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2

Argue by contradiction: For every $k$ there exists $u_k$ with $\| u_k\|_{L^2(B)}=1$ and $$ \| \nabla u_k \|_{L^2(B)} + \| u_k\|_{L^2(\partial B)} \leq 1/k. $$ Then $u_k\to 0$ weakly in $H^1$, this contradicts the fact that the $u_k$ have unit $L^2$ norm by Rellich's compactness theorem.


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Q1. Non-zero constant functions DO NOT BELONG to $W_0^{1,2}(\Omega)$. Q2. The space $W_0^{1,2}(\Omega)$ is defined as the «completion» of $C_0^\infty(\Omega)$ with norm $$ \|u\|_{W^{1,2}_0(\Omega)}=\left(\int_\Omega\big(u^2(x)+\big|\nabla u(x)\big|^2\big)\,dx\right)^{1/2}. $$ Thus, indeed, every element of $W_0^{1,2}(\Omega)$ can be ...


2

Thanks for Jose27's help, I finally figured this out. First step: Since $u$ has compact support and $c(0)=0$, and $c(x)$ is continuous, $c(u(x))$ is continuous and has compact support and so it is in $L^2(R^n)$. Second step: Take Fourier transform of $$-\Delta u+c(u(x))=f(x)$$ Then you can get $$|\xi|^2\hat u(\xi)+\widehat{c(u)}(\xi)=\hat f(\xi)$$ It ...


2

The trace operator is the continuous extension of the densely defined operator $T: C^1_0(\mathbb{R}^n) \to H^1(\mathbb{R}^{n-1} \times\{0\})$. This $T$ is in particularly linear, and hence commutes with partial differentiation tangential to $\mathbb{R}^{n-1}\times \{0\}$. Thus, appealing to any of the standard approximation procedures you see that indeed, ...


2

Let $h \in L^r$. Let $r'$ be the conjugate exponent. Set $w = (1 + sign(h))/2$. Thus $\int_\Omega h \cdot w = \|h^+\|_{L^1}> 0$. Then take $u$ to be a suitable non-negative $C^\infty_0$ approximation of $w$, such that $\|u - w\|_{L^{r'}} = \varepsilon$. Therefore $$ \int h \cdot u = \int h \cdot w + \int h \cdot (u-w) \ge \int |h^+|- \varepsilon ...


2

Pick a finite open cover of charts $\{U_j\}_{j=1}^N$ with diffeomorphisms $s_j: \Bbb R^n \to U_j$. Let $\varphi_j$ be a partition of unity subordinate to the $U_j$. We put a norm on $C^\infty(M)$ as follows: $$\|u\|_{H^s}^2 = \sum_{j=1}^N \|\varphi_j(s_j(x))u(s_j(x))\|_{H^s}^2,$$ where the norm in the sum is the $H^s$ norm on $\Bbb R^n$. The point being ...


1

There are examples that are simple to state (in $n$ dimensions): $$f(x) = |x|^{\alpha-n}\chi_{|x|\le 1},\qquad \frac{n}{2}<\alpha\le \frac{n+1}{2}$$ But proving that $f\notin H^{1/2}$ is a bit awkward; it's best done by using the divided-difference-integral characterization of $H^{1/2}$ or a related, simplified formulation in Exercise 33 here. Then ...


1

No, not in general. Consider the following example: Take $D=(-1,1)$ and $u\colon (0,1)\to\mathbb{R},\,x\mapsto 1-x^2$. Then $u$ is in $H^1_0((0,1))$. However, $u$ can not be approximated by smooth functions with support in $(-1,0)$. This works more general for open intervals $I$: By the Sobolev embedding theorem, the functions in $H^1(I)$ are continuous, ...


1

If you assume a $C^1$-boundary rather than just Lipschitz, we can use Morrey's inequality. (It might be possible to use this inequality for Lipschitz domains, I am not sure.) So there exists $M>0$ such that, for all $u\in W^{1,\infty}(\Omega)$, we have $$\|u\|_{C^{0,1}(\Omega)}\le M\|u\|_{W^{1,\infty}(\Omega)}$$ and in particular $u$ is Lipschitz on ...


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The answer is yes. By definition, a $L^p$ function takes finite values (in $\mathbb R$ or $\mathbb C$) on a set of full measure (if you are in doubt, check your go-to reference for functional analysis and re read carefully the definition).


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Any Lipschitz function that is not $C^1$ will do, since convergence in the $W^{1,\infty}$ norm implies uniform convergence (for the function and its derivatives) in $C^1$.


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For $0<s<1$ this is true in all dimensions $n$, and follows by writing the norm as in integral of divided differences [Leoni, A first course in Sobolev spaces, 14.8] $$ \|f\|_{H^s}^2 \approx \|f\|_{L^2}^2+\iint\frac{|f(x)-f(y)|^2}{|x-y|^{n+2s}}\,dx\,dy \tag{1} $$ Indeed, (1) shows at once that for any Lipschitz function $\varphi$ fixing $0$ we have ...


1

Note that $(x+y)^2\le 2(x^2+y^2)$. More generally, if $p\in [1,\infty)$, we have that $$(x+y)^p\le 2^{p-1}(x^p+y^p), \ \forall\ x,y\ge 0.$$


1

From your definitions, you directly obtain $\|u\|_{H^{-1}}\le \|u\|_{L^2}$. To see this, take $u\in L^2(\Omega)$, and define $$ f(\phi) = \int u \phi. $$ Now setting $f_0=u$, $f_i=0$ for $i=1\dots n$, we immediately find $$ \|u\|_{H^{-1}}\le \|u\|_{L^2}. $$


1

$$ \int_0^1 x^{pn} = \frac{1}{pn +1} x^{pn+1} |_0^1 <\infty =\frac{1}{pn+1}$$ and $$ \int_0^1 (n x^{n-1} )^p = \frac{n^p}{p(n-1)+1} $$ So for $1\leq p< \infty $, $x^n \in W^{1,p}$. (1) But norm of $x^n$ goes to $\infty$ when $p> 1$. Hence in this case it does not converge (2) Let $p=1$. The norm of $x^n$ goes to $1$. Note that the limit is ...


1

Rewrite your function as $u(x_1, x_2)=1-x_1\lvert x_1\rvert$. This makes evident that $u\in C^1$.


1

Forget about the unnecessary detail that complicates things. A Lipschitz continuous functions, if it is differentiable$^{[1]}$, has a bounded derivative. The bound on the derivative is the Lipschitz constant. This follows trivially from the definition and from your favorite version of the mean value theorem. In this case we have a sequence which is ...


1

The argument is incorrect. Suppose $\phi \in C^\infty(\bar{\Omega}) \subset W^{k,p}(\Omega)$. As you say, we can define a function $$\phi_n(x) = \begin{cases} \phi(x), & x \notin N \\ 0, & x \in N \end{cases}$$ Then $\phi_n$ is a function that vanishes on $N$ and is almost everywhere equal to $\phi$. However, $\phi_n$ typically will not be ...


1

In the end we don't want the gradient term on the RHS of the inequality: $$\int_U|Du|^p\,dx\le C\left(\int_U |u|^\frac{p}{2} |D^2u|^\frac{p}{2}\right)^\frac{2}{p} \left(\int_U |Du|^p \right)^\frac{p-2}{p},$$ dividing by $\left(\int_U |Du|^p \right)^\frac{p-2}{p}$, we obtain $$\left(\int_U |Du|^p \right)^{\frac{2}{p}}\le C\left(\int_U |u|^\frac{p}{2} ...


1

(I pick $p=2$ for simplicity). It is natural to require the time derivative $u' \in L^2(0,T;X^*)$ because typically you have a parabolic equation of the form $u' + Au =0$ where $A$ is an elliptic operator like $A=-\Delta$. Then $\langle -\Delta u, v \rangle := \int \nabla u \nabla v$ implies that $-\Delta u$ lies in the dual space of $L^2(0,T;X)$, i.e., ...


1

For a counterexample to your claim, notice that $C^\infty_c(\Omega)$ is a subspace of $H^1_0(\Omega)$ (although it is not closed), and its dual is $D'(\Omega)$ (the space of all distributions on $\Omega$) which is not a subset of $H^{-1}(\Omega)$. Note that to use the Hahn-Banach theorem, you need a sublinear function which is defined on the whole space and ...


1

This is only correct for $u''=0$, since $$ -\frac13\int_0^1 (u'(x)^3)' dx = -\int_0^1 u'(x)^2 u''(x)dx, $$ the latter integral is not equal to $\int_0^1 u''(x)u(x)dx$ (except for special cases $u=0$ or $u''=1$), while the first integral is not zero in general.


1

This is similar to Thomas' comment: No, it can't work. Indeed, for each $u_n \in H_0^1$, we have $\int \nabla u_n = 0$, but $\int \nabla u$ might not be zero. It is zero iff $u \in H_0^1$, since $u(T) = 0$.


1

Take $w$ a solution to the problem $\text{div}\sigma \nabla w = 0$ and $w=g$ on $\partial \Omega$, let also $v$ be a solution of $\text{div} \sigma\nabla v = f$ with zero boundary conditions. Obviously we can write $u=v+ w$, so we obtain the appropriate estimates for the last ones. $v$ is the easiest, since by definition of weak solution we have $$ ...



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