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In the definition of $Z$, you probably want $|v|>N$ instead of $v>N$. Also, in item 1, the definition of $B$, you have the Sobolev norm of $x$, so it's better to use norm notation for that. Let's also not use subscripts in superscripts... say, $p<q$ and the embedding is $H^q\to H^p$. The $H^q$ norm is given by $$\|f\|_{H^q}^2 = ... 3 "Locally" is ambiguous here f is locally Lipschitz in \Omega if and only if f \in W^{1,\infty}_{loc}(\Omega) The validity of this claim depends on interpretation of "locally Lipschitz". Does it mean every point of \Omega has a neighborhood in which f is Lipschitz, or there is L such that every point of \Omega has a neighborhood in ... 3 I think smoothness is the wrong term to focus on; the difference concerns the continuity of f. To get first-order classical derivative of f, we would need f\in W^{2,p} with p>n; compare with item 4 below. The fact that f has a weak derivative makes it locally absolutely continuous on almost every line. If the derivative is also in ... 3 Forming the convolution of the (scaled) bump function with indicator functions you get "qausi-indicator functions" in \mathscr D, in particular, there are \psi_n\in\mathscr D(\mathbb R) which are positive and equl to 1 on [-n,n]. It is then easy to see that the elements of E^{loc} are those distributions which, on every compact set, have the same ... 2 Integration in polar coordinates (or spherical, in higher dimensions). Write z=x+\rho\theta where \theta\in S^{n-1} is a unit vector and \rho\in [0,r]. Then for any integrable f we have$$ \int_{B(x,r)} f(z)\,dz = \int_0^r \int_{S^{n-1}} f(x+\rho \theta) \rho^{n-1}\,d\theta \,d\rho \tag{1} $$(Compare with n=2 case, when this is the usual polar ... 2 It seems to me that one can easily extend the definition so that, for each f\in L^1, for almost all a, \int\delta(x-a)f(x)\,dx is defined; just define it to be the limit in the Lebesgue density theorem. But you seem to want the quantifiers in the other order: For almost all a, for all f\in L^1, \dots. I see no reasonable way to get this. That ... 2 (Too long for a comment) In my opinion, this is an interesting curiosity and I think the correct answer is what Andreas has said - simply define \langle \delta_a,f\rangle as the Lebesgue limit. This might save you some space if you need to say something like "the value of f at x" in the context of an L^1 function (for which function values are not ... 2 Observe that$$\int_{\partial B(0,r)} u^2 \nu \cdot \frac x{|x|^2} \, dS = \frac{1}{r}\int_{\partial B(0,r)} u^2 \, dS $$because \nu is pointing in the direction of x, which makes \nu\cdot x = |x|. After you plug this into the equation before "therefore", all that's left to do is to rearrange the term. The integral with u^2/|x|^2 goes to the left ... 2 Yes. x \rho_t(x) is smooth and compactly supported, and any such function f is in H^s for every s. The easiest way to see this may be this: Let k be any integer larger than s. Verify from the definition that H^k \subset H^s. (Note that (1+|\xi|^2)^s \le (1+|\xi|^2)^k). Using the fact that the Fourier transform takes differentiation to ... 2 The trace theorem, as stated by Evans: Assume U is bounded and \partial U is C^1. Then there exists a bounded linear operator T:W^{1,p}(U)\to L^p(\partial U) such that Tu=u_{|\partial U} if u\in W^{1,p}(U)\cap C(\overline{U}) \|Tu\|_{L^p(\partial U)}\le C\|u\|_{W^{1,p}(U)} for all u\in W^{1,p}(U), with C depending only on ... 2 This proof cover all cases: a,b finite or not. In the case that a,b are not finite, u(a) is understood as \lim_{x\to-\infty}u(x). Analogous for b. Also, if a,b are not finite, then we weill consider locally things, i.e. BV_{loc}((a,b)). I am also assuming that H_0^1((a,b)) is the closure of C_0^1((a,b)) with respect to the H^1((a,b)) ... 1 No, you can't have p<1 there. Take the constant function u\equiv \lambda. Your inequality becomes$$\|\lambda\|_{2^*}\le \|\lambda\|_1^p$$which (if p< 1) fails when \lambda is large enough. When you imagine an inequality you'd like to be valid, consider how it scales when u is multiply by a positive number, or (when working on vector ... 1 First, let a\in\mathbb{R} be arbitrary. As you noted, this implies$$ \limsup_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}\leq\lim_{n\rightarrow\infty}\left\Vert f-a\right\Vert _{L^{p}\left(\Omega_{n}\right)}=\left\Vert f-a\right\Vert _{L^{p}\left(\bigcup\Omega_{n}\right)}, $$and thus$$ \limsup_{n\rightarrow\infty}\left\Vert f\right\Vert ...

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Poincaré inequality holds for every subspace of $W^{1,p}(\Omega)$ which has compact embedding in $L^p$ and does not contain constants. Let me be more precise. Theorem. Let $\Omega$ be an open, Lipschitz, bounded, connected set in $\mathbb R^d$ and let $p \in [1,+\infty)$. Let $W \subset W^{1,p}$ be a subspace which has compact embedding in ...

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An element of $L^{1}$ is an equivalence class of functions which are equal a.e.. Elements of $L^{1}$ don't have pointwise values. Point values can make sense for specific cases, such as when there is a function in the equivalence class which happens to be continuous--that's because if there is such a function in the equivalence class, then there cannot be a ...

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Let $X$ be a Banach space and $u\in L^1(0,T,X)$. We say that $u'\in L^1(0,T,X)$ the weak derivative of $u$ if $$\int_0^T u(t)\phi'(t)dt=-\int_0^T u'(t)\phi(t)dt,\forall\ \phi\in C_0^\infty(0,T).$$ According to this definition, the problem is not that $u'(t)$ does not belong to $H^1$, it does belong to $H^1$. The question is, if it belong to $L^\infty$, ...

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First, you apply Hölder to get $$\|u(t)\| \leq \|u(s)\| + \int_0^T\|u'(\tau)\|\,\mathrm{d}\tau \leq \|u(s)\| + T^{1-1/p}\|u'\|_{L^p(0,T;X)} .$$ Then, taking the $p$-th power of both sides, and integrating with respect to $s$, we have $$T\|u(t)\|^p \leq c\int_0^T\|u(s)\|^p\,\mathrm{d}s + cT\cdot T^{p-1}\|u'\|_{L^p(0,T;X)}^p = c\|u\|_{L^p(0,T;X)}^p + ... 1 Start with an example. So, look at \Omega = [0,\pi]\subset \mathbb{R}^{1}, where you have an orthonormal basis of eigenfunctions for -\Delta=-\frac{d^{2}}{dx^{2}} given by \{ e_{n}(x)=\sqrt{2/\pi}\sin(nx)\}_{n=1}^{\infty}. To solve the equation in this case, write the solution as$$ u(x,t) = \sum_{n=1}^{\infty}a_{n}(t)e_{n}(x) $$Then$$ ...

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Only large frequencies matter for smoothness. For every $M$, the part of Fourier transform with $\{\xi:|\xi|\le M\}$ contributes a real-analytic term to the function. You know that integrability of $|\xi|^\alpha \hat u(\xi)$ implies certain smoothness of $u$. So you want to show that this product is integrable for every $\alpha$. On every ball \$\{|\xi|\le ...

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