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

And as I emphasized last time, a Hilbert space is a special case of a Banach space.

So recall that a Hilbert space is a vector space H equipped with a...

Well, we take a complex vector space. There are also real Hilbert spaces,

but we look always at complex Hilbert spaces equipped with a then sesquilinear inner product

which induces a norm with respect to which H is complete.

So we're guaranteed that every Cauchy sequence converges.

It's very easy to observe. Well, I already said it.

Because you have this norm defined on a Hilbert space, namely by virtue of the norm of psi,

so we'll usually denote the elements of a Hilbert space by some psi, phi, drala,

because this is defined like this, and this is indeed a norm, makes this into a Banach space in particular.

If you take this norm, makes the Hilbert space into a Banach space.

And thus we know from our work we did on Banach spaces, we know that everything we know about Banach spaces already applies.

For instance, we know that if we look at the set of bounded linear maps from the Hilbert space into itself.

Well, usually we last lecture we had here Banach spaces, but the Hilbert space is in particular a Banach space.

We know that this space equipped with the operator norm.

So let me maybe indicate this norm by an H just for a while, because it's the norm on the space H.

And here we have the operator norm that we defined last time.

That's a different norm than this one. It's the one that's defined on this space.

We know that this is a Banach space.

So you see even if we start with Hilbert spaces as a special case, you could say why would we ever look at Banach spaces?

Well, already the construction of the set of all bounded linear maps from H into itself carries the norm as a Banach space, but not a Hilbert space.

However, there is a special type of Hilbert space and that because H is a Banach space, I know what H star is.

Remember what H star was? So this is point A, point B. H star was what?

The dual space.

Yes, the dual space, but how is it defined?

It's all linear maps from H to the real numbers or complex numbers.

To complex numbers? No, not quite, not all linear maps?

All bounded linear maps. Please don't forget this. This is all bounded linear maps from H to C.

So you quoted the linear algebra notion. If you don't have an inner product, if you don't have a norm, then you can't talk about boundedness or not.

But if you do, we explicitly restrict to bounded functionals.

Bounded functions are also continuous. That follows, comes on the problem sheet.

We know that this space here, equipped with the corresponding operator norm, which it is, is also a Banach space.

Because it's just a special case of this situation if I had a different Hilbert space G here, because the C is also a Hilbert space.

Okay, so already learned that. Remark, we will see that the dual H with the norm it carries is even a Hilbert space.

And somebody already said it last time, that can only be, so that means that there exists an inner product on this H star,

such that the norm on H star, which we already have by virtue of the construction of the operator norm, such that this norm is given by an inner product, by the square root of an inner product.

Okay, and we will see. Okay.

So, but that prompts the more general question. So in order to see this, so you see, we're abstractly guaranteed it's a Banach space, but the question is, is it even a Hilbert space?

In order to see this more generally, note the theorem.

The theorem is the following. If a norm on a vector space is induced

by a sesquilinear inner product,

if and only if, and if I say induced, I could mean many, many things, but I really mean induced like this, okay?

You could induce it maybe otherwise, if and only if the parallelogram identity,

of F plus G, plus the norm of F minus G equals twice the norm of F plus twice the norm of G, holds.

And of course for all F and G in the vector space. So if this parallelogram identity holds, then the norm really comes from an inner product.

In this case, when it does come from an inner product, the inner product can be explicitly constructed.

Up here, constructed as, well, you want to learn how an F with a G, what the inner product is,

because we have a sesquilinear map, it works as follows, it's one fourth bracket.

You take F plus G norm minus F minus G norm. So the plus goes with the plus and the minus goes with the minus.