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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}$$so that w_n' = n^{1/4} 1_{(0, n^{-1})}. It's easy then to see that w_n' \to 0 and w_n \to 0 in L^2, so w_n \to 0 in W^{1,2}(U). Taking w = ... 0 This will not hold. Take your favorite sequence \{v_n\}\subset H^1(M) and v \in H^1(M) such that \{v_n\} converges towards v in L^2(M), but not in H^1(M). Now, set u_n(t,x) = v_n(x) and u(t,x) = v(x). This yields u_n \to u in L^\infty(\mathbb{R};L^2(M)), but \partial_{x_j} u_n \not\to \partial_{x_j} u in the same space. 1 Forget about the unnecessary detail that complicates things. A Lipschitz continuous functions, if it is differentiable^{[1]}, has a bounded derivative. The bound on the derivative is the Lipschitz constant. This follows trivially from the definition and from your favorite version of the mean value theorem. In this case we have a sequence which is ... 0 Take a look at the proof of the trace inquality, for example in Evans book https://books.google.com.br/books?id=Xnu0o_EJrCQC&printsec=frontcover&dq=evans+partial&hl=pt-BR&sa=X&ei=UOgyVfvRJMm1sATUjYH4DA&ved=0CB0Q6AEwAA#v=onepage&q=evans%20partial&f=false when he apply Young's inequality you could apply Young's inequality with ... 0 You can apply the L^p estimates to get bounds in the W^{2,p} space, take a look at section 9.5 of Gilbarg and Trudinger's book. 1 In the end we don't want the gradient term on the RHS of the inequality:$$\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} ...

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 ...

0

If $f\in L^p(\mathbb{R}^n;\mathbb{R}^m)$, then $f=(f_1,\ldots, f_m)$ with $f_j\in L^p(\mathbb{R}^n)$ for $j=1,\ldots, m$, therefore $$|f|=\sqrt{f_1^2+\ldots+ f_m^2}\leq \sqrt{n}\max_j |f_j|$$ is in $L^p(\mathbb{R}^n)$ (because the maximum of $L^p$ functions is in $L^p$). On the other hand, as $$|f_j|\leq\sqrt{|f_1|^2+\ldots+|f_m|^2}=|f|^2$$ if the latter is ...

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

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

0

It's because the function Ψj has compact support, since {Ψj} is a partition of unity.

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 $$\|u\|_{H^{-1}}\le \|u\|_{L^2}.$$

0

I do believe so, since $|e^{ix\xi}|\le 1$, so $$\int_{\Bbb R^n}(1+|\xi|^s)^{-2}e^{ix\xi}\,d\xi\le\int_{\Bbb R^n}(1+|\xi|^s)^{-2}\,d\xi<\infty.$$

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

$$\int_0^1 x^{pn} = \frac{1}{pn +1} x^{pn+1} |_0^1 <\infty =\frac{1}{pn+1}$$ 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 ...

0

The answer is yes, and it follows immediately from the usual Poincare inequality: We know that for any function of zero average we have $$\| g\|_{W^{1,1}} \leq C\| \nabla g \|_{L^1}.$$ Then we know that for such a $g$, if $T$ denotes the trace operator, $$\| Tg\|_{L^1(\partial \Omega)} \leq C\| g\|_{W^{1,1}} \leq C\| \nabla g \|_{L^1}.$$ Now just plug ...

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

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

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).

0

The solution is based on Solution attempt 2: Trying to establish that $U$ is weakly closed. Suppose $U \ni f_n \rightharpoonup f$ in $H_P^1$. By the Rellich–Kondrachov theorem, $H_P^1$ is compactly embedded in the space of $P$-periodic continuous functions $C_P$ with the uniform norm. Hence, as $\{f_n\}_n$ is weakly convergent and therefore bounded, there ...

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 ...

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 $$\int h \cdot u = \int h \cdot w + \int h \cdot (u-w) \ge \int |h^+|- \varepsilon ... 0 u(x)=|x| for -1<x<1? (Weak) u'(x)=H(x)=1[x>0]+0[x<0] (where [] is Iverson bracket), that's discontinuous, so no uniform (L^\infty) convergence of continuous (let alone smooth) function sequence. 0 For example take u(x):=|x| on \mathbb{R}. I assume that you can prove that u\in W^{1,\infty}(\mathbb{R}) (more generally, it is not so difficult to see that W^{1,\infty}(\mathbb{R}) consists exactly of the Lipschitz functions). The weak derivative of u is v(x)=\text{sgn}(x) (which equals 1 on \mathbb{R}^+ and -1 on \mathbb{R}^-). Now ... 1 Any Lipschitz function that is not C^1 will do, since convergence in the W^{1,\infty} norm implies uniform convergence (for the function and its derivatives) in C^1. 2 The trace operator is the continuous extension of the densely defined operator T: C^1_0(\mathbb{R}^n) \to H^1(\mathbb{R}^{n-1} \times\{0\}). This T is in particularly linear, and hence commutes with partial differentiation tangential to \mathbb{R}^{n-1}\times \{0\}. Thus, appealing to any of the standard approximation procedures you see that indeed, ... 2 Q1. Non-zero constant functions DO NOT BELONG to W_0^{1,2}(\Omega). Q2. The space W_0^{1,2}(\Omega) is defined as the «completion» of C_0^\infty(\Omega) with norm$$ \|u\|_{W^{1,2}_0(\Omega)}=\left(\int_\Omega\big(u^2(x)+\big|\nabla u(x)\big|^2\big)\,dx\right)^{1/2}. $$Thus, indeed, every element of W_0^{1,2}(\Omega) can be ... 0 You can characterize W^{1,2}_0 as "the 'compactly supported' L^2 functions that have L^2 weak derivatives." Since the weak derivatives of a constant function are all zero, hence L^2, the answer to your question is the answer to, "are constant functions compactly supported L^2?" This is never true unless the constant is zero. 1 For 0<s<1 this is true in all dimensions n, and follows by writing the norm as in integral of divided differences [Leoni, A first course in Sobolev spaces, 14.8]$$ \|f\|_{H^s}^2 \approx \|f\|_{L^2}^2+\iint\frac{|f(x)-f(y)|^2}{|x-y|^{n+2s}}\,dx\,dy \tag{1} $$Indeed, (1) shows at once that for any Lipschitz function \varphi fixing 0 we have ... 1 There are examples that are simple to state (in n dimensions):$$f(x) = |x|^{\alpha-n}\chi_{|x|\le 1},\qquad \frac{n}{2}<\alpha\le \frac{n+1}{2}$$But proving that f\notin H^{1/2} is a bit awkward; it's best done by using the divided-difference-integral characterization of H^{1/2} or a related, simplified formulation in Exercise 33 here. Then ... 0 The restriction part is bounded by the trace theorem. Now it remains to prove that P_e is bounded. This follows from the common "Cauchy-Schwarz with 1" trick:$$|P_e(\phi)| = \left | \int_e \phi \cdot 1 \right | \leq \| \phi \|_{L^2(e)} \| 1 \|_{L^2(e)} \leq \| \phi \|_{H^1(e)} m(e)^{1/2}$$where m(e) is the measure of the edge e. 2 Argue by contradiction: For every k there exists u_k with \| u_k\|_{L^2(B)}=1 and$$ \| \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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