Tag Info

Hot answers tagged

3

I am not completely sure at the moment how the supports of $\widehat{p_1 u} + \dots + \widehat{p_m u^m}$ and of $\widehat{u}$ could be connected, but if you consider $$ \widehat{u} = \delta_x, $$ i.e. $u = e^{2\pi i \langle x, \cdot\rangle}$, then $u^2 = e^{2\pi i \langle 2x, \cdot\rangle}$, i.e. $$ \widehat{u^2} = \delta_{2x}, $$ so that even ${\rm ...


2

Here is what I think: Let $P=\sum_{|\alpha|\leq k}a_{\alpha}x^{\alpha},$ then $$\widehat{P}=\widehat{\Big(\sum_{|\alpha|\leq k}a_{\alpha}x^{\alpha}\Big)}=\sum_{|\alpha|\leq k}a_{\alpha}\widehat{x^{\alpha}}=\sum_{|\alpha|\leq k}a_{\alpha}\frac{\partial^{\alpha}\delta_0}{\partial x^{\alpha}}$$ Thus supp$(\widehat{P})\subset$ supp$(\sum_{|\alpha|\leq ...


1

The hint would be to write $$\phi(1/k) = \phi(0)+\frac{\phi'(0)}{k}+\frac{\phi(\xi_k)}{2k^2},\quad \xi_k\in(0,1/k).$$ After some manipulations you will obtain that $u$ is indeed a distribution of at most second order.


1

First of all, note that $u^n$ is not well-defined for tempered distributions, because you can't multiply them. Therefore we need suppose that $u$ is function. One of extreme cases is $u= x^p$ for $p\in \Bbb N$. Easy to see that $\sum_{|\alpha|\leq k}\frac{\partial^{\alpha}\widehat{u^n}}{\partial x^{\alpha}}$ will be supported in the point $\{0\}$. Another ...


1

By the $\Gamma$ reflection formula it follows that: $$I(a)=\int_{-\infty}^{+\infty}\frac{x^{a-1}}{x^2+1}\,dx = \pi\sin\frac{\pi a}{2}$$ for any $a$ in the strip $0<\Re(a)<2$, so the analytic continuation gives $I(2)=0$, as expected.


1

Separating your variables appropriately, we have $$\int_0^T \delta(t-t_j)e^{-(s+\mu\lambda^2)t}\,dt\int_0^l \delta(x-R)\varphi(x)\,dx.$$ Since $0 < R < l$, the sifting property of the Dirac delta gives us $\varphi(R)$ for the second integral. The first is very similar since $0 < t_j < T$.


1

Yes, $D'(\Omega)$ is complete but not induced by a norm, basically you have a bunch of seminorms; it is (what is known) as a LF space. A more conceptual viewpoint on the topology on $D'(\Omega)$ is that it is a certain locally convex direct limit of spaces $C^\infty_0(\Omega')$. A good reference for the general framework is Bourbaki's espaces vectoriels ...


1

First of all, the wiki article on distributions is a good source to start with. This type of convergence is called the weak-* convergence - more on it here. The space of distributions is complete under this notion of convergence, but this space is not normed. It is a locally convex space. It is a multinormed space. For a topological space to be complete ...



Only top voted, non community-wiki answers of a minimum length are eligible