In particular the RAGE theorem shows the connections between long time behavior and spectral types. Finally, Chapter 6 is again of central importance and should be studied in detail. The chapters in the second part are mostly independent of each others except for the rst one, Chapter 7, which is a prerequisite for all others except for Chapter 9. If you are interested in one dimensional models Sturm-Liouville equa- tions , Chapter 9 is all you need.

If you are interested in atoms, read Chapter 7, Chapter 10, and Chap- ter In particular, you can skip the separation of variables Sections If you are interested in scattering theory, read Chapter 7, the rst two sections of Chapter 10, and Chapter Chapter 5 is one of the key prereq- uisites in this case. Gerald Teschl Vienna, Austria February, Part 0 Preliminaries Chapter 0 A rst look at Banach and Hilbert spaces I assume that the reader has some basic familiarity with measure theory and func- tional analysis. For convenience, some facts needed from Banach and L p spaces are reviewed in this chapter.

A crash course in measure theory can be found in the appendix. If you feel comfortable with terms like Lebesgue L p spaces, Banach space, or bounded linear operator, you can skip this entire chapter. However, you might want to at least browse through it to refresh your memory. Warm up: Metric and topological spaces Before we begin I want to recall some basic facts from metric and topological spaces. I presume that you are familiar with these topics from your calculus course.

A good reference is [8].

A point x of some set U is called an interior point of U if U contains some ball around x. If x is an interior point of U, then U is also called a neighborhood of x. Note that a limit point must not lie in U, but U contains points arbitrarily close to x. Moreover, x is not a limit point of U if and only if it is an interior point of the complement of U. A set consisting only of interior points is called open. O That is, O is closed under nite intersections and arbitrary unions.

In general, a space X together with a family of sets O, the open sets, satisfying i iii is called a topological space. A family of open sets B O is called a base for the topology if for each x and each neighborhood U x , there is some set O B with x O U. If B O is a base for the topology, then every open set can be written as a union of elements from B. If there exists a countable base, then X is called second countable. In the case of R n or C n it even suces to take balls with rational center and hence R n and C n are second countable.

A topological space is called Hausdor space if for two dierent points there are always two disjoint neighborhoods. Note that dierent metrics can give rise to the same topology. For example, we can equip R n or C n with the Euclidean distance as before, 0.

## Mathematical Methods in Quantum Mechanics: With Applications to Schrdinger Operators

Warm up: Metric and topological spaces 5 or we could also use. Hence the topology is the same for both metrics. We can always replace a metric d by the bounded metric. The complement of an open set is called a closed set.

That is, closed sets are closed under nite unions and arbitrary intersections. CC,UC C, 0.

## Gerald Teschl's articles on arXiv

Let X be a metric space, then the interior of U is the set of all interior points of U and the closure of U is the set of all limit points of U. A sequence x n. Clearly the limit is unique if it exists this is not true for a semi-metric. Both R n and C n are complete metric spaces.

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A point x is clearly a limit point of U if and only if there is some sequence x n U converging to x. Hence Lemma 0. A closed subset of a complete metric space is again a complete metric space. Note that convergence can also be equivalently formulated in terms of topological terms: A sequence x n converges to x if and only if for every neighborhood U of x there is some N N such that x n U for n N.

In a Hausdor space the limit is unique. A metric space is called separable if it contains a countable dense set.

Lemma 0. Let X be a separable metric space. Every subset of X is again separable. The only problem is that AY might contain no elements at all. To see that B is dense choose y Y. Let X be a metric space. The following are equivalent i f is continuous at x i. Warm up: Metric and topological spaces 7 Proof. The last item implies that f is continuous if and only if the inverse image of every open closed set is again open closed.

Note: In a topological space, iii is used as denition for continuity.

However, in general ii and iii will no longer be equivalent unless one uses generalized sequences, so called nets, where the index set N is replaced by arbitrary directed sets. In particular, the projections onto the rst x, y x respectively onto the second x, y y coordinate are continuous. If we consider RR we do not get the Euclidean distance of R 2 unless we modify 0. In the case of metric spaces this clearly agrees with the topology dened via the product metric 0.

A cover of a set Y X is a family of sets U. A subset K X is called compact if every open cover has a nite subcover. A topological space is compact if and only if it has the nite intersection property: The intersection of a family of closed sets is empty if and only if the intersection of some nite subfamily is empty. A rst look at Banach and Hilbert spaces Proof. By taking complements, to every family of open sets there is a cor- responding family of closed sets and vice versa. Moreover, the open sets are a cover if and only if the corresponding closed sets have empty intersec- tion.

A subset K X is called sequentially compact if every sequence has a convergent subsequence. Let X be a topological space.

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XY is an open cover for X. We show that XY is open. By the denition of Hausdor, for every y Y there are disjoint neighborhoods V y of y and U y x of x. By compactness of Y , there are y 1 ,. For every x, y X Y there is some x, y such that x, y O x,y. By denition of the product topology there is some open rectangle U x, y V x, y O x,y. Hence for xed x, V x, y yY is an open cover of Y. Hence there are nitely many points y k x such V x, y k x cover Y. Since nite intersections of open sets are open, U x xX is an open cover and there are nitely many points x j such U x j cover X. For every n there is a ball B n x n which contains only nitely many elements of K.

However, nitely many suce to cover K, a contradiction. In a metric space compact and sequentially compact are equivalent. Then a subset is compact if and only if it is sequentially compact. Warm up: Metric and topological spaces 9 Proof. Since if this were false we could construct a sequence x n X. In particular, we are done if we can show that for every open cover O. So it remains to show that there is such an. Since X is sequentially compact, it is no restriction to assume x n converges after maybe passing to a subsequence. Please also recall the Heine-Borel theorem: Theorem 0.

In R n or C n a set is compact if and only if it is bounded and closed. By Lemma 0.