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This just uses the definition of weak derivative. The trick is to notice that if $u,\phi$ are both in $L^2(\mathbb R)$ then $$\int_{\mathbb R} \frac{u(x+h) - u(x)}{h} \phi(x) \, dx = \int_{\mathbb R} u(x) \frac{\phi(x-h) - \phi(x)}{h} \, dx \tag{\ast}.$$ If (2) holds, you can use the dominated convergence theorem and ($\ast$) to see that $$... 3 You need to prove that uv\in H^1(\mathbb{R}), or equivalently, that uv\in L^2(\mathbb{R}) and there is g\in L^2(\mathbb{R}) such that$$\int_\mathbb{R}(uv)\varphi'=-\int_\mathbb{R}g\varphi,\ \forall\ \varphi\in C_0^\infty(\mathbb{R}).\tag{1}$$The first question is: does uv\in L^2(\mathbb{R})? The answer is yes, because H^1(\mathbb{R}) is a subset ... 2 No, this is not true. You only have$$(W^{1,p}_0(\Omega))^*=W^{-1,p}(\Omega) $$However, the dual space of W^{1,p}(\Omega) is not identified, although it is smaller then W^{-1,p}(\Omega). For more information, please read 10.4 in Leoni's book, it has a complete treatment of dual of Sobolev space. 2 The usual way is with cutoff functions. Let \phi \in C^\infty_c be a smooth compactly supported function which equals 1 on the unit ball, and let \phi_n(x) = \phi(x/n). Then set f_n = f \phi_n. Now f_n is C^\infty_c, converges to f pointwise (indeed f_n(x) = f(x) as soon as n \ge |x|), and |f_n| \le \|\phi\|_{\infty} |f| so by dominated ... 2 The derivative of u is$$u'(x)= \begin{cases} 2x\cos\frac{1}{x} + \sin\frac{1}{x}&0<x\leq 1\\ 0 & x=0 \end{cases}.$$In particular, note that since \left|h\cos\frac{1}{h}\right|\leqslant |h|,$$u'(0) = \lim_{h \rightarrow 0} \frac{h^2\cos\frac{1}{h}-0}{h}=\lim_{h \rightarrow 0} h\cos\frac{1}{h}=0.$$So u has a bounded derivative and ... 2 Fix some r_0 so that B(x_0,r_0)\subset U. If you know that (0,r_0)\ni r\mapsto u_{x_0,r} is differentiable and satisfies$$ \left|\frac{d}{dr}u_{x_0,r}\right|\leq Cr^{\frac{\varepsilon}{p}-1}, $$you can use the fundamental theorem of calculus to observe that$$ u_{x_0,r}=u_{x_0,r_0}-\int_r^{r_0}\frac{d}{ds}u_{x_0,s}ds $$for all r\in(0,r_0) and the ... 2 It may be helpful to notice that if function f is lipschitz and f(0)=0 then f\circ u is sobolev with \partial_i (f\circ u)(x)=f'(u)\partial_i u. Hence, you could use$$f(x):= \begin{cases} x & |x|<k\\ k & x>k\\ -k & x<-k \end{cases} $$as your truncation function. Of course f satisfies all conditions I wrote above and of ... 2 The function x\mapsto \|x\|^{2s-4}x_i^2 is homogeneous of degree 2s-4+2=2s-2. So it is integrable on the unit ball if and only if 2s-2>-n. This is a special case of general fact about homogeneous functions. (I.e., those with f(tx)=t^df(x) for all x and all t>0.) Indeed, suppose f is homogeneous of degree d and is not zero a.e. ... 2 Let B=B(0,1/2) and we recall the definition of weak partial derivative. We say a function g\in L^1_{\text{loc}}(B) is weak derivative of f if for any \phi\in C_c^\infty (B) we have$$\int_B f\partial_i \phi\,dx=-\int_B g\,\phi\,dx $$So we need to find out what is g in your question. Let us suppose x\neq 0 and we compute the classical ... 2 Uniformly continuity is not really a very restrictive condition: All continuous functions restricted on \Omega' \subset \subset \Omega is automatically uniform continuous. To write down a counterexample, consider the Cantor's staircase function f, which is a monotone function on (0,1) and is uniform continuous. But f is not in W^{1, 1}(0,1) as it ... 1 I think it is. Let 1<p<\infty be given and notice that the mapping T from W^{1,p}(\Omega)\to L^p(\Omega, R^{N+1}) via$$ T[u]\to (u,\nabla u) $$is isomorphic and closed. Together with the fact that L^p(\Omega, R^{M}) is uniformly convex for any M\geq 1, here we are interested in the case M=N+1, hence we know that W^{1,p} is uniformly ... 1 Look in Chapter 5 of Partial Differential Equations by Evans. Of course the inequality requires |x - x_0| \le r but the proof in Evans may have a stricter requirement. 1 I think @PhoemueX's answer will lead you to a general situation. It works, of course. But here let me provide you a quick and insight example on R^1. Take I=(0,1) and define u_n in following way: For each fixed n, we partition I into n's small subinterval with length 1/n. Let's name those interval by I^n_i:=(i/n,i+1/n) for ... 1 Another example. Let \Omega=(0,1)\subset\mathbb{R} and f_n(x)=\sin(n\,\pi\,x)/n. Then$$ \|f_n\|_2\le\frac1n\to0\text{ as }n\to\infty. $$On the other hand$$ \|f'_n\|_2^2=\pi^2\int_0^1\cos^2(n\,\pi\,x)\,dx=\frac{\pi^2}{2}. $$1 These bump functions, or test functions, are extremely important in distribution theory. They can be constructed using partitions of unity. I didn't find any good references online in this context, but if you can get hold of Hörmanders The Analysis of Linear Partial Differential Operators I'', then it's an excellent reference. 1 First of all, f(x)=|x|^r is a very good smooth function and hence you should expect that classical derivative equal to weak derivative a.e., if there is any. Now, the only thing you worry about is the singularity at 0. However, as weak derivative, we never care a value at a single point. That is, we could always define \nabla f(x)=\nabla |x|^r if ... 1 There are two ways we could fix this. The fastest way is noticing that U is an extension domain and we could extend u to \bar{u}\in W^{1,p}(R^N) such that \bar{u}=u inside U. Next, we could use (v_n)\subset C_c^{\infty}(R^N) to approximate \bar{u} in W^{1,p} by the fact that W_0^{1,p}(R^N)=W^{1,p}(R^N). Then v_n restrict to U will do ... 1 Here is what I thought. I am not very sure but I am happy to discuss with you. The way we cast \Delta^2u=0 into a weak formulation, so that we could use Lax-Milgram, tells us that H_0^2 is a suitable space. Suppose we have a nice solution already, then we test \Delta^2u=0 with a C^\infty function v and see what happens. We have$$\int_\Omega ...

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Let $C_n=q_1^n|_\Sigma-q_2^n|_\Sigma$ be the the sequence of constant functions, which as you alread have noted, converge to $q_1|_\Sigma-q_2|_\Sigma$ in $H^{1/2}(\Sigma)$. Therefore, $C_n \to q_1|_\Sigma-q_2|_\Sigma$ in $L^2(\Sigma)$, or equivalently, $$\int_\Sigma |C_n-(q_1|_\Sigma-q_2|_\Sigma)|^2d\Sigma\to 0.\tag{1}$$ There are some ways to prove now ...

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(This is just a comment without too many details, and I'm not sure whether it will help you, but it is too long to fit into a comment, so I used the answer field) If you want to do something like that on a manifold, then you need to say what $|.|$ is, i.e. you need a metric, so you are probably referring to Riemannian manifolds. If you have a compact ...

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This is a quick observation I am not sure this is what you want. If you go on and compute second derivative, you will have $$\partial_j\partial_i f(x)=\phi''(|x|/n)\frac{1}{n^2}\frac{x_ix_j}{|x|^2} +\phi'(|x|/n)\frac{\delta_{ij}|x|-x_ix_j}{|x|^3}\frac{1}{n}$$ Hence, by induction, you could have $$|D^\alpha f(x)|\leq \sum_{i=1}^k ... 1 I think you probably mean \xi > \eta? The overall inequality is$$ |u(\xi) - u(\eta)|^2 \leq (b-a)\int_a^b (u')^2~dx. $$Write this as$$ u(\xi)^2 - 2 u(\xi)u(\eta) + u(\eta)^2 \leq (b-a)\int_a^b (u')^2~dx. $$Integrate in \xi from a to b. u(\eta)^2 does not depend on \xi, so we obtain a factor of (b-a). Similarly with the integral on the ... 1 You have showed that u'''=0, hence u\in H^3(I). Moreover, all weak derivatives of order greater than 3 are zero, too. Hence u\in H^k(I) for all k. The function v(x) = \frac16 x |x|^2 is in H^2(I)\setminus H^3(I). It holds v''(x)= |x|, and T_u(\phi''')=-\phi(0). 1 Formally (and by this I mean: applied in smooth functions) the p-Laplace operator is defined by$$\Delta_pu=\operatorname{div}(|\nabla u|^{p-2}\nabla u).$$If h(x)=\frac{u(x_0+dx)}{d}, we see from the previous equality that$$\Delta_p h(x)=\frac{\operatorname{div}(|\nabla u(x_0+dx)|^{p-2}\nabla u(x_0+dx)}{d^{p-1}},$$so if \Delta_p u=f, it is ... 1 Fix \delta>0 such that the Hausdorff pre-measure \mathcal H_\delta^{n-1}(A_\epsilon^N) is at least \frac12 \mathcal H ^{n-1}(A_\epsilon^N). For each x\in A_\epsilon^N, pick r_x<\delta/2 so that$$\|D(f-M)^+\|( B(x,r_x))>\epsilon r_x^{n-1}$$The balls  B(x,r_x) form a Besicovitch cover of A_\epsilon^N; the Besicovitch covering theorem ... 1 so convergence in Lp implies there exists subsequence unk converging pointwise a.e. to u, and convergence in L∞ implies unk converges uniformly (and hence pointwise) a.e. to some v, so that v=u a.e.. 1 For your first question, no H_0^1(\mathbb{R}^n) is not a Hilbert space with the inner product (u,v)=\int \nabla u \cdot \nabla v. Its completion however is, and is usually denoted by \mathcal{D}^{1,2}(\mathbb{R}^n) or L^{1,2}(\mathbb{R}^n) and consists of functions in L^{2^*}(\mathbb{R}^n) with integrable gradient, where 2^*=2n/(n-2) is the usual ... 1 This is a long comment. Quote from Hlawka's functional inequality: Moreover, Witsenhausen showed that the space L^p(0, 1) is a Hlawka space for 1\le p\le 2. Therefore, one can see that all Banach spaces having the property that all its finite dimensional subspaces can be embedded linearly and isometrically in the space L^p([0, 1]), with some 1\le ... 1 First of all, the problem should be$$u(x):=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right) $$but not$$u(x)=\ln\left(\ln\left(\frac{1}{1+|x|}\right)\right)$$as you stated. Next, we have$$ \partial_i u(x) = = \frac{1}{\ln(1+\frac{1}{|x|})}\frac{x_i}{|x|^3}\frac{1}{1+\frac{1}{|x|}}$$Hence we have$$ |\nabla u| \approx ...

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