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New answers tagged sobolev-spaces

1

A priori, an element $[f] \in W^{1,2}(\mathbb{R}) \subset L^2(\mathbb{R})$ is only an equivalence class of functions in $L^2(\mathbb{R})$ (where two functions are identified if the set on which they differ is of measure zero). However, one can show that if $[f] \in W^{1,2}(\mathbb{R})$, then there is a (unique) $\overline{f} \in [f]$ that is continuous. The ...

2

As you probably know, elements of $W^{1,2}$ aren't functions but equivalence classes of functions (up to almost everywhere equality). A representative is then just an element of this equivalence class. The Sobolev embedding theorem asserts that every $f\in W^{1,2}(\mathbb{R})$ has a continuous representative. Hopefully, this explains both 1. and 2.

1

Let $u$ be a smooth cutoff, supported on $B(0,1)$, and set $$u_n(x)=\frac{1}{n^2}u\left(\frac{x}{n}\right).$$ Then, the change of variables $x=yn$ shows that $$\int_{\mathbb R^2}|u_n(x)|\,dx=\int_{\mathbb R^2}\frac{1}{n^2}\left|u\left(\frac{x}{n}\right)\right|\,dx=\int_{\mathbb R^2}\frac{1}{n^2}|u(y)|n^2\,dy=\|u\|_1,$$ while $$\int_{\mathbb ... 0 u\in H^{-s}, therefore by definition of H^{-s}, u\in S', the space of tempered distributions (I did not realize this fact when asking the question). Therefore (1+|\xi|^2)^s\bar{\hat{u}}=0 in the sense of tempered distributions. If we multiply by (1+|\xi|^2)^-s, and since multiplying by this function with this crescence rate let S' stable, we find ... 0 This can never happen if \Omega is bounded. As a first remark, the extension operator E:H^\tau(\Omega)\to H^\tau(\mathbb R^n) is not unique. If E is an extension operator with the desired properties and \phi is a compactly supported smooth function with \phi\equiv1 on \Omega, then Fu:=\phi Eu also satisfies the desired properties. It is easy ... 1 You have$$\int_{\Bbb R^d} u(x)\phi(x)dx = \int_{\Bbb R^d} \hat u(\xi)\bar{\hat \phi}(\xi)d\xi =\int_{\Bbb R^d} \hat u(\xi) (1+|\xi|^2)^{-s/2} \bar{\hat \phi}(\xi)(1+|\xi|^2)^{s/2}d\xi. $$Now you have \hat u(\xi) (1+|\xi|^2)^{-s/2} is a L^2 function. Take, for example, arbitrary \phi\in S, therefore, functions of the form \bar{\hat ... 2 No, the function$$v(x) = \begin{cases} -1, & x < 0, \\ 0, & x = 0, \\ 1, & x > 0, \end{cases}$$is not weakly differentiable. Indeed, we have$$- \int v(x) \phi'(x) dx = 2 \phi(0)$$for all smooth \phi with compact support. Note that there is no functions w so that$$\int w(x) \phi(x) dx = 2 \phi(0)$$for all \phi\in C^\infty ... 1 First note that the integral you've written (that the author assumes is zero) is precisely the inner product \langle \phi, u\rangle_{H^s}. Now pick an element u in the orthogonal complement of D in H^s. Our goal is to prove that u is zero, and hence \overline D = H^s. (This implies the result by the standard fact that for a Hilbert space, ... 2 Smooth functions are dense in L^2, and you can pick an arbitrarily small (in whatever norm you like) smooth function with specified boundary values on I. So if f_n \to f, f \in L^2, you can replace f_n with f_n-g_n, where say \|g_n\|_{H^1} \leq \varepsilon. So f_n - g_n \to f in L^2, and f_n-g_n is a smooth function that vanishes on the ... 2 By divergence theorem$$ \int_{\partial \Omega }u\cdot n \ \phi=\int_\Omega div(u\phi) = \int_\Omega \phi div(u) + \nabla \phi \cdot u. $$This gives$$ \left|\int_{\partial \Omega }u\cdot n \ \phi\right|\le \|\phi\|_{L^2(\Omega)}\|div(u)\|_{L^2(\Omega)} + \|\nabla \phi\|_{L^2(\Omega)}\|u\|_{L^2(\Omega)} \le \|\phi\|_{H^1(\Omega)}\|u\|_{H(div,\Omega)}. $$2 Remark: u(t)\in W^{1,p}(I) holds only for a.e. (x_2,\dots,x_n). No, the slices need not converge in any sense. Take the standard example that L^1 convergence does not imply convergence a.e. Make these functions f_n Lipschitz continuous by replacing rectangles by trapezoids in their graphs. Define u_n(x,y)=f_n(x) for x,y\in [0,1]. Now u_n\to ... 1 Since v is in W_0^{1,2}(\Omega), extending it by zero outside of the domain gives a function in W^{2,2}. Thus, we can replace the multiply-connected domain \Omega (with potentially bad boundary) by a ball B. The existence of \psi such that \nabla \psi=(-v_2,v_1) is a form of the Poincaré lemma about closed form being exact. By definition, ... 0 Any linear functional f on W^{1,p}(U), 1 \le p < \infty can be identified with (v_0,\ldots,v_n) \in L^{q}(U)^{n+1}, 1/q + 1/p = 1, via$$f(u) = \int_U u \, v_0 + \sum_{i=1}^n u_{x_i} \, v_i \, \mathrm{d}x.$$This shows the equivalency of your conditions. And this identification can be shown by using T : W^{1,p}(U) \to L^p(U)^{n+1},$$u ...

2

Yes, at least for $\,p=6\,$. This is according to the Sobolev Embedding Theorem which states \begin{align} \begin{cases} k>l \\ 1\le q<p<\infty \\ \big(k-l\big)\,q<n \\ \dfrac{1}{p} = \dfrac{1}{q} - \dfrac{k-l}{n} \end{cases} \implies W^{k,q}\big(\mathbb R^n\big) \subseteq W^{l,p}\big(\mathbb R^n\big) \end{align} In your case $\, n=3,\, k=2,\, ... 0 The Heat semigroup has an explicit form as convolution with respect to a Gaussian distribution$G(t,x)$. That is,$H(t)f=\int G(t,x-y)f(y)dy$. This kernel is positive and$\int G(t,y)dy = 1$for all$t > 0$. It follows that$H(t)$is a contractive semigroup on every$L^{p}(\mathbb{R}^{n})$for$1 \le p \le \infty$, but it's only$C^0$for$1 \le p < ...

3

Since the answer was given by Willy Wong and Phillip Andreae as comments, I just recap their inputs, giving exact references to the results in the mentioned books. Partial Differential Equations - Basic theory gives an introduction on Sobolev spaces and the main embedding theorems also on compact Riemannian manifolds with smooth boundary, regarded as ...

0

The $L^2$-norm is weakly sequentially lower semicontinuous, then you can prove the desired inequality with the liminf provided you show that $\nabla u_n \rightharpoonup \nabla u$ in $L^2$. The fastest way to achieve this is probably to show that $\{u_n\}$ is bounded in $H^1(\Omega)$ (and then use the sequential Banach-Alaoglu theorem together with uniqueness ...

2

We have the definition $$u_{\varepsilon}(x) = \int_{\Omega} \varphi_{\varepsilon}(x-y)\, u(y)\,dy.$$ For $u_{\varepsilon}(x)$ to be well-defined, that integral needs to exist. Since measurability is given, the problem is whether $$\int_{\Omega} \varphi_{\varepsilon}(x-y)\,\lvert u(y)\rvert\,dy < +\infty.$$ If we required $u \in L^1(\Omega)$, or just ...

0

I can not get an upper bounded of the total variation of $u_\epsilon$ Sure, because there isn't one. The condition you stated allows the functions $u_\epsilon$ to have total variation of size $\epsilon^{-1/2}$; you can obtain such functions by adding small zigzags to $u_0$ (small in the $L^2$ norm, but large in the $BV$ norm). Then the sequence ...

2

To answer your second question: If $u$ belongs merely to $H_0^1(0,\ell)$, you have no chance to define $u_{,x}(0)$ and $u_{,x}(\ell)$ in a reasonable manner. So your weak definition is not well-defined. To give an example: It is easy to verify that $u$ defined by $$u(x) = x^{2/3} \, (1-x)$$ belongs to $H_0^1(0,\ell)$ and its weak derivative is $$u_{,x}(x) = ... 2 The strong formulation implies the weak formulation; this follows after integrating by parts. To show that the weak formulation actually implies the strong formulation for functions u with two derivatives that have sufficient regularity, note that the weak formulation implies that$$\int_0^lu_{xx}v=-\int_0^lv_xu_x=-\int_0^lv,$$which implies that ... 0 If you do not have any information about the boundary values of u, then you only get local regularity u\in H^2_{loc}(\Omega), but no information about the smoothness up to the boundary (check for instance Evans' book). 1 I found three different ways to prove your inequality, but they are essentially the same. For convenience, let t = 1. For s \ge 0, define$$f(s) = \frac12 \, \| s \, u -u_0\|_{L^2}^2 + |s \, u|_{TV}.$$Then,$$0 = f'(1) = (u - u_0, u) + |u|_{TV}.$$The optimality conditions of your problem yield$$0 \in u - u_0 + \partial |\cdot|_{TV}(u).$$Here, ... 1 For (a) we do the same as in the (known) case m=1, namely, we define$$ W^{1,p}(U, \def\R{\mathbf R}\R^m) := \left\{u \in L^p(U, \R^m) : \forall 1\le i \le n\; \exists v_i \in L^p(U, \R^m) \; \forall \phi \in C^\infty_c(U) : \int_U \partial_i\phi \cdot u\, dx = -\int_U \phi \cdot v_i\, dx \right\} $$we denote the function v_i by \partial_i u, and for ... 1 Note that u(x)=(1-|x|^2)^\alpha\in C^{\infty}(B(0,1)), and u=0 on \partial B(0,1). It follows, that generalized derivative (Sobolev derivative) coincides with usual derivative$$ \frac{\partial u}{\partial x_i} = \alpha(1- |x|^2)^{\alpha-1}2x_i. $$To prove u \in W_0^{1,p} (B(0,1)) one needs to show that norm is finite$$ \|u \|_{W_0^{1,p} (B(0,1))} ...

1

No, in two dimensions the $L^2$ norm of the first derivative does not control the supremum of the function. A standard example is $$u(x,y)=\log\log (x^2+y^2)$$ (multiplied by a smooth cutoff function) which is unbounded, yet has square integrable gradient: $$|\nabla u(x,y)| = \frac{1}{r \log r},\quad r=\sqrt{x^2+y^2}$$ The above is concisely expressed ...

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