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2

By the fundamental theorem of calculus, we have that $$u_\delta(x)=u_\delta (c)+\int_c^xu_\delta'(t)dt,$$ Can you conclude now?

4

(1) Yes, your argument is correct. The fact that composition with $T^{-1}$ preserves Sobolev classes also needs to be proved, but the proof is immediate from consideration of what this composition does to Cauchy sequences (wrt $W^{1,p}$ norm) of smooth functions. (2) Yes, and this generalization is one of fundamental results for the theory of Sobolev ...

1

The space $W_0^{1,\infty}(\Omega)$ consists of functions that tend to $0$ at the boundary of $\Omega$; the limit is understood in the classical way because $W^{1,\infty}$ functions have a continuous representative. See the discussion here. When $\Omega=\mathbb R^n$, the role of boundary is played by the point at infinity. Therefore, the zero extension of ...

1

The problem is that $C_0^\infty(\mathbb{R}^n)$ is not dense in $W^{1,\infty}(\mathbb{R}^n)$. Particularly, a function in $W^{1,p}(\mathbb{R}^n)$ for $1\leq p<\infty$ must decay "far outside". As an example, choose $v=1$ which is clearly in $W^{1,\infty}(\mathbb{R}^n)$, but does not have a zero trace or is in the completion of $C_c^\infty(\mathbb{R}^n)$ ...

1

Functions in $H_1$ are absolutely continuous (or more precisely have representatives that are absolutely continuous). Since $H_s^{loc} \subset H_1^{loc}$ for $s > 1$ an affirmative answer to your question would imply that functions in $C(\mathbb R)$ have AC representatives, which they do not.

4

Since $(V(f))' = f$, it suffices to see that $\lVert V(f)\rVert_{L^2} \leqslant C\lVert f\rVert_{1,2}$. But that is a direct consequence of the continuity of the Volterra operator on $L^2([0,1])$, \begin{align} \int_0^1 \lvert V(f)(t)\rvert^2\,dt &=\int_0^1\left\lvert \int_0^t f(s)\,ds\right\rvert^2\,dt\\ &\leqslant \int_0^1 \left( \int_0^t \lvert ... 3 The estimate |Vf(t)| \leq \int_0^1|f| \leq \left(\int_0^1|f|^2\right)^{1/2} $$gives \|Vf\|_{L^2}\leq\|f\|_{L^2}. Since (Vf)'=f, this gives$$ \|Vf\|_{1,2}^2 = \|Vf\|_{L^2}^2 + \|(Vf)'\|_{L^2}^2 \leq 2\|f\|_{L^2}^2 \leq 2\|f\|_{1,2}^2. $$This gives continuity V:H^1\to H^1 (and L^2\to H^1). If you have trouble bounding the value of a function by ... 3 By one hand, if \Omega\subset \mathbb{R}^N is any open set then, C_0^\infty(\Omega) is dense in L^p(\Omega), as you can see, for example, in Brezis chapter 4. Once$$C_0^\infty(\Omega)\subset C^\infty(\Omega)\cap L^p(\Omega)\subset L^p(\Omega),$$the result follows. On the other hand, if -\infty<a<b<\infty, you can see in chapter 8 of ... 0 Let Lf = -if' be defined on the linear space \mathcal{D}(L) of absolutely continuous functions f \in L^{2}[a,b] for which f(a)=f(b) and f' \in L^{2}[a,b]. L is symmetric on its domain, i.e., (Lf,g)=(f,Lg) for all f,g\in\mathcal{D}(L). It is not hard to show that (L-\lambda I) is surjective for \lambda \ne n\frac{b-a}{2\pi} for n=0,\pm ... 2 This proof cover all cases: a,b finite or not. In the case that a,b are not finite, u(a) is understood as \lim_{x\to-\infty}u(x). Analogous for b. Also, if a,b are not finite, then we weill consider locally things, i.e. BV_{loc}((a,b)). I am also assuming that H_0^1((a,b)) is the closure of C_0^1((a,b)) with respect to the H^1((a,b)) ... 1 Let X be a Banach space and u\in L^1(0,T,X). We say that u'\in L^1(0,T,X) the weak derivative of u if$$\int_0^T u(t)\phi'(t)dt=-\int_0^T u'(t)\phi(t)dt,\forall\ \phi\in C_0^\infty(0,T).$$According to this definition, the problem is not that u'(t) does not belong to H^1, it does belong to H^1. The question is, if it belong to L^\infty, ... 2 Yes. x \rho_t(x) is smooth and compactly supported, and any such function f is in H^s for every s. The easiest way to see this may be this: Let k be any integer larger than s. Verify from the definition that H^k \subset H^s. (Note that (1+|\xi|^2)^s \le (1+|\xi|^2)^k). Using the fact that the Fourier transform takes differentiation to ... 1 Only large frequencies matter for smoothness. For every M, the part of Fourier transform with \{\xi:|\xi|\le M\} contributes a real-analytic term to the function. You know that integrability of |\xi|^\alpha \hat u(\xi) implies certain smoothness of u. So you want to show that this product is integrable for every \alpha. On every ball \{|\xi|\le ... 3 Forming the convolution of the (scaled) bump function with indicator functions you get "qausi-indicator functions" in \mathscr D, in particular, there are \psi_n\in\mathscr D(\mathbb R) which are positive and equl to 1 on [-n,n]. It is then easy to see that the elements of E^{loc} are those distributions which, on every compact set, have the same ... 1 First, let a\in\mathbb{R} be arbitrary. As you noted, this implies$$ \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 ...

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 ...

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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 $$\|f\|_{H^q}^2 = ... 1 Start with an example. So, look at \Omega = [0,\pi]\subset \mathbb{R}^{1}, where you have an orthonormal basis of eigenfunctions for -\Delta=-\frac{d^{2}}{dx^{2}} given by \{ e_{n}(x)=\sqrt{2/\pi}\sin(nx)\}_{n=1}^{\infty}. To solve the equation in this case, write the solution as$$ u(x,t) = \sum_{n=1}^{\infty}a_{n}(t)e_{n}(x) $$Then$$ ...

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