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4

$$ \int_{t_1}^{t_2} \delta(D - x(t)) dt = \int_{t=t_1}^{t_2} \delta(D - x(t))\frac{dx(t)}{|x'(t)|} = \sum_{t:x(t)=D\wedge t\in[t_1,t_2]}\frac{1}{|x'(t)|}. $$


3

The identification of functions with distributions uses the pairing $$ (f,\phi) = \int f\phi \, dx $$ Where $f$ is at least locally integrable, and $\phi$ is compactly supported smooth. We choose these $\phi$ as test functions so that (a) every locally integrable function defines a distribution in the sense of a continuous functional, and (b) by using the ...


3

Let $u_s=\frac{1}{\pi s} e^{-r^2/s^2}$. Observe that if $\frac{1}{s} u_s\rightarrow u$ as $s\rightarrow 0$, then $u_s\rightarrow 0$ as $s\rightarrow 0$. It follows, since $u_s\rightarrow \delta$ and $\delta\neq 0$, that $\frac{1}{s}u_s$ does not converge.


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

Since you know the upper ($H$) and lower ($L$) bounds of your index range, the "ideal" index distribution would be discrete-uniform on $[L,H]$. Unfortunately, the standard deviation, skew, or kurtosis will only partially characterize what you are looking for. I have actually had to deal with the same issue as you (that of finding a maximally "even" ...


1

The famous "Impossibility Result" of Laurent Schwartz precisely says: There is no associative algebra over $\mathbb{R}$ containing $\mathcal{D}'$ as a vector subspace and the constant function $1$ as unity element, having a differential operator acting like the differential operator on $\mathcal{D}'$ and the algebra multiplication of continuous functions ...


1

Let $\phi$ a test function such that $\phi(0) = 1$ then $$\frac{1}{\pi s^2} \int_{\mathbb R}e^{-\frac{r^2}{s^2}} \phi(r)dr = \frac{1}{s}\frac{1}{\pi s} \int_{\mathbb R}e^{-\frac{r^2}{s^2}} \phi(r)dr $$ and if $s\rightarrow 0^+$ $$\frac{1}{\pi s} \int_{\mathbb R}e^{-\frac{r^2}{s^2}} \phi(r)dr \rightarrow \phi(0) = 1$$ $$\frac{1}{s} \rightarrow \infty$$ so ...


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 ...


1

Remark that $\textrm{supp}(\phi u) \subset \textrm{supp}(\phi) \cap \textrm{supp}(u)$, which is a compact set since $\textrm{supp}(\phi)$ is compact and $\textrm{supp}(u)$ is closed. Thus, $\phi u \in \mathcal{E}'$ i.e. it is a compactly-supported distribution.


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 $L^\infty$, ...


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.


1

EDIT: A lot of what I said in my original post is wrong, as it turns out :( The $L^1$ example below shows that your formulation of the Fatou property is not as useful as it first appears. The Fatou property as you formulate it is not true in general. As a counterexample, consider $$ C_0 := \{f : \Bbb{R} \to \Bbb{C} \mid f(x) \to 0 \text{ as } |x|\to\infty ...



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