Domain of an operator in functional analysis

I used to think that if we say $f$ is a function from a set $X$ to $Y$ then this implied that $f$ was defined on all of $X$. Because the definition of function is that it's a set $\{(x,y) \mid \text{ for every } x \in X \text{ there is exactly one } (x,y) \text{ where } y \in Y \}$.

Now I'm reading the definition of closed graph of a linear operator:

So a linear operator is somehow not a function. But then what is it? Why does it make sense to say that $T$ is a map from $X$ to $Y$ when it's not even defined on all of $X$?

In response to comment:

Edit

Perhaps my question is: why and when does it make sense to say $T:X \to Y$ is a linear operator when $T$ is not even defined on all of $X$?

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How are linear operators defined in your book? Sometimes authors use function to cover both total function and partial function, so this may have something to do with it. Though I must admit that I've not seen this distinction made before in the context of functional analysis $-$ I can't think what use partial functions would be in this context! – Clive Newstead Jul 30 '12 at 12:31
@CliveNewstead I added the definition given in my lecture notes. – Rudy the Reindeer Jul 30 '12 at 12:36
I suspect it has something to do with the following, taken from Wikipedia: "The restriction on the domain is needed due to the existence of closed unbounded linear operators. The differentiation operator on $C([0,1])$ is a prototypical counter-example." Although it doesn't say with what norm. Probably, $\|\cdot\|_\infty$. But I don't understand why one does not define the differentiation operator on $C^1([0,1])$ instead. – Rudy the Reindeer Jul 30 '12 at 12:39
A linear operator can be seen as a pair $(D_T,T)$, where $D_T$ is a subspace of $X$ and $T\colon D_T\to Y$ is a linear map. – Davide Giraudo Jul 30 '12 at 12:39
Re: Edit. Apart from very sloppy people no-one says "$T\colon X \to Y$ is a linear operator" if $T$ isn't everywhere defined. The domain of definition is a crucial part of the data -- the "same" operator can behave differently depending on the domain, see Paul's answer. People might say "$T$ is a densely defined operator on $X$" if the domain $D$ is understood but such abuse of language usually goes no further. Typically the (formal) operator $T$ is clear from the problem at hand, but neither $D_T$, $X$, nor $Y$ is fixed a priori. They're suitably chosen depending on what you want to do. – t.b. Aug 15 '12 at 15:37

This is an admittedly confusing abuse of terminology which your book appears to make even more confusing by using somewhat nonstandard terminology. The following is more standard but also confusing in its own way. For what follows let $X$ and $Y$ be Banach spaces.

A bounded linear operator from $X$ to $Y$ is a function $T: X \to Y$ which is linear and norm continuous.

An unbounded linear operator from $X$ to $Y$ is a pair $(T,D_T)$ where $D_T$ is a linear subspace of $X$ and $T: D_T \to Y$ is a linear map.

• An unbounded linear operator from $X$ to $Y$ is not in general a function from $X$ to $Y$, and the definition does not claim it is. The phrase "from $X$ to $Y$" is part of what is being defined.

• Because unbounded operators are not functions, extra care must be taken: two unbounded linear operators from $X$ to $Y$ cannot necessarily be added, and given unbounded linear operators from $X$ to $Y$ and from $Y$ to $Z$ they do not necessarily have a composition.

• "Unbounded" really means "not necessarily bounded"; every bounded linear operator is an unbounded linear operator.

• Some authors require that the domain $D_T$ of $T$ is a dense subspace of $X$. With this convention one can conceive of operators which are neither bounded nor unbounded. When this is not part of the definition, the assumption is added via the phrase "densely defined".

• Adjoints can be very confusing: there are densely defined unbounded operators such that the largest possible domain of definition for the adjoint is $\{0\}$.

The only real way to sort through all of this language is to look at some examples and results. The theory was invented by Von Neumann in the 20's to make precise sense of what physicists were doing in quantum mechanics. They had stumbled onto the idea that classical notions like position and momentum can be profitably viewed as linear operators on Hilbert space which satisfy certain relations. Mathematicians noticed, however, that there are no bounded operators on Hilbert space which satisfy these relations, and in particular the physicists' use of spectral theory had no mathematical basis. So Von Neumann invented the language of unbounded operators to make sense of the operators that physicists were doing. In particular he generalized the spectral theorem to certain classes of unbounded operators.

Maybe it would help to look at a quick example. Consider the operator $D = i \frac{d}{dx}$, a linear map $C_{per}^\infty[0,1] \to C_{per}^\infty[0,1]$, the space of smooth periodic functions on the interval $[0,1]$. Notice that $D$ has a rich supply of eigenfunctions: setting $e_n(x) = e^{-2 \pi i n x}$, we find that $e_n$ is an eigenfunction of $D$ with eigenvalue $2 \pi n$. The set of functions $\{e_n\}$ forms an orthonormal basis for the Hilbert space $L^2[0,1]$, so even though $D$ isn't an operator on $L^2[0,1]$ we can in a sense diagonalize it and use Hilbert space techniques to study it.

By contrast, consider the same operator $D = i \frac{d}{dx}$, this time viewed as a linear map $C_0^\infty(0,1) \to C_0^\infty(0,1)$, the space of smooth functions on the interval $[0,1]$ which vanish at $0$ and $1$. Now the only eigenfunctions for $D$ are constant functions, so it is not clear how one would diagonalize $D$. Secretly we know that we should just enlarge the domain of $D$ to include all periodic functions, but this isn't immediately obvious. Part of Von Neumann's accomplishment was to understand abstractly why $D$ with the domain $C_{per}^\infty[0,1]$ is "good" but $D$ with the domain $C_0^\infty(0,1)$ is not.

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I disagree with your second and third bullet points. You can always add and compose unbounded operators. You just need to be careful about the domains of definition (and you might get something substantially smaller than what you started with). – t.b. Jul 30 '12 at 14:24
How do you add two unbounded operators with disjoint domains? – Paul Siegel Jul 30 '12 at 14:27
Being linear subspaces, the domains of definition $D_S$ and $D_T$ always have $0$ in common and the sum of $S$ and $T$ is the usual sum but with domain $D_{S+T} = D_S \cap D_T$. The composition of two operators is defined as the usual definition on $D_{TS} = D_S \cap S^{-1}D_T$ (both these domains might be zero, in particular non-dense, of course). Both operations are associative when defined this way but they need not distribute over each other. Sorry: it was really just meant as a nitpick (and for some reason I read the second bullet point as two bullet points). It's a very nice answer, +1. – t.b. Jul 30 '12 at 15:03
I see, you're right. I should have written something like "the sum / composition of two densely defined operators can be trivial in the sense that they are only defined on the $\{0\}$ subspace". That together with your remark that these notions of addition and composition don't distribute over each other helps make my point correctly that addition and composition of unbounded operators is pathological. – Paul Siegel Jul 30 '12 at 17:48
I should have made this more clear. It is typical to require $X$ and $Y$ in the definition of unbounded operator to be Banach spaces (Hilbert spaces are even better), and spaces of smooth functions basically never are. But in both examples we can take $X = Y = L^2[0,1]$ (Hilbert spaces); the difference between the examples is that the domain of definition is $C^\infty_{per}[0,1]$ in the first example and $C^\infty_0(0,1)$ in the second (both are subspaces of $X = L^2[0,1]$). The examples show that the domain you choose for the operator matters a great deal. – Paul Siegel Aug 2 '12 at 12:40

A linear operator $T$ is actually a pair $(D_T,T)$, where $D_T$ is a subspace of $X$ and $T\colon D_T\to Y$ is a linear map. So two linear operators $S$ and $T$ are considered to be equal if they have the same domain $D$, and $Sx=Tx$ for all $x\in D$.

This allows to consider the concept of extension: $(D_T,T)$ is an extension of $(D_S,S)$ if $D_S\subset D_T$ and for all $x\in D_S, Sx=Tx$.

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