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4

Using Hölder, we deduce that $$\int\nabla f\cdot v\ dx:= \int f\mathrm{div }v\ dx\leq\|f\|_{L^2}\|\mathrm{div }v\|_{L^2} \leq\|f\|_{L^2}\|v\|_{H^1} $$ for $v\in H_0^1(\Omega)$, so $\nabla f\in H^{-1}$.


3

Let $u \in H^1_0(\Omega)$ be any function, then $u + c \in H^1(\Omega) \setminus H^1_0(\Omega)$ (provided $\Omega$ is bounded) And yes, you need the function in $H^1_0$ for this to work. The intuition is easy (picture one dimension): if you know how big the derivative is (how fast the function is growing), you can know more or less which level will the ...


3

For $u$ to be a classical solution, both $u$ and the coefficients in the partial differential operator $P$ in the LHS of your first formula need to be regular enough so that $P$ does not map u away from $C^0(\Omega)$, for in this case you can integrate by parts the LHS of your second formula. The path from weak solutions to classical solutions relies on two ...


2

You know that if $$ u(0) = \int_{0}^{1}u'v_0'+uv_0 dx,\;\;\; u\in H^1(0,1), $$ then $$ 0=\int_{0}^{1}\varphi' v_0'+\varphi v_0 dx, \;\; \varphi\in\mathcal{C}_{c}^{\infty}(0,1), $$ which implies that $v_0'$ has a weak derivative and $v_0''=v_0 \in L^2$. That's enough to imply to $v_0$ is twice absolutely continuous, and \begin{align} ...


2

It depends on your definition of $\|f\|_{H^{-1}}$. Let us equip $H_0^1$ with the scalar product $$(u,v)_{H_0^1} = \int \nabla u \nabla v \, \mathrm dx.$$ Then, the weak formulation of Poisson's equation is $$(\phi,v)_{H_0^1} = f(v) \quad\forall v \in H_0^1.$$ Hence, the solution $\phi \in H_0^1$ is just the Riesz representative of $f \in H^{-1} = (H_0^1)'$. ...


2

D1X, this fact has been said here What is precisely the definition of Elliptic Partial Differential Equation? We remember that if $\Omega \subset \mathbb{R}^n$ is a limited open with boundary $\partial \Omega$ of class $C^1$, the following are true Green's identity (with $u,v \in C^2(\Omega)\cap C^1(\overline{\Omega})$) \begin{align*} \displaystyle ...


2

Let $B\subset \mathbb R^3$ denote the open unit ball and consider the function $$ w(x)=|x|^{-\lambda}-1,\quad \lambda <\frac12.$$ This function is clearly not continuous but $w\in H^1_0(B)$. Define $f=-\Delta w$ and consider the problem $$ \begin{cases} -\Delta u = f, & \text{on }B\\ u\in H^1_0(B) \end{cases}$$ This equation has the unique ...


2

First, complement (P2) with the "boundary condition" $|u(x)|\rightarrow0$ as $|x|\rightarrow\infty$. Next, remind that the set $C_0^\infty{(\mathbb{R}^N)}$ of all compactly supported smooth functions defined in $\mathbb{R}^N$ is dense in $H^1(\mathbb{R}^N)$. Now, $$\displaystyle\int_{\Omega}\nabla u \nabla v = - \displaystyle\int_{\Omega} u\Delta v + ...


2

Following Daniel Fischer comments I'm trying to post an answer: Let $G = (a,b)$ We have that $$\lVert u\rVert_{L^\infty(G)}^2 = \lVert u^2\rVert_{L^\infty(G)} \le \int_a^b 2 |u(t)u'(t)|dt \le 2 \lVert u\rVert_{L^2(G)}\lVert u'\rVert_{L^2(G)}$$ where the first inequality is justified by the fact that $u^2$ is absolutely continuos and $u(a) = 0$ so that we ...


1

Let $\Omega \subset \mathbb{R}^n$ open. If $u \in \mathscr{C}(\Omega)$, and the distributional derivatives of $u$ are given by integration against continuous functions, that is, there exist $v_1,\dotsc, v_n \in \mathscr{C}(\Omega)$ such that for all $\varphi \in \mathscr{D}(\Omega)$ we have $$- \int_{\Omega} u(x)D_k\varphi(x)\,dx = \int_{\Omega} ...


1

The equality $$ \int_{\Omega \setminus F} (\dots) \, dx = \int_{\Omega^{(i)}} \int_{ \{x_i \in \mathbb{R} \, : \, (x_1, \dots, x_i, \dots, x_N) \in \Omega \setminus F \}} (\dots) \, dx_i \, \dots \, dx_{i-1} \, d{x_{i+1}} $$ is Fubini's theorem. It has nothing to do with the functions being integrated, or with the assumption on $F$. It's just doing ...


1

Let us assume that $f_k$ converges to $F \in D'(\Omega)$ in the sense that $$\int_\Omega f_k \, v \, \mathrm{d}x \to F(v) \qquad\forall v \in D(\Omega).$$ Now, since $W^{1,q}(\Omega)$ is reflexive, you get $f \in W^{1,q}(\Omega)$ and a subsequence such that $f_{n_k} \rightharpoonup f$ in $W^{1,q}(\Omega)$. In particular, $f_{n_k} \rightharpoonup f$ in ...


1

The footnote 31 on page 305 explains the logic of the choice of $h$: the support of $w$ is at some distance from the top and side surface of the cylinder $ Q_+$, but it need not be separated from the bottom surface, which is $x_N=0$. This is how the chosen partition of unity works: in order to cover the domain by finitely many sets, we need these sets to ...


1

Actually, you have $H_\perp^1 = \{0\}$, the set containing only the zero function. Indeed, suppose that $u \in H^1(\Omega)$ satisfies $\int_\Omega u \, v \, \mathrm{d}x = 0$ for all $v \in H_0^1(\Omega)$. Then, using the density of $H_0^1(\Omega)$ in $L^2(\Omega)$ you get $\int_\Omega u \, v \, \mathrm{d}x = 0$ for all $v \in L^2(\Omega)$ and this shows $u = ...


1

The following is an answer by Terry Tao, which is similar to Winther's comment. If $f$ is dimensionless, then $D^\alpha f$ has the units of $L^{-|\alpha|}$, and so $\int_\Omega |D^\alpha f|^2\ dx$ has the units of $L^n \times (L^{-\alpha})^2$. As $\Omega$ has units of $L^n$, each summand in $\|f\|_{m,\Omega}^2$ has the units of ...


1

Means that $\partial_{x_j} u \in L^p(\Omega)$ $\forall j=1,...,n$, i.e. $\nabla u \in L^p(\Omega) \times \cdot \cdot \cdot \times L^p(\Omega)$.


1

An identity that is originally proved for smooth functions can be extended to a Sobolev space provided that both sides are continuous with respect to the Sobolev norm. This is a general topological fact: if two continuous functions agree on a dense subset, then they agree everywhere. In the case of linear expressions like $u\mapsto \int \varphi \nabla u$, ...


1

This is a consequence of the Sobolev Embedding Theorems, which is explained, e.g., here (and it's not trivial, but well known. If this is a textbook what you are reading it should be mentioned somewhere).


1

Take $u \equiv 1$. Then, $u \in H^1(\Omega) \setminus H_0^1(\Omega)$ and your inequality fails (as long as $\Omega \ne \emptyset$).


1

Let $u \in H^2 \cap H^1_0$. According to ellipticity we know that $$ C_0 \|\nabla u \|_{L^2}^2 = \int_\Omega C_0 \partial_i u \partial_i u \le \int_\Omega a^{ij} \partial_i u \partial_j u. $$ Next we integrate by parts, using the fact that $u=0$ on $\partial \Omega$: $$ \int_\Omega a^{ij} \partial_i u \partial_j u = \int_\Omega -\partial_j(a^{ij} ...


1

You can use the Fourier transform for a nice class of functions to solve for the resolvent. \begin{align} (\lambda I-\Delta)^{-1}f & = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\frac{\hat{f}(\xi)}{\lambda+|\xi|^2}e^{ix\cdot\xi}d\xi \\ & = ...


1

Do you know the weak derivative of $\gamma(nu)$? Using this, you can show for $V = \gamma(n u)$ $$\int_I u' v' \ge 0$$ and, then, the claim follows by simply $n \to \infty$ (and the dominated convergence theorem). What about $v = \gamma(n \, u)^q$ for suitably chosen $q$?


1

Keep in mind that there is a constant $C$ with the property that $\|v\|_\infty \le C \|v\|_{H^1}$ for all $v \in H^1(I)$. If $f \in L^1(I)$ then $$\left| \int_I fv \right| \le \|f\|_1\|v\|_\infty \le C \|f\|_1 \|v\|_{H^1}.$$


1

Yes. A corollary to the Rellich Kondrachov theorem says that for all $p\ge 1$, we have $W^{1,p}$ compactly contained in $L^p$. Evans PDE book has a nice treatment of this. EDIT: Actually, if I recall correctly, for $\Omega \subseteq \mathbb R^n$, to prove that $W^{1,p}(\Omega)$ is compactly contained in $L^p(\Omega)$, we only need the Rellich Kondrachov ...


1

You can show your regulary if $g \in W^{1,p}(\Omega)$, $p > d \ge 2$, by using a trick by Stampacchia. First, we note $$a(u - g,v) + \int_\Omega f(u) v \, \mathrm{d} x = a(-g,v).$$ Now, let $K > 0$ and set $$(u - g)_K = \begin{cases} u - g & \text{if } |u - g| \le K, \\ \operatorname{sign}(u-g) \, K & \text{else}.\end{cases}$$ Then, $(u-g)_K ...


1

Short answers. Q1: yes. Q2: no, but yes for almost every $r$. Q3: sharp forms of trace theorem also exists, and the target space in it (a Besov space) has lower regularity. Explanation: Q1. You can apply trace theorem on a smaller disk. Q2-3. The reason we may lose regularity when restricting to a surface is that the surface may pass through a singularity ...


1

For part (1), to not stack subscripts too deeply, let us assume that the full sequence $(u_n)$ converges uniformly to the continuous function $u\in C([0,1])$. If $1 < p < \infty$, by the reflexivity of $W^{1,p}(I)$ we know that $(u_n)$ has a subsequence $(u_{n_m})$ converging weakly to some $v \in W^{1,p}(I)$. Then $(u_{n_m}')$ converges weakly to ...


1

Concerning 3., do not estimate the integrals $\int f_j(x) \phi(x+n)dx$ in your way but approximate $f_j$ by functions with compact support (just multiply with an indicator function). If $f_j$ has compact support the integral is $0$ for $n$ large enough. Concerning 2. the limit of a convergent subsequence would be necessarily $0$ because of 3. But the norms ...



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