (1+\lvert\xi\rvert^2)^\frac{s}{2}\ge 1$. This, and Plancherel's theorem. EDIT: This actually addresses another question, namely, "When $s\ge 0$, is it true that $u\in H^s$ implies $u\in L^2$?". Indeed, when $s<0$, it might well happen that $u\in H^s$ is not a function. (The Dirac delta $\delta$ is such that $\delta\in H^{-s}$ for $s> ...


0

One word: scaling. Consider $u_\lambda(x) = u(\lambda x)$ where $\lambda>0$. The Lebesgue norm scales as $$\|u_\lambda\|_p = \lambda^{-n/p}\|u\|_p$$ while the norm of the gradient scales as $$\|\nabla u_\lambda\|_p = \lambda^{1-n/p}\|\nabla u\|_p$$ This immediately shows that $\|\nabla u_\lambda\|_p$ can control $\|u\|_q$ only when $\frac{n}{q} = ...


1

An "abstract" construction can be done as follows: Take $u$ the solution of the problem $-\Delta u= 1$ in $\Omega$ and $u=0$ on $\partial \Omega$. If $\Omega$ is smooth enough then $u\in H^2(\Omega)$ by elliptic regularity. On the other hand, each partial derivative is harmonic so if $u\in H_0^2(\Omega)$, this would mean that the partial derivatives have ...


1

Assume $c=0$. The desired result follows easily when we consider the Gagliardo norm on $H^{\frac 12}$ and from the identity $|u^+(x)-u^+(y)| \leq |u(x)-u(y)|$ for any $u \in H^{\frac 12}$. Am I right?


3

You have all the ingredients ready: To show that $\ell$ is bounded, we need $|\ell(f)| \le C||f||_{H^1}$ for some $C$. Now there is $\zeta$ (depending on $f$) so that $$\ell(f) = f(c) = f(\zeta) + \int_\zeta^c f'(x)dx = \int_a^b f(x) dx + \int_\zeta^c f'(x)dx$$ Then $$|\ell (f)| \le \int_a^b |f(x) |dx + \int_\zeta^c |f'(x)| dx \le \cdots$$


0

This is a classical result in singular integrals. Let me prove a similar inequality in the one dimensional case (for the several dimension case, you can read the paper by Kato and Ponce, Well-posedness of the Euler and NS equations in Lebesgue spaces, Revista Matemática Iberoamericana, 1986). We want to prove the bound $$ \|Hf\|_{L^\infty}\leq ...


1

$C_c^\infty$, the space of $C^\infty$ functions with compact support is dense in $W^{1,2}$. Let $\{\phi_n\}$, $\{\psi_n\}$ be sequences in $C_c^\infty$ converging to $\phi$ and $\psi$ respectively in $W^{1,2}$. We have $$ \int_{\mathbb{R}}\phi_n\,\psi'_n=-\int_{\mathbb{R}}\phi'_n\,\psi_n. $$ Taking limits as $n\to\infty$ we get $$ ...


0

Since constant functions have zero $H^{1/2}$ seminorm, it follows that $$\vert T u \vert_{H^{1/2} (\partial M)} = \vert (T u) -c \vert_{H^{1/2} (\partial M)} = \vert T (u -c) \vert_{H^{1/2} (\partial M)}\tag{1}$$ for every $c\in\mathbb{R}$. (Trace operator commutes with adding a constant, because the trace of a constant function is that constant function.) ...


0

A possible projection $u \in W_0^{1,p}(\Omega)$ of $v \in W^{1,p}(\mathbb{R}^n)$ is given by \begin{equation*} u = \operatorname{argmin}_{\tilde u \in W_0^{1,p}(\Omega)} \| \tilde u - v \|_{W^{1,p}(\Omega)}. \end{equation*} This $\operatorname{argmin}$ should exist in the reflexive case $1 < p < \infty$. Moreover, this projection is linear in case $p ...


3

You can formalize this: Take two Hilbert spaces $H\subset V$ such that The inclusion is dense (i.e. the image of $H$ in $V$ is dense in the topology of $V$) and, The inclusion is continuous (i.e. $H$ has a stronger norm than $V$). Then, if we identify $V$ with its dual (let's assume the spaces are real, though you can do this in general), we have ...


3

Your examples are lacking squares (think of homogeneity), but apart from that they are both valid norms for $H^4(I) \cap H^2_0(I)$, as is the normal $L^2$-norm. What you really want in most situations is not just any norm, but a norm with which the space is complete. This is not true for the $H^2$-norm: There holds by definition $$C^\infty_0(I) \subset ...


1

Use integration by parts and the Riesz representation theorem. For the one inclusion, if $u\in H^4(I) \cap H^2_0(I)$, we can integrate by parts twice to obtain $$a(u,v) = \int\limits_I u''(x)v''(x)\,dx = \int\limits_I u^{(4)}(x)v(x)\,dx$$ and see that $$\lvert a(u,v)\rvert \leqslant \lVert u^{(4)}\rVert_{L^2(I)}\cdot \lVert v\rVert_{L^2(I)}.$$ For the ...


1

Okay so (strong) measurability of $f$ is equivalent to weak measurability and $f$ being a.e. separably valued. Since $H^{-1}(\Omega)$ is separable we get the equivalence you mentioned. Now to show weak measurability we have to show that for every $L\in H^{-1}(\Omega)^*$ the map $t\mapsto \langle f(t), L\rangle$ is measurable. So far this is exactly what you ...


1

I think you have the inclusion backwards: For instance the function $f(t)=|t|^{1/2}$ is Holder continuous in, say, $\Omega=(-1,1)$ but $f'(t)=t^{-1/2}\notin L^2(\Omega)$, therefore the inclusion $C^{0,1/2} \to W^{1,2}$ fails. To prove that every $u\in W^{1,p}$ has a (locally absolutely) continuous representative see here.


0

Well, by definition the Fourier transform is an isometry between $H^s(\mathbb{R}^n)$ and $L^2(\mathbb{R}^n, (1+|\xi|^2)^sd \xi)$.


1

Since $u' \in H_0^1(I)$, you can apply Poincaré to $u'$. This gives $$\|u'\|_{H^1} \le C \, \|u''\|_{L^2}.$$ Can you conclude?


0

Maybe you want to take a look to this paper: http://arxiv.org/abs/1104.4345


2

In general, no. Take $U = (0,1) \subset \mathbb{R}^1$, $p=2$ and $F(z,w,x) = z^4$. Then take $$w_n(x) = \begin{cases} n^{1/4} x, & 0 < x < n^{-1} \\ n^{-3/4}, & x \ge n^{-1} \end{cases}$$ so that $w_n' = n^{1/4} 1_{(0, n^{-1})}$. It's easy then to see that $w_n' \to 0$ and $w_n \to 0$ in $L^2$, so $w_n \to 0$ in $W^{1,2}(U)$. Taking $w = ...



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