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2

Yes, you must show that the tempered distribution $(1+|\xi|^2)^{s/2}\,\hat{u}$ is given by integration against an $L^2$ function. The multiplication by $(1+|\xi|^2)^{s/2}$ does not change whether or not it is an a.e. pointwise function, but it does affect square integrability. For example, the Fourier transform of Dirac $\delta$ is (up to a constant) the ...

1

$\delta$ function is not strictly a function. If used as a normal function, it does not ensure you to get to consistent results. While mathematically rigorous $\delta$ function is usually not what physicists want. Physicists' $\delta$ function is a peak with very small width, small compared to other scales in the problem but not infinitely small. So what I ...

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$D((0,T)\times \Omega)\subset L^2((0,T);H^1(\Omega))\subset L^1_{loc}((0,T);H^1(\Omega))$. Define $$f(v) = \int_0^Tdt \langle f(t,\cdot),v (t,\cdot)\rangle_{H^{-1};H^1},$$ then $$|f(v)|\le \int_0^Tdt \| f(t,\cdot)\|_{H^{-1}}\|v (t,\cdot)\|_{H^1} \le | K |\int_0^Tdt \| f(t,\cdot)\|_{H^{-1}}\sqrt{\|v (t,\cdot)\|_{L^\infty}^2+\|\nabla_x v ... 2 I'd like to offer another approach to building \tilde h. First step: h is C^1 and non-zero on K, hence there exists \epsilon>0 such that h(x)\ne0 whenever dist(x,K)\le\epsilon. Second step: let's take a function$$\phi(x) = \begin{cases}c\exp\left(-\frac{1}{1-|x|^2}\right), &|x|<1,\\0,&\text{otherwise.}\end{cases}$$where c is ... 2 This result holds at least for \alpha>0. I will describe the method that allows to make an analysis even in the case of arbitrary dimension. Denote \vec r=(x,y), r=\|\vec r\|. First element: let the function f_\epsilon\in L^1_{loc}(\Bbb R^2) be given by$$f_\epsilon = \begin{cases}r^\alpha,&r>\epsilon,\\\epsilon^\alpha,&r\le\epsilon. ...

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The Fourier transform of a tempered distribution is a distribution. Let $\hat f$ be the Fourier transform of $f\in\mathcal S'(\Bbb R)$. Your equation leads to $$(1+4\pi|\xi|^2)\hat f =0$$ in the sense of distributions. If we have an equation $gT=0$, where $g\in C^\infty(\Bbb R)$, $T\in D'(\Bbb R)$, then $supp T\subset \{x\in \Bbb R: g(x)=0\}$. In our case ...

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Let $\phi\in\mathcal{D}(\mathbb{R})$ de a test function. Then $$T_n(\phi)=\sum_{k=1}^N\frac1{k^2}\,\phi\Bigl(\frac1k\Bigr).$$ Since $\phi$ is bounded, the series is absolutely convergent, that is $$\lim_{N\to\infty}T_n(\phi)=\lim_{N\to\infty}\sum_{k=1}^N\frac1{k^2}\,\phi\Bigl(\frac1k\Bigr)=\sum_{k=1}^\infty\frac1{k^2}\,\phi\Bigl(\frac1k\Bigr).$$ Define ...

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It seems that the first inclusion (without a closure) is false. Try the following counterexample in $\mathcal{D}'(\mathbb{R})$. Consider the sequence $u_n=\delta_{1/n}$, there $\delta_a$ denotes the Dirac function defined by $\delta_a(f):=f(a)$.

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If $T :\mathscr D(\mathbb R^n)\to \mathbb R$ satisfies $$\lvert T(f)\rvert \le \|f\|_{L^2(\mathbb R^n)},$$ then $T$ is a functional defined in a dense subset of $L^2(\mathbb R^n)$, and as it is bounded (equivalent, continuous) there, it extends continuously in the whole of $L^2(\mathbb R^n)$. And by Riesz representation, there exists a unique $g\in ... 2 Your hypothesis leads to $$T:D\to\mathbb C,$$ linear continuous functional in the sense of$L^2(\mathbb R)$.$D$is dense in$L^2$, therefore by continuity we can extend$T$to the whole$L^2$. By Riesz representation theorem, there exists an$L^2$function$g$such that$T(f) = (g,f)_{L^2}$. Moreover, it implies that$T$is given by$T_{\bar g}$and ... 2 The issue with this question is you haven't specified what kind of derivative you would like. However, the first one can be seen in a simpler way, as follows: For$x$non zero, the function is either$f(x) = x$, or$f(x) = -x$, for which the derivatives can be taken easily, either they're$+1$for$x>0$or$-1$for$x<0$(which agrees with the answer ... 2 To do so, I used the family of functions$f_m(x)=\sqrt{x^2+m^2}$, which satisfies: $$\lim_{m\to 0} f_m(x)=f(x),$$ uniformly in$x$, since the maximum distance between the succession and the limit function is$|m|$itself (for$x=0$). Then: $$\frac{\mathrm{d}}{\mathrm{d}x}f(x)= \frac{\mathrm{d}}{\mathrm{d}x}\lim_{m\to 0} f_m(x) =\lim_{m\to 0} ... 2 In general, if X is a locally convex topological vector space of uncountable dimension (as a linear space), then the weak^* topology on X^* is not first countable. Proof. In the weak^* topology a sub-base of the neighborhoods of 0 is obtained by sets of the form$$ W_{x,\varepsilon}=\{x^*\in X^*: |x^*(x)|<\varepsilon\}, \quad ... 1 Vobo's comment is correct: the space of solutions of$xT=0$does not include the shifted Dirac functions. Indeed, for$T=\delta_k$we have$xT=k\delta_k$which is nonzero if$k\ne 0$. To describe the space of solutions, suppose$xT=0$. Every test function$\varphi$vanishing at$0$can be written as$\varphi(x)=x\psi(x)$, see Quotient of two smooth ... 1 What you wrote does make perfect sense, as integration by parts on a sufficiently smooth manifold$S$is permissible. Note that you have made a little mistake, and this is the correct version$$\left\langle \frac{d}{dt}u - \Delta_\Gamma u, \varphi \right\rangle_{\mathscr D^*(S), \mathscr D(S)} = -\left\langle u, \frac{d}{dt}\varphi \right\rangle - \langle ... 1 This is indeed the definition of distributional derivatives. The book "Heat Kernel and Analysis on Manifolds" by Grigoryan (I can even access that part as a free preview on books.google.com) contains formula (7.30): locally integrable function$u(t,x)$satisfies the heat equation on a manifold$N=(0,\infty)\times M\$ in a distributional sense if and only if ...

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