4

When you solve the Schrodinger equation, you solve for the possible energies of the system. These are the eigenvalues of the Hamiltonian operator. Just like a single eigenvalue can correspond to multiple eigenvectors, a single energy can correspond to multiple eigenfunctions of the Hamiltonian. This is called degeneracy. An eigenvalue (energy) is called ...


3

Define$$\eta(z)=\begin{cases}e^{\frac1{\lvert z\rvert^2-1}}&\text{ if }\lvert z\rvert<1\\0&\text{ otherwise.}\end{cases}$$


3

$f(x)=e^{-\frac 1 {1-x}}$ for $x<1$ and $0$ for $x \geq 1$ defines a smooth function which is positive on $(-\infty,1)$ and $0$ outside it. So $f(\|x\|^{2})$ is a smooth function on $\mathbb R^{2}$ which is positive for $\|x\|<1$ and $0$ elsewhere. For any other disk in $\mathbb R^{2}$ use an appropriate affine transformation.


2

From the inequality we know that $f$ is invertible and that $f^{-1}$ is Lipschitz continuous. Take any closed set $A\subset K$, then it is compact because of compactness of $K$, and $f^{-1}(A)$ is then also compact, hence closed, so that $f$ is continuous. Hence $f$ and $f^{-1}$ are continuous, and so $f$ is a homeomorphism between $K$ and some subset of $K$....


2

The answer (under the conventional metrics, etc.) is yes! The set of positive integers under the usual metric is what is known as a "discrete space." Given any discrete space $X$ and any topological space $Y,$ we have for all functions $f:X\to Y$ that $f$ is continuous on $X,$ since every subset of $X$ is open.


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