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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 ...


5

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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