A What happens when an operator maps a vector out of the Hilbert space?

[WWGD]

Well, IIRC, the " default" unbounded operators are the differential ones. And symmetry has a Spectral theorem to go along with it. Though we can define an unbounded operator in any infinite-dimensional space.

 

[SergejMaterov]

Differential operators are unbounded because they’re only defined on a proper dense subset of the Hilbert space, not on every vector.
If you insist on “full” domain, the only unbounded ones you get are pathological, non‑continuous algebraic beasts with no symmetry or spectral story.

 

[WWGD]

Differential operators are unbounded because they’re only defined on a proper dense subset of the Hilbert space, not on every vector.
If you insist on “full” domain, the only unbounded ones you get are pathological, non‑continuous algebraic beasts with no symmetry or spectral story.

Including multiplication operators? Sorry, I am a bit rusty.

 

[SergejMaterov]

Exactly so—multiplication operators fit the same pattern as the differential ones:
Multiplication by an unbounded function is fundamentally not everywhere defined, so it doesn’t contradict Hellinger–Toeplitz. Any physically meaningful unbounded operator—differential or multiplicative—lives on a proper dense domain.

 

[dextercioby]

[...]
If you insist on “full” domain, the only unbounded ones you get are pathological, non‑continuous algebraic beasts with no symmetry or spectral story.

Yes, this is what I'm after: a non-continuous operator whose domain is the whole of \mathcal{L}^2 (\mathbb{R}).

 

[gentzen]

Yes, this is what I'm after: a non-continuous operator whose domain is the whole of $\mathcal{L}^2 (\mathbb R)$.

Zorn's lemma is your friend:

Actually, you immediately have unbounded linear operators on a normed spaces as soon as you have a Hamel basis, and as you know, in general the existence of a Hamel basis on a linear space is ensured by the Zorn lemma. Then, if $(x_i)$ is any Hamel basis and $(y_i)$ is any family of vectors indicized on the same set, there is a unique linear map sending $x_i$ to $y_i$, and it is certainly unbounded if e.g. the $y_i$ are chosen in such a way that $|y_i|/|x_i|$ is unbounded.

 

[WWGD]

The example I remember is that of $l^2$, the sequence space. Choose an orthonormal Hamel basis $\{e_i\}$ , define $T(e_i):=ie_i$ and extend by linearity.

 

[A. Neumaier]

I've always had an issue with that, because you see, the theorem of Hellinger & Toeplitz precisely says that a symmetric operator everywhere defined on a Hilbert space must be bounded/continuos.

Yes, I was thinking of Hermitian operators; I corrected my statement.

 

[WWGD]

How do we define symmetric operators in infinite-dimensional (inner-product)spaces, is it <Tx,y>=<x, Ty>?

 

[dextercioby]

How do we define symmetric operators in infinite-dimensional (inner-product)spaces, is it <Tx,y>=<x, Ty>?

Yes, for whatever x and y from their maximal domain.