The following content has been provided by the University of Erlangen-Nürnberg.

So good morning, welcome to the second lecture of the quantum theory course.

As we said last time, a Hilbert space underlies quantum mechanics,

and the Banach space is a generalization of a Hilbert space.

And it's a slight generalization.

And the reason why we're looking at it is not because we want to be general for generalities' sake,

it is rather that if we look at Hilbert spaces, very quickly we will look at operators, linear maps on Hilbert spaces,

and such spaces like for instance the bounded linear maps, which I will define today, they constitute a vector space again.

It will be a normed vector space, it will be a complete normed vector space, but it will not be a Hilbert space, it will be a Banach space.

So even if we start with Hilbert spaces as we would like to focus on in quantum mechanics, in line two we need Banach spaces anyway.

So we start with them, and obviously everything that's true for Banach spaces is true for Hilbert spaces too.

It's a generalization of a Hilbert space, but also appears from constructions that start with a Hilbert space.

So what is a Banach space definition? If you have a Banach space, it's a vector space,

so it's a vector space V with a vector space addition and S multiplication, and it's additionally equipped with a norm.

So it's a vector space with a norm. Now what is a norm? A norm is a map that takes an element of the vector space into the real numbers.

So even if we are looking at this complex vector space, nevertheless a norm takes you into the real numbers,

and it satisfies the following properties. If you look at an element v, small v in capital V, let me call it f,

let me call the elements of a Banach space f, and you multiply with a number from the field over which the vector space is taken,

let's in our case always take the complex numbers, then this is the absolute value of the complex number times the norm of this vector.

The second condition is that it satisfies the triangle inequality, so f plus g norm will always be less or equal to f norm plus g norm.

And the third property is that the norm is always non-negative for all f in V, and equality is assumed if and only if the f itself is the zero vector in this vector space.

So Banach space is a vector space with a norm which is complete. So there is again this completeness property, we already had for Hilbert spaces,

which is complete with respect to this norm, and I remind you to be complete means that if you look at any Cauchy sequence in the space,

you may already conclude that the sequence converges. Every convergent sequence is a Cauchy sequence, but the converse does not hold.

Okay, now a standard topic in mathematics is you study a structure by studying maps between different instances of that structure,

and it's not different here. So study maps of the following type, A, that start in a vector space V, well actually a map starts in a set,

and it goes to a set, and the set may have further structure, and we go as far as saying the set that is the target set here, this should be a Banach space.

So it comes with an addition, an s-multiplication, and the norm, which in order to distinguish it from other norms is called the norm down w.

Now well in the future I will increasingly not write down the plus and the s-multiplication and so on, and we would like to be as general as admitting

that the domain of the maps we're looking at are only normed spaces. So this is only supposed to be a normed space, which means it also has an addition,

an s-multiplication, and the norm, but we do not require completeness. For the target we require completeness, for the domain we don't.

So this is slightly weaker assumption than if I assume this also needs to be a Banach space, and that will serve us very well that we do it in this generality.

So we study such maps, and so we have the definition, a map A is bounded,

A is said to be bounded if the following is true. If you take the A and you apply it to any f from the domain,

then you take the norm and you normalize this by the f up here, and now you look at all the f's from the domain and you take the supremum of these numbers here,

so this is the smallest upper bound, and if this is finite, so that's the condition, if this is finite then the A is said to be bounded.

Now it's often useful to rewrite this, you see, because of the, well it's not the linearity of the norm, but the linearity up to then taking an absolute value of the norm,

if you scale the f down to a unit vector, meaning that the f has norm 1, then we have a 1 down here and the scaling factors cancel,

so we often look at only all the f's that have, that are unit vectors with respect to the norm, and we only need to look at the A f.

Now obviously here you need to take the norm on the target and here you take the norm on the domain, so this is also the norm on the target and this here is the norm on the domain.

A is said to be bounded if that is the case. Now if we have such a bounded operator, we have the definition,

a bounded, no, the operator norm, the operator norm, and that it's a norm in principle needs to be shown, but it's straightforward, the operator norm of a bounded operator.

For bounded operator is the norm like so, and I do not write anything here, so if it doesn't have an index, it's the operator norm,

and increasingly I will not write indices anywhere on norms because it's clear from context. This can only be the norm on the domain,

this can only be the norm on the target, and this can only be the operator norm, but maybe for a while it's good to focus on that.

So this is then the supremum of all unit f on here of the A f here. So if this number here is finite, then we call it the norm of the bounded operator.

Now why do we look at bounded operators? And the answer is they are very important technically, but in fact in quantum mechanics we will need,

and actually all the requirement to do these deep mathematics in functional analysis is that we need to deal also with unbounded operators.

But the bounded operators are an important step to there because many of the proofs for unbounded operators you always try to reduce to a sequence

of something of bounded operators and then you deal with that. And for unbounded operators there is no such norm.