New answers tagged

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The analogue to the top proposition is true, and it is irrelevant whether or not $X$ is a finite or infinite-dimensional Banach space. The only thing that matters is that you can construct a Borel measure $\mu$ on open $U \subset X$ and that the function $f$ in question is $\mu$-measurable. This is because: The chain rule is valid for Frechet derivatives. ...


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It is true that $\|\psi_{n}\star f-f\|_{L^2}\rightarrow 0$ and $n\rightarrow \infty$ for any $f \in L^2$. If you start with the wedge $f$ as described, then $\|\psi_{n}\star f - f\|\rightarrow 0$ and $$ \|(\psi_{n}\star f)'-f'\|=\|\psi_{n}'\star f-f'\|=\|\psi_{n}\star f'-f'\|\rightarrow 0. $$ This is because, even though $f$ has only a piecewise ...


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The reason is that both sides of the inequality are continuous with respect to the $H^1$ norm. When you have two continuous function $f,g$ on some space, and $f\le g$ holds on a dense subset, then it follows that $f\le g$ everywhere. The continuity of $u\mapsto \int_{\mathbb{R}^3} |\nabla u(x)|^2 dx$ is immediate from the definition of $H^1$ norm. Let ...


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It is well known that for functionals of the form $J(u) = \int_{\Omega}f(x,\nabla u)\,dx$ a sufficient condition for sequential lower semicontinuity is that $\xi \mapsto f(x,\xi)$ is convex. Proofs of this result are a little bit involved, for a reference look at Direct Methods in the Calculus of Variation by B. Dacorogna or Modern Methods in the Calculus ...


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We have, by definition of $H^{t_\sigma}(\def\R{\mathbf R}\R^d)$, that $$ \def\norm#1{\left\|#1\right\|}\def\F{\mathscr F} \norm{u}_{H^{t_\sigma}} = \norm{ (1 + \def\abs#1{\left|#1\right|}\abs\cdot^2)^{t_\sigma/2}\F u}_{L^2} $$ We have for $\frac 1\alpha + \frac 1\beta = \frac 12$ by Hölder \begin{align*} \norm{u}_{H^{t_\sigma}} &= \norm{ (1 + ...


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Let $M=\|u\|_{\infty}$. Since $G'$ is continuous, there exists a constant $K$ such that $|G'(s)|\leq K$ for all $s\in [-M,M]$. Thus $|G'\circ u|\leq K$; it follows that $|G'\circ u|^p$ is bounded by $K^p$. The desired conclusion follows from Hölder's inequality as you explained with a little correction (you have to change parenthesis by norms): ...


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For $\varphi \in C^{\infty}_c$ one defines $\langle\partial_{x_1}f,\varphi\rangle = -\int_{\Omega}f\partial_{x_1}\varphi$. Extending this definition to $H^1_0(\Omega)$ we obtain a linear functional on $H^1_0(\Omega)$. Continuity, i.e. boundedness, follows from Holder's inequality (together with Poincare's inequality, depending on the norm you define on ...


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Q1/ Yes. If you know the basic fact that $H^2(0,1)$ is dense in $H^1(0,1)$, then just take $v\in V$, then there is $u_n$ a sequence of $H^2$ converging to $v$. Then $\tilde{u}_n:=u_n-u_n(0)$ belongs to $H^2(0,1)\cap V$ and as $u_n(0)\to_{n\to\infty} v(0)=0$ (I guess you know that $H^1$ convergence implies uniform convergence, in dimension 1), you have ...


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Even though Matt has solved the original problem properly, I feel obligated to give a proper solution to the fixed version. This is just for completeness: Step 1: By scaling, we may suppose that $t=1$. This is a calculation just like the one done by Matt above. Step 2: By translating $f$ if necessary, we may restrict ourselves to the evaluation of $|T_1 ...


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So we're in dimension $n=3$ according to the problem statement. I agree with you that we can get the inequality without the $1/t$ factor basically by Cauchy-Schwarz. But am I being silly or does the stated inequality not hold by scaling? Suppose the inequality were true. Let $f_{t}=f(\cdot/t)$. Then ...


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Is might be of interest that the $d$-fold tensor product of $W^k_2(\mathbb{R})$ is not only dense in $W^k_2(\mathbb{R}^d)$ (for any k and d) but in the space $W^k_{2,\text{mix}}(\mathbb{R}^d):=\{ f : \partial_{\alpha}f\in L^2 \quad \forall |\alpha|_{\infty}\leq k\}$(equipped with the obvious norm). Note that this space has a stronger norm than ...


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To answer your edit: once you have found the correct domain of definition for the Laplacian, you have that $(I-t\Delta)^{-1} = \sum \limits _{k=0} ^\infty t^k \Delta^k$. Since $\int (\Delta u) v = \int u (\Delta v)$ (on test functions, at least), you may use induction and prove that $\int (\Delta^k u) v = \int u (\Delta^k v)$, so $\int (\sum \limits _{k=0} ...


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In the sense of (1), it can be defined for $u_i,v\in L^1_{\text{loc}}(\Omega)$. More generally, it can be defined for arbitrary distributions. There are $u \in L^2(\Omega; \mathbb{R}^d)$, such that there is no $v \in L^1_{\text{loc}}(\Omega)$ with $v = \operatorname{div} u$. But $u$ has always a distributional divergence. If each component of $u$ is weakly ...


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The answer is yes if $u$ has zero-boundary values and $\Omega$ is sufficiently regular. This follows from $H^2$-regularity of \begin{align}-\Delta u &= f \text{ in } \Omega \\ u &= 0 \text{ on }\partial\Omega\end{align} with $f = -\Delta u \in L^2(\Omega)$. Otherwise it may not hold.


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Here is a suggestion for a proof. Some details are missing, but I am confident that these can be filled. Let us assume that there is a functional $J : H_0^1(\Omega) \to \mathbb{R}$, such that $J$ is Fréchet-differentiable at $u \in H_0^1(\Omega)$ with $$J'(u)\,v = \int_\Omega \nabla u \cdot \nabla v - f(u,\nabla u)\,v\,\mathrm{d}x$$ for all $v \in ...


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I think it's implicitly assumed that $k\ge 2$; otherwise the gradient need not be in $L^2$. To begin with, all the three limits of $\{\phi_j\}$ are the same. Indeed, let $\phi$ be its weak limit in $W^{1,k}$. Since the embedding into $L^2$ is compact (and compact operators map weakly convergent sequences to strongly convergent ones), we have $\phi_j\to ...


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Take $v_\epsilon \equiv 1$. Then, your assumption is satisfied for all $a,b > 0$. But $$\frac1{(2t(\epsilon))^2} \int_{-t(\epsilon)}^{t(\epsilon)}1 \mathrm{d}s = \frac1{2t(\epsilon)} \to \infty$$ as $\epsilon \to 0$. After the question has completely changed, still a similar construction is possible: $$v_\epsilon(t) = \max\{1, n_\epsilon \, |t|\}$$ with ...


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Hint: the derivative of $(I+t\Delta)^{-1}$ w.r.t. $t$ is $-(I+t\Delta)^{-1}\Delta(I+t\Delta)^{-1}$.


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You probably know that the Fourier transform takes derivatives to polynomials, ie $F (\partial_i u) = \xi_i \cdot Fu$. So, if all $m$-th order derivatives are in $L^2$ you get that $(1 + |\xi|^2)^{m/2} Fu$ is in $L^2$. This provides an easy way to define Sobolev spaces for $\mathbb{R}^d$. One gets a natural generalization to arbitrary $m \in \mathbb{R}$. ...


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Its the distributional divergence, that is for $v \in L^2(\Omega)$ we have $v = \operatorname{div}u$ iff $$\int_\Omega v \, \varphi \, \mathrm{d}x = -\int_\Omega u \cdot \nabla \varphi \, \mathrm{d}x \quad\forall \varphi \in C_0^\infty(\Omega).$$ To be compared with https://en.wikipedia.org/wiki/Integration_by_parts#Higher_dimensions.


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Rewrite the inequality that you want to prove as $$\|u\|_{p}^{p/2}\|(-\Delta)^{\sigma/4}|u|^{\frac{p+m-1}{2}}\|_{2}\geq C\|u\|_{\frac{N(2p+m-1)}{2N-\sigma}}^{\frac{2p+m-1}{2}}$$ Set $v:=|u|^{\frac{p+m-1}{2}}$. The LHS above becomes $$\|v\|_{q}^{\alpha}\|(-\Delta)^{\sigma/4}v\|_{2}$$ with $q=\frac{2p}{p+m-1}$ and $\alpha=\frac{p}{p+m-1}$. Apply the NGN ...


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There are a few related concepts here: 1) A bump function is a term for a smooth function with compact support. The set of all bump functions forms a vector space. If these functions are on $\mathbb{R}^n$, then it is often denoted $C_c^\infty(\mathbb{R}^n)$. In distribution theory, this is what is most commonly referred to when one refers to test functions, ...


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A mollifier is a function $f$ that you convolve with another function $g$ to get a function which is "close" to $g$ but "nicer". For instance $f$ might be a general $L^1$ function and $g*f$ might be a smooth, compactly supported approximation to $f$. Really a mollifier is not one function but a sequence, or even sometimes a one-parameter continuous family. ...


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The assumption is implicitely used here: \begin{align*} \int_{\mathbb{R}^n}\int_{\Omega'} \tilde{u}(x-y) J_{\varepsilon}(y) D^{\alpha}\phi(x){\rm d}x {\rm d}y &= \int_{B_\varepsilon(0)}\int_{\Omega'} \tilde{u}(x-y) J_{\varepsilon}(y) D^{\alpha}\phi(x){\rm d}x {\rm d}y \\ &= (-1)^{\lvert \alpha \rvert} \int_{B_\varepsilon(0)}\int_{\Omega'} ...


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No, you can't. Consider $\Omega$ being a unit circle on $\mathbb{R}^2$. Then: $$ \int\limits_{\Omega} \sum_{|\alpha|=2} | D^{\alpha}u | = \int\limits_{x^2 + y^2 < 1} \left( \left| \frac{\partial^2 u}{\partial x^2} \right| + \left| \frac{\partial^2 u}{\partial x \partial y} \right| + \left| \frac{\partial^2 u}{\partial y^2} \right| ...


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It will also depend on the dimension of $\Omega$. Roughly speaking, the embedding theorem works the same for fraction ordered Sobolev space. Please check Theorem 6.14 from this book, and some related theorems from there as well.


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Since $v(0) = v(a) = 0$, you have $$v(x) = \int_0^x v'(x) \, \mathrm{d} x = -\int_x^a v'(x) \, \mathrm{d} x.$$ Hence, $$|v(x)| \le \int_0^x |v'(x)| \, \mathrm{d} x$$ and $$|v(x)| \le \int_x^a |v'(x)| \, \mathrm{d} x.$$ Thus, $$|v(x)| \le \min\{\int_0^x |v'(x)| \, \mathrm{d} x, \int_x^a |v'(x)| \, \mathrm{d} x\} \le \frac12 \int_0^a |v'(x)|\,\mathrm{d}x.$$


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if $p\in[1,6)$, we have $\nabla u_n\to \nabla u$ weakly in $L^2$ implies $u_n\to u$ strongly in $L^p$ (this is sobolev compact embedding). Hence, you have l.s.c. of $F$. if $u_n\to u$ strongly in $H^1$, it implies that $u_n\to u$ strongly in $L^6$. (Note that if $u_n\to u$ weakly in $H^1$, you only have $u_n\to u$ strongly in $L^p$ for $p<6$, like in ...


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For $f\in H^1(\Omega)$, you already know that the weak derivatives $f_\alpha := \partial^\alpha f$ exist on all of $\Omega$. Thus, you only need to show $f_\alpha \in L^p(\Omega)$. For this, it suffices to show (why?) $f\alpha \in L^p(\Omega_1)$ and $f_\alpha \in L^p(\Omega_2)$, which follows (apparently) from Sobolev embedding.


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I have found the answer in Kobayashi & Nomizu, volume 1, page 124. If $K$ is a tensor of type $(r,s)$, then one may construct a new tensor $\nabla K$ of type $(r, s+1)$, defined by $$(\nabla K) (X_1, \dots, X_s, Y) = (\nabla _Y K) (X_1, \dots, X_s)$$ and thus define inductively $\nabla ^k K$ as $\nabla (\nabla ^{k-1} K)$. Choosing now $K$ to be $f$, a ...


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At the beginning of this section, Evans makes the assumptions $$ a^{ij},b^i,c\in L^\infty(U), f\in L^2(U). $$ The following inequalities are crucial: if $w\in H^1(U)$, then $$ \|w\|_L^2(U) \leq \|w\|_{H^1(U)}, \|w_{x_i}\|_{L^2(U)} \leq \|w\|_{H^1(U)}. $$ This is pretty much immediate from the definition of the Sobolev norm. You may need to tack on ...


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Remember that $f \in W^{1,1} (\gamma_1)$ means that $f, f' \in L^1 (\gamma_1)$ where $f'$ is understood in the weak sense. Remember also that $f'$ in the Sobolev sense in the same as $f'$ in the distributional sense, therefore let us compute the latter one. Finally, we shall use that $\gamma_1 = \dfrac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2}$. Note, ...


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The functions $U$, $U_x$ and $U_y$ are locally bounded, except possibly at $(x,y)=(0,0)$. If $B_1$ is the unit ball centred at $(0,0)$, We have $$ \int_{B_1} \lvert u\rvert = \int_0^{2\pi}\int_0^1 r\lvert u(r\cos\vartheta,r\sin\vartheta)\rvert\,dr\,d\vartheta= \int_0^{2\pi}\int_0^1 r\lvert U(r)\rvert\,dr\,d\vartheta=2\pi\int_0^1 r\lvert U(r)\rvert\,dr. $$ ...


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As stated, existence is trivial: $u\equiv 0$ is a solution. When a nontrivial solution exists, it is not unique since one can multiply it by a constant. But this happens only for some specific $\lambda$: those that are in the spectrum of the Dirichlet Laplacian. You can't infer this kind of structure from Lax-Milgram. Rather, sine Fourier series should be ...


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You can't just choose the function to be identically zero outside $U$ in general, because this can cause the weak derivative to fail to be $L^p$ because of "bad regularity" at the boundary. For instance when $n=1$, all $W^{k,p}$ functions are actually continuous, so extending $f(x)=1$ on $(0,1)$ to be identically zero elsewhere certainly does not give a ...


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It's false. For a counterexample consider $n = k = p = 1$, $U = (0,1)$ and $f \equiv 1$. It is possible to extend Sobolev functions under certain regularity assumptions on the boundary of the set $U$, but it is delicate. Any book on Sobolev spaces addresses the issue.



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