Does there exist $u\in H^2(\Bbb{R}^n)$ so that $-\Delta u=\lambda u$. Of course if such $u$ exists, it must be a smooth function by elliptic estimate and bootstrap strategy. So we need to find $u$ so that $-\Delta u=\lambda u$ in classical sense with norm constrain. I don't think such an eigenfunction exist, but I can't prove it. Thanks for advance for anyone who could offer help.
70156
I know very well that Laplacian in bounded domain has a discrete spectrum. How about Laplacian in $\mathbb{R}^n$?(not in some fancy-shaped unbounded domain, but the whole domain)  Where can I find such results?  Moreover, is there a counterpart of Hilbert-Schmidt theorem for Laplacian in $\mathbb{R}^n$? Hilbert-Schmidt asserts there is a countable set of eigenfunctions $\phi_n$ so that $x=\sum \langle x,\phi_n\rangle \phi_n,\forall x\in H$.  Is there a similar theorem saying $x=\int_0^\infty \langle x,\phi_\lambda\rangle \phi_\lambda\,\mathrm{d}\lambda,\forall x\in H$ where $\phi_\lambda$ is the eigenfunction of Laplacian to spectral value $\lambda$?
71687
Let $B(0;1)=\{x\in \mathbb{R}^N ;|x|\leq 1\}$,  the ball of $\mathbb{R}^N$ equipped with the euclidian scalar product  $$x \cdot y=x_1y_1+...+x_Ny_N,\ \ \ x=(x_1,...,x_N),\ \ y=(y_1,...,y_N)\ \ \ |x|=\sqrt{x\cdot x}.$$ Let $\alpha\mbox{ , } \beta \mbox{ two real numbers, } \alpha\geq 0\mbox{ , } a \in \partial B(0,1)$ fix. $$h(x)=\log\left((1-x\cdot a)^{\alpha}+|x-a|^{\beta}\right)$$ Can we find $\alpha$ and $\beta$ such that $h$ is harmonic in $B(0,1)$?
27861
Hm, this just occurred to me while answering another question:   If I write the Hamiltonian for a harmonic oscillator as $$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ then wouldn't one set of possible basis states be the set of $\delta$-functions  $\psi_x = \delta(x)$, and that indicates that the size of my Hilbert space is that of $\mathbb{R}$.  On the other hand, we all know that we can diagonalize $H$ by going to the occupation number states, so the Hilbert space would be $|n\rangle, n \in \mathbb{N}_0$, so now the size of my Hilbert space is that of $\mathbb{N}$ instead.  Clearly they can't both be right, so where is the flaw in my logic?
24062
I have trouble proving the following problem (Evans PDE textbook 5.10. #15). Could anyone kindly help me solving the problem? I know that I should somehow use Poincaré inequality but I still cannot solve it.   Fix $\alpha&gt;0$ and let $U=B^0(0,1)\subset \mathbb{R}^n$. Show there exists a constant $C$ depending only on $n$ and $\alpha$ such that $$ \int_U u^2 dx \le C\int_U|Du|^2 dx,  $$ provided that $u\in W^{1,2}(U)$ satisfies $|\{x\in U\ |\ u(x)=0\}|&gt;\alpha$.
108649
Let $A$ and $B$ be $m \times n$ and $n \times m$ real matrices.  I proved that if $\lambda$ is a nonzero eigenvalue of the $m \times m$ matrix $AB$ then it is also an eigenvalue of the $n \times n$ matrix $BA$. But I am having trouble finding an example showing that this need not be true if $\lambda=0$  Thank you in advance
16668
I wonder  whether there is a function $f\colon\Bbb R\to\Bbb R$ with the folowing characteristic?   for every two real numbers $\alpha,\beta,\alpha\lt\beta$,  $$\{f(x):x\in(\alpha,\beta)\}=\Bbb R$$   I can't  say such a function does not exist, neither can I construct a example  Thanks a lot!
117752
Let $\lambda_1, \lambda_2$ be the eigen value and $f_1 , f_2$ be the coressponding eigen  functions for the homogeneous integeral equation  $$ \phi(x) - \lambda \int_0^1 (xt +2x^2) \phi(t) dt = 0 $$  Then   $\lambda_1 = -18 - 6 \sqrt{10} , \lambda_2 = -18 + 6 \sqrt{10}$ $\lambda_1 = -36 - 12 \sqrt{10} , \lambda_2 = -36 + 12 \sqrt{10}$ $\int_0^1f_1(x)f_2(x) dx = 1$ $\int_0^1f_1(x)f_2(x) dx = 0$   I have solved simply and get the values of $\lambda_1 \ \ and \ \ \lambda_2$ belong to (1).  Please tell me about (3) and (4) option.  Thank you.
19065
There was this question on one of the whiteboards at my company, and I found it intriguing. Maybe it's a dumb thing to ask. Maybe there is a simple answer that I couldn't see. Anyway, here it is:   Does there exist a non-trivial, monotonically increasing function such that $f'(x) = f(f(x))$ (in $\mathbb{R}$)?   I checked a bunch of functions, from elementary to special (gamma, digamma, zeta, Riemann, Lambert ...) and none seems to work (not surprisingly). I managed to convince myself that a function expressible as a power series would not work, regardless of convergence issues, because the derivative lowers the degree of polynomials, when the composition raises it. The Dirac delta or some sort of generalized function looked promising for a while, but the Dirac delta is not monotonically increasing anyway, and I'm not very familiar with generalized functions. I tried to use the Fourier transform on both sides, but it seems the Fourier transform is difficult for f(f(x)) (at least for me). I thought about somehow seeing "taking the derivative" as a differential operator, finding its (infinite) matrix in some basis (which one?), do the same thing to the RHS and show that the 2 matrices could not be identified (reducing the problem to a linear algebra problem) - that didn't work. Nothing on the geometric front either. I thought about trying to prove that there is no such function by deducing a contradiction, but didn't manage that. My hunch is that no such function exists, based on the completely invalid and semi-meaningless idea that differentiation pulls f in one direction, and composition in the other. Any idea?
6156
I noticed the following:  $$[L_{+},L^2]=0,\qquad [L_{+},L_3]\neq 0,\qquad  [L^2,L_3]=0.$$  This would suggest, that $L^2,L_+$ have a common system of eigenfunctions, and so do $L^2,L_3$, but $L_+,L_3$ don't. How is that possible?
21409
Given some potential $V$, we have the eigenvalue problem   $$ -\frac{\hbar^2}{2m}\Delta \psi + V\psi = E\psi $$   with the boundary condition   $$ \lim_{|x|\rightarrow \infty} \psi(x) = 0 $$   If we wish to seek solutions for   $$ E &lt; \lim_{|x|\rightarrow \infty}V(x), $$   I find textbooks like Sakurai and others assert that (without proof) set of all such eigenvalues $E$ that admits nontivial solutions is discrete, which gives us liberty to index them with natural numbers (principal quantum numbers). Can anyone help me with a little elaboration of this mathematical fact ? As a student of mathematics, I know that compact operators have discrete spectrums, so for the operator $ T : L^2 \rightarrow L^2 $given as  $$ T(\psi) = -\frac{\hbar^2}{2m}\Delta \psi + V\psi. $$   Are there methods to show that $T$ is compact? Or possibly other ways to establish discreteness of its eigenspectrum?
75505
Let $\Omega\subset\mathbb{R}^n$ be a bounded domain. And let $L^2\left(\Omega\right)$ be the space of equivalence classes of square integrable functions in $\Omega$ given by the equivalence relation $u\sim v \iff u(x)=v(x)\, \text{a.e.}$ being a.e. almost everywhere, in other words, two functions belong to the same equivalence classes if they only are different in a zero meassure set.  This space is a Hilbert space with the norm is:  $$ \Vert v \Vert ^2_{L^2 \left( \Omega \right) }= \int_\Omega v(x)^2\,\mathrm{d}x $$  Let $H^1\left(\Omega\right)=\left\{ v\in L^2\left( \Omega \right);\,\vert\mathbf{grad}(v)\vert \in L^2\left( \Omega \right ) \right\}$ be also a Hilbert space and its norm :  $$ \Vert v \Vert ^2_{H^1 \left( \Omega \right) }= \Vert v \Vert ^2_{L^2 \left( \Omega \right) } + \Vert \vert\mathbf{grad}(v)\vert \Vert ^2_{L^2 \left( \Omega \right) } $$   I've been asked to prove that:  $$ \Vert v \Vert ^2_{H^1 \left( \Omega \right) } \le C\left( \Omega\right) \left( \Vert \vert\mathbf{grad}(v)\vert \Vert ^2_{L^2 \left( \Omega \right) } + \int_\Gamma \left ( \gamma v(x) \right) ^2\,\mathrm{d}\sigma \right) \quad \forall v\in H^1 \left( \Omega \right) $$  where $\Gamma = \partial \Omega $, $\Omega $ is smooth enough, $\mathrm{d}\sigma$ is a meassure on the boundary of $\Omega$, and the operator $\gamma: L^2\left( \Omega\right )\mapsto L^2\left( \Gamma\right )$ is the trace operator. ( if it is needed I can define it also )  Sincerely, I just don't know how to begin, and maybe someone could give me some clue about it.(by the way the $C$ is a positive constant depending only on the domain $\Omega$).  Thank you very much!
82848
I always heard the eigenfunctions of a  form a complete basis. Where can I find a proof in infinite dimension space? Presumably readable for physicists.
27594
If $U$ is a bounded domain in $\mathbb R^n$ whose boundary is smooth, and $f$ is a smooth function on $U$ whose partial derivatives of all orders have a continuous extensions to $\bar U$. For an arbitrary domain $V \supseteq \bar U$, is there a smooth function $\tilde f$ on $V$ extending $f$?
117533
Let $\Omega$ a bounded domain, connexe and regular, and let $f \in L^2(\Omega).$ Let the variational problem: Find $u \in H^1(\Omega)$ such  $$\int_{\Omega} \nabla u \nabla v dx + (\int_{\Omega} u dx)(\int_{\Omega} v dx) = \int_{\Omega} f v dx, \forall v \in H^1(\Omega)$$ 1- Prouve that this variational problem admits a unique solution in $H^1(\Omega).$  2- Deduce the boundary problem associate to this variational problem ( study the two cases $u \in  H^2(\Omega)$ and $u \notin H^2(\Omega)$.  I dont't understand in the question 2, why we mus study the two cases $u \in H^2$ and $u \notin H^2$?
45885
Suppose $A=uv^T$ where $u$ and $v$ are non-zero column vectors in ${\mathbb R}^n$, $n\geq 3$. $\lambda=0$ is an eigenvalue of $A$ since $A$ is not of full rank. $\lambda=v^Tu$ is also an eigenvalue of $A$ since $$Au = (uv^T)u=u(v^Tu)=(v^Tu)u.$$  Here is my question:     Are there any other eigenvalues of $A$?   Added:  Thanks to Didier's comment and anon's answer, $A$ can not have other eigenvalues than $0$ and $v^Tu$. I would like to update the question:     Can $A$ be diagonalizable?
58764
Let $A$ be a nonsingular $n \times n$ matrix, $\|\cdot\|$ be any natural norm, and $K_{p}(A)=$ $\|A\|_{p}\left\|A^{-1}\right\|_{p} .$ Let $\lambda_{1}$ be the smallest and $\lambda_{n}$ be the largest eigenvalues of the matrix $A^{t} A$ (b) Show that $K_{2}(A)=\sqrt{\frac{\lambda_{n}}{\lambda_{1}}}$. (Hint: One should use the $\|A\|=\sqrt{\rho\left(A^{t} A\right)}=\sqrt{\rho\left(A A^{t}\right)}$ relation) For part $(b)$, it seems minimal eigenvalue of matrix $A$ will be the largest eigenvalue of the inverse of matrix $A$, so we have to use this idea but could not understand formally why square roots are needed for $K_{2}(A)$ and how spectral radius is necessary for this part?!
74985
For a compactly supported distribution $f$, I want to show the non-surjectivity of the map $H:f \mapsto f*f \in \mathcal{E}'$.     My attempt: I have chosen to show that this is not the case by showing that $\text{image } (H)$ does not contain $ \delta_0'$. For instance I restrict myself to $d=1$.   Suppose that $H$ is surjective. Applying the Fourier transform on the convolution, I get $(\hat{f}(z)) ^2 = i z$. The theorem of Paley-Wiener-Schwartz guarantees that $(\hat{f}(z)) $ should be an entire function. However, as it is well-known, no such entire function exists:   If this were true, then $e^{i \pi /4}\hat{f}(z)$ would be the square root function, which cannot be well defined on $\mathbb C$, and even less so entire.    Q1: Is there another way of achieving the result (i.e. with a choice significantly different than $\delta_0'$, and with an approach different than the one using the Fourier transform)?  Q2: Can I generalize this without much effort in the case of $d \ge 1$ (probably using the tensor product)?    Edit: The source is .