##L^2## square integrable function Hilbert space

[cianfa72]

TL;DR Summary

About the definition of $L^2$ square integrable functions and the Hilbert vector space dimensionality

Hi,
I'm aware of the $L^2$ space of square integrable functions is an Hilbert space.

I believe the condition to be $L^2$ square-integrable actually refers to the notion of Lebesgue integral, i.e. a measurable space $(X,\Sigma)$ is tacitly understood. Using properties of Lebesgue integral, one can show that the set of square integrable functions on any $(X,\Sigma)$ measurable space actually fulfills the axioms of vector space.

Then one introduce the inner product $< ,>$ that induces a norm and a metric/distance. Next step proves that every Cauchy sequence converges to an element of the vector space itself (i.e. the completeness condition is met).

Therefore $L^2$-space turns out to be an infinite dimensional Hilbert space.

Is its dimension either countable or uncountable infinity ?

 

[martinbn]

It can aldo be finite dimenssional if the $X$ is a finite set with the counting measure.

Also you need equvalent classes of functions, not just functions.

 

[cianfa72]

Also you need equivalent classes of functions, not just functions.

Which is the equivalence relation and why we need to consider equivalent classes w.r.t. it and not just functions themselves?

 

[martinbn]

Which is the equivalence relation and why we need to consider equivalent classes w.r.t. it and not just functions themselves?

Functions that differ on a set of measure zero should be identified. For example if a function is non zero on a zero measure set it will have zero norm, but it will not be the identically zero function.

 

[cianfa72]

For example if a function is non zero on a zero measure set it will have zero norm, but it will not be the identically zero function.

Ah ok, so leveraging on this equivalence relation such a function on $(X,\Sigma)$ is identified with the zero function on $X$. This way the inner product $<,>$ turns out to be positive definite and actually induces an (effective) norm.

Edit: similar definition of $L^2$ spaces as Hilbert spaces can be done using Riemannian integral?

 

[martinbn]

Ah ok, so leveraging on this equivalence relation such a function on $(X,\Sigma)$ is identified with the zero function on $X$. This way the inner product $<,>$ turns out to be positive definite and actually induces an (effective) norm.

Edit: similar definition of $L^2$ spaces as Hilbert spaces can be done using Riemannian integral?

No, because you will not have a complete space.

 

[cianfa72]

Ok, so in the context of Quantum mechanics, for a particle without spin, I believe one takes the Hilbert space to be $L^2$ on $(X,\Sigma)$, $X=\mathbb R^3$ and $\Sigma$ the Borel sigma algebra on $\mathbb R^3$ (or perhaps the Lebesgue-measurable sigma-algebra ?).

Then each particle's state $\vec{\psi}$ is an element of the equivalence classes of square-integrable functions on $(X,\Sigma)$ and it can be given "in components" via the wave function $\psi(x,t)$ at any given $t$ (using the inner product with eigenfunctions of position or momentum operator).

 

[cianfa72]

Btw, how do we know actually exists a basis for the Hilbert space $L^2$ ?

In other words: we can show the axioms for an Hilbert space are fullfilled, however I'am not sure that it automatically follows that a basis (a system of linearly independent generators) actually exists.

 

[martinbn]

Btw, how do we know actually exists a basis for the Hilbert space $L^2$ ?

In other words: we can show the axioms for an Hilbert space are fullfilled, however I'am not sure that it automatically follows that a basis (a system of linearly independent generators) actually exists.

Every Hilbert space has a basis. Consider subsets consisting of orthonormal vectors ordered by inclusion. By Zorn's lemma there are maximal sets. Any of those are bases. If not pick an element from the orthocompliment and add it to the set, you get a bigger one.

 

[cianfa72]

Every Hilbert space has a basis. Consider subsets consisting of orthonormal vectors ordered by inclusion. By Zorn's lemma there are maximal sets. Any of those are bases. If not pick an element from the orthocompliment and add it to the set, you get a bigger one.

Ah ok, the above follows from the fact that Hilbert space is a vector space. Namely as limit case include in the basis all the vector's space elements that happen to be linearly independent.

 

[WWGD]

There's also the fact that the inner product gives rise to its norm, in the sense that $<f,f>=||f||^2$, this is a defining trait of Hilbert Spaces.

 

[WWGD]

There's also the fact that to define a norm ||.|| , we must have ||f||=0 iff f=0. As Martin Bn points out, that's why we identify functions that are 0 a.e. with the 0 function.