# Tag Info

4

Essentially, the problem is $u''+\lambda u=0$ on $[-\pi,\pi]$ with periodic boundary conditions. (Note that this interval has the same length as the circumference of the circle; this choice of parametrization ensures that the Laplace-Beltrami operator on the circle directly corresponds to the Laplacian on this interval.) The solutions to the DE itself are ...

3

Not a solution, but a sketch. You need to find a manageable expression for $\Delta f$ if $f:S^1 \rightarrow \mathbb{R}$. One option is to work with a (almost onto) chart of $S^1$, e.g. by using an arclength parametrized curve $c:(-\pi,\pi)\rightarrow S^1$ and by looking at the resulting equation for $f\circ c$ instead. The fact that $f$ is assumed to be ...

2

Your proposed approach is fine: plug in $g$ into both sides of your equation and you will get $$g'' = \sum_{n\geq 1} -a_n n^2 \cos(nx) - b_n n^2 \sin(nx)$$ $$\lambda g = \frac{\lambda a_0}{2} + \sum_{n\geq 1} a_n \lambda \cos(nx) + b_n \lambda \sin(nx).$$ There is something special and fortuitous about the basis you have chosen for $g$: compute ...

2

You are making this much to difficult because you are ignoring what you were told. You were told that the differential operator was $d^2/dx^2$, the "Laplacian", but only in one dimension. The general solution to $d^2y/dx^2= 0$ is the linear function $y= ax+ b$. The solution to $d^2G/dx^2= \delta(x- x')$, then, is a "broken line: $G(x, x')= px+ q$ for ...

2

Consider the map $f$ defined by $x \mapsto \frac{Ax}{\sum_i (Ax)_i}$ defined on the (topological) disk $D$ that consists of vectors $x$ satifying $x_1 \ge 0, x_2 \ge 0, \ldots, x_n \ge 0, x_1 + x_2 + \ldots + x_n = 0$ (i.e., $D$ is the standard simplex in the positive octant). Then $$f : D \to D$$ is a continuous map of a closed disk to itself (this ...

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It seems as if you confused what to assume and what to prove in the second part. If you want to show the existence of $\pi$, you cannot include it in the definition of $H$. But if you get your arguments sorted out, it's quite obvious. Let $\pi=G_2^{-1}\circ G_1^{-1}$ and $f\colon \sigma(S)=\sigma(T)\to \mathbb{C},\,z\mapsto z$. By definition, $G_1(f)=S$ and ...

1

Suppose you have a basis $V_W$ for $W$ and a basis $V_{W^{\perp}}$ for $W^{\perp}$. Using Gram-Schmidt's method, you find a orthonormal basis $V^{'}_W$ for $W$ and a orthonormal basis $V^{'}_{W^{\perp}}$ for $W^{\perp}$. Now, you know that $W \cap W^{\perp} = \emptyset$, so given $u_1 \in V^{'}_{W}$ and $u_2 \in V^{'}_{W^{\perp}}$, you have that they are ...

1

The case where $\mathcal{N}(T-\lambda I)\ne \{0\}$ is covered. So assume $\mathcal{N}(T-\lambda I)=\{0\}$ and $\lambda\in\sigma(T)$. Because $T-\lambda I$ is normal, then $$\|(T-\lambda I)x\|=\|(T^*-\overline{\lambda}I)x\|,\;\;x\in H,$$ which also implies that $\mathcal{N}(T^*-\overline{\lambda}I)=\{0\}$. Therefore,  ...

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