Old title: Components of unbounded endomorphism wrt Schauder basis? Changed for discoverarbility.
In finite-dimensional linear algebra, given an endomorphism $T$ and a basis $\boldsymbol{e}_i$, we can easily find a matrix representation of $T$: $$ T(v) = T(\sum_i v^i \, \boldsymbol{e}_i) \overset{*}{=} \sum_i T(v^i \, \boldsymbol{e}_i) = \sum_i v^i T(\boldsymbol{e}_i) =: \sum_{i,j} v^i \, T_{ij} \, \boldsymbol{e}_j $$
Let us now assume that $X$ is a separable Banach space with a Schauder basis $\{\boldsymbol{e_i}\}_{i \in \mathbb{N}}$ and $T: X \to X$ is an unbounded linear operator. If we try to replicate the previous equation, the sum will become infinite, and since for unbounded operators $T(\lim_n v_n) = \lim_n T(v_n)$ doesn't hold, the equality $(\overset{*}{=})$ above doesn't hold.
This would seem to indicate that unbounded operators don't have a coordinate representation in a Schauder basis, at least not in the same sense as bounded operators do. Or is there a simple generalization that I just don't see?
If we now consider $X$ to be a separable Hilbert space and $\boldsymbol{e}_i$ an orthonormal basis, we know that it's isomorphic to $\ell_2$ with basis: $$ \begin{pmatrix}1\\0\\0\\\vdots\end{pmatrix}, \begin{pmatrix}0\\1\\0\\\vdots\end{pmatrix}, \begin{pmatrix}0\\0\\1\\\vdots\end{pmatrix}, \;\dots $$ This means that either the endomorphisms on the space of „infinite column vectors“ aren't always „infinite matrices“, or that my previous conclusion about the „non-representability“ of unbounded operators was wrong. Which of these is true?
In summary, these are my questions:
- Is it possible to represent an unbounded operator using its coordinates wrt. a Schauder basis of a separable Banach/Hilbert space?
- If yes, how? If no, are there other practical ways to represent it? (Eg. using Hamel basis? Or as an integral kernel on $L^p$?)
- Are all operators on $\ell_2$ „infinite matrices“?
Claim: Given a Schauder basis $\{ \boldsymbol{e}_j \} \subset X$ and a linear operator $T: \mathrm{D}(T) \subset X \to X$, the vectors $T(\boldsymbol{e}_j)$ aren't always sufficient to describe the action of the operator on a general vector of $X$.
Or in other words: There exists a non-zero linear operator, whose matrix is zero.
Proof: Let $X = L^2([0, 2\pi])$ be the space of all square-integrable functions equipped with the Fourier basis $\{ \boldsymbol{e}_j \} = (1,\, \sin x,\, \cos x,\, \sin 2x,\, ...)$. We now use the distribution theory to construct an unbounded linear operator $T$.
Let us take $\mathrm{D}(T)$ to be the space of bounded piece-wise continuous functions. Since $\mathrm{D}(T) \subset L^1_{\mathrm{loc.}}(\mathbb{R})$, all vectors $f \in X$ can be considered regular distributions on $\mathcal{D}(\mathbb{R})$ and assigned a distributional derivative $Df$. The result will be a sum of a regular function and (singular) delta distributions in discontinuities: $Df = (Df)_{\mathrm{reg.}} + (Df)_{\mathrm{sing.}}$. Finally we chose a test function $\varphi \in \mathcal{D}(\mathbb{R})$ that is nonzero on $[0, 2\pi]$ (ie. $\varphi(x)=1$ on $[0, 2\pi]$). We define: $$ T(f) = (Df)_{\mathrm{sing.}}[\varphi] \; \boldsymbol{e}_1 $$ This operator is zero for all continuous functions, including the entire basis $\{ \boldsymbol{e}_j \}$, but it is generally nonzero for discontinuous functions.
Corollary: Since, given a Schauder basis, there's a natural isomorphism $\iota: L^2 \to \ell_2$: $$ \underbrace{f}_{L^2} = \sum_j \underbrace{c_j}_{\ell_2} \boldsymbol{e}_j \: , \quad \iota(f) = \{ c_j \} $$ we can construct a non-zero operator on $\ell_2$ whose matrix is only zeros: $$ S(a_j) = \iota \, T \, \iota^{-1} \; a_j \quad \neq \quad \sum_{j,k} S_{jk} \, a_j \, \boldsymbol{e}_k = 0 $$
Claim: Let $X$ be a separable Banach space and $T: \mathrm{D}(T) \subset X \to X$ a densely defined closed operator. Given a Schauder basis $\{ \boldsymbol{e}_j \} \subset \mathrm{D}(T)$ (such a basis always exists), we construct the coordinates $T_{jk}$ of the operator $T$: $$ T(\boldsymbol{e}_j) =: \sum_k T_{jk} \, \boldsymbol{e}_k $$ Now we define an operator $T_0: \mathrm{D}(T_0) \to X$ using those coordinates: $$ T_0(v) = T_0( \, \sum_j v^j \boldsymbol{e}_j \, ) = \sum_{j,k} T_{jk} \, v^j \, \boldsymbol{e}_k \: , \qquad \mathrm{D}(T_0) = \{ v \in X \,|\, \text{the sum converges}\} $$ Then $T_0$ is also closed and $T_0 \subset T$.
Proof: The Schauder basis in question exists, because $\mathrm{D}(T)$ is dense in $X$ – therefore it is a generating set in the sense of countable linear combinations. We can throw away its linearly dependent vectors until we're left with a basis.
The defining feature of closed operators is: $$ T \text{ is closed} \;\wedge\; x_n \to x \;\wedge\; T(x_n) \to y \quad \implies \quad x \in \mathrm{D}(A) \;\wedge\; T(x) = y $$ To show that $T_0 \subset T$, we chose an arbitrary $v \in \mathrm{D}(T)$ and define: $$ x_n = \sum_{j=1}^n v^j \, \boldsymbol{e}_j \: , \quad x_n \to v $$ Then: $$ T(x_n) = \sum_{j=1}^n v^j \, T(\boldsymbol{e}_j) = \sum_{j=1}^n \sum_k T_{jk} \, v^j \, \boldsymbol{e}_k $$ If that sum converges for $n\to\infty$, by the definition of $T_0$, it holds that $v \in \mathrm{D}(T_0)$ and the result is equal to $T_0(v)$. However, from the closedness criterion, we also know that $v\in\mathrm{D}(T)$ and the result is equal to $T(v)$. If the sum doesn't converge, by definition $v\notin\mathrm{D}(T_0)$, but it doesn't tell us anything about $T$. Thus, $T_0 \subset T$.
It remains to prove that $T_0$ is also closed. We assume $① \Leftrightarrow x_n \to x$ and $② \Leftrightarrow T_0(x_n) \to y$ and want to prove $T_0(x) = y$. If we expand the expressions, we get: $$ ① \; \implies \; \sum_{j=1}^m (x_n)^j \, \boldsymbol{e}_j \xrightarrow{n} \sum_{j=1}^m x^j \, \boldsymbol{e}_j \\[8pt] ② \; \Longleftrightarrow \; \sum_{j,k} T_{jk} \, (x_n)^j \, \boldsymbol{e}_k \xrightarrow{n} y $$ To simplify the matters, we introduce a double-sequence $\Phi_{mn}$: $$ \Phi_{mn} = \sum_{j=1}^m \sum_k T_{jk} \, (x_n)^j \, \boldsymbol{e}_k $$ The assumption $②$ is equivalent to $\lim_{n\to\infty} \lim_{m\to\infty} \Phi_{mn} = y$. The assumption $①$ implies that $\lim_{n \to \infty} \Phi_{mn}$ exists – as it turns out, the convergence is even uniform wrt. $m$ (is it really tho?). Then it follows from the Moore-Osgood Theorem, that $\lim_{m\to\infty} \lim_{n\to\infty} \Phi_{mn}$ exists and is equal to $y$, too.