# Tag Info

0

This is not true, even for this particular $F$. To see it, let $Q:H^1_0(\Omega)\to\mathbb{R}$ be the linear functional, defined by $$Q\phi=\int_\Omega \nabla w\nabla \phi.$$ If $w\neq 0$ then, the image of $Q$ is $\mathbb{R}$. Therefore, there is $v\in H_0^1(\Omega)$ such that $Qv\neq 0$. Note that for this particular $v$, we have that $Tv=0$, because ...

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

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 ...

0

First of all, I think you mean $H_0^1$ is the closure of $C^1$, not $H^1$. Secondly, your PDE will solved by using second existence theorem or directly by Lax-Milgram theory by using Bilinear operator. For more information, I suggest you to read our bible Evans book, chapter 6. Look for First existence theorem and second existence theorem

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

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 $T_u(\phi''')=-\phi(0)$.

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

The weak derivative only needs to be defined almost everywhere. If the set of singularities has measure $0$ then you can define it to be whatever you like at those singularities. Remember, a weak derivative is defined in terms of integration, so you may throw out sets of measure zero as you wish.

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.

0

Define $T:W^{s,p}(\Omega)\to L^p(\Omega)\times L^p(\Omega\times\Omega)$ by $$T(u)=\left(u,\frac{|u(x)-u(y)|}{|x-y|^{N/p+s}}\right),$$ where $W^{s,p}(\Omega)$ is equipped with the norm given by you and $L^p(\Omega)\times L^p(\Omega\times\Omega)$ is equipped with the norm $$\|(u,v)\|=\left(\int_\Omega |u|^p+\int_{\Omega\times\Omega}|v(x,y)|^p\right)^{1/p}.$$ ...

0

I think that you forgot $\|\xi\|^2$ in the definition the uniformly positive definite matrix. With this adjustment what you want to prove is straightforward.

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.

4

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 $$... 2 Uniformly continuity is not really a very restrictive condition: All continuous functions restricted on \Omega' \subset \subset \Omega is automatically uniform continuous. To write down a counterexample, consider the Cantor's staircase function f, which is a monotone function on (0,1) and is uniform continuous. But f is not in W^{1, 1}(0,1) as it ... 1 This is a long comment. Quote from Hlawka's functional inequality: Moreover, Witsenhausen showed that the space L^p(0, 1) is a Hlawka space for 1\le p\le 2. Therefore, one can see that all Banach spaces having the property that all its finite dimensional subspaces can be embedded linearly and isometrically in the space L^p([0, 1]), with some 1\le ... 1 This is a quick observation I am not sure this is what you want. If you go on and compute second derivative, you will have$$\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 ...

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 ...

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 ...

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 ...

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

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

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 $$\int_\Omega ... 0 I realized that u=0 on \partial\Omega and \nabla u\cdot \nu=0 imply \nabla u=0 on \partial\Omega for smooth u. So taking H^2_0 as space where we seek the weak solution, i.e. intuitively requiring \nabla u=0 on \partial\Omega, is NOT more than \nabla u\cdot \nu =0 1 First of all, f(x)=|x|^r is a very good smooth function and hence you should expect that classical derivative equal to weak derivative a.e., if there is any. Now, the only thing you worry about is the singularity at 0. However, as weak derivative, we never care a value at a single point. That is, we could always define \nabla f(x)=\nabla |x|^r if ... 0 Sorry to bother here but I double the answer by @user127096. Let me quote the result from H. Brezis's book. This book states that, in Proposition 9.20, for v\in H^{-1}(\Omega), there \exists \, v^0,v^1, \dots , v^n in L^2(U) such that$$\langle v,u\rangle=\int_Uv^0u+\sum_{i=1}^{n}v^{i}u_{x_i} \,dx.$$and MOREOVER: (1):  \, v^0,v^1, \dots , v^n ... 1 Fix \delta>0 such that the Hausdorff pre-measure \mathcal H_\delta^{n-1}(A_\epsilon^N) is at least \frac12 \mathcal H ^{n-1}(A_\epsilon^N). For each x\in A_\epsilon^N, pick r_x<\delta/2 so that$$\|D(f-M)^+\|( B(x,r_x))>\epsilon r_x^{n-1}$$The balls  B(x,r_x) form a Besicovitch cover of A_\epsilon^N; the Besicovitch covering theorem ... 0 No, we only define the dual of (H_0^1(M)) is H^{-1}(M), but not H^1(M). Although the dual of H^1(M) does exist, and it is smaller then H^{-1}(M). If you want an intuitive explanation, I recommend you to read the definition by Fourior transform then you will see the dual of H^s is H^{-s}, not by definition but by computation. 1 (This is just a comment without too many details, and I'm not sure whether it will help you, but it is too long to fit into a comment, so I used the answer field) If you want to do something like that on a manifold, then you need to say what |.| is, i.e. you need a metric, so you are probably referring to Riemannian manifolds. If you have a compact ... 0 I think your "result" is not true. Take \Omega = (0,1) for simplicity. Then,$$\int_0^1 f \, v' \, \mathrm{d} x = 0\quad\forall v \in H_0^1(\Omega)$$just implies that f is constant. 0 You can take any orthonormal sequence \{f_n\} in H^1. Then you have f_n \rightharpoonup 0, but \|f_n\|_{H^1} = 1. Using the compact embedding from H^1 to L^2, you have f_n \to 0 in L^2. More generally, you can construct a counterexample by using any weakly but not strongly convergent sequence in H^1. 1 Another example. Let \Omega=(0,1)\subset\mathbb{R} and f_n(x)=\sin(n\,\pi\,x)/n. Then$$ \|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}. $$1 I think @PhoemueX's answer will lead you to a general situation. It works, of course. But here let me provide you a quick and insight example on R^1. Take I=(0,1) and define u_n in following way: For each fixed n, we partition I into n's small subinterval with length 1/n. Let's name those interval by I^n_i:=(i/n,i+1/n) for ... 2 No, this is not true. You only have$$(W^{1,p}_0(\Omega))^*=W^{-1,p}(\Omega) $$However, the dual space of W^{1,p}(\Omega) is not identified, although it is smaller then W^{-1,p}(\Omega). For more information, please read 10.4 in Leoni's book, it has a complete treatment of dual of Sobolev space. 0 Actually you can get, for a fixed \epsilon>0, that for any u\in W^{m,p}(\Omega)$$ \|{u}\|_{W^{m,p}}\leq \epsilon\|{D^\alpha u}\|_{L^p}+C_{\epsilon}\|{u}\|_{L^q}, $$where |{\alpha}|=m. Usually we just take q=p but if domain is good enough and by embedding we could extend q to where embedding would go. This theorem states that the extreme term ... 0 We have \|\partial^\beta u\|_{L^p(U)} \leq \|u\|_{W^{k-1,p}(U)} Also, by the theorem, W^{k,p}(U) \rightarrow W^{k-1,p}(U) is compact. Now, suppose there is no C_\epsilon. Then we get a sequence u_n \in W^{k,p}(U) (normalize so these are all norm 1) violating it with n in place of C_\epsilon. That is, \|u_n\|_{W^{k-1,p}} \geq \epsilon + n ... 1 so convergence in Lp implies there exists subsequence unk converging pointwise a.e. to u, and convergence in L∞ implies unk converges uniformly (and hence pointwise) a.e. to some v, so that v=u a.e.. 0 You can use the fact that L^p spaces are a Radon-Riesz spaces, for p>1. and the case p=1 i don't have a proof but you can use the fact that each of the L^p-spaces, 1 ≤ p < ∞, has the property that each sequence on the unit sphere that converges almost everywhere converges also in norm. 1 Let C_n=q_1^n|_\Sigma-q_2^n|_\Sigma be the the sequence of constant functions, which as you alread have noted, converge to q_1|_\Sigma-q_2|_\Sigma in H^{1/2}(\Sigma). Therefore, C_n \to q_1|_\Sigma-q_2|_\Sigma in L^2(\Sigma), or equivalently,$$\int_\Sigma |C_n-(q_1|_\Sigma-q_2|_\Sigma)|^2d\Sigma\to 0.\tag{1}$$There are some ways to prove now ... 2 The function x\mapsto \|x\|^{2s-4}x_i^2 is homogeneous of degree 2s-4+2=2s-2. So it is integrable on the unit ball if and only if 2s-2>-n. This is a special case of general fact about homogeneous functions. (I.e., those with f(tx)=t^df(x) for all x and all t>0.) Indeed, suppose f is homogeneous of degree d and is not zero a.e. ... 0 First notice that, by Hölder's inequality, we have cu\in L^p for 1/p=1/q+1/6 (since u\in L^6 by the Sobolev embedding). Next your inequality gives u\in W^{2,p} for this same p. Applying Sobolev embedding again, we get \nabla u \in L^{p*} where as usual 1/p^* =1/p-1/n. If we substitute we get$$ \frac{1}{p^*} = \frac{1}{q} +\frac{1}{6} ...

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 $$|\nabla u| \approx ... 1 I think you probably mean \xi > \eta? The overall inequality is$$ |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. $$Integrate in \xi from a to b. u(\eta)^2 does not depend on \xi, so we obtain a factor of (b-a). Similarly with the integral on the ... 2 The derivative of u is$$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 ...

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