Are there corner cases when using Hilbert basis and bounded operators?

Question

It seems that with a Hilbert basis and a bounded operator, you can do all the calculations while ignoring convergence issues, as if you were in a finite dimensional case.

For example if $a_{ij}$ are the coefficients of operator $A$ in the basis, then $\langle f(x),e_1\rangle=\sum_{i=0}^\infty x_ia_{1i}$ like the formula for a component of a vector image by a linear application in the finite dimensional case.

Or for example the product of two bounded operators would be (never seen it but I guess it works) : $c_{ij}=\sum\sum a_{ik}b_{kj}$ My question : are there still "traps" where treating these sums like finite sums in linear algebra, won't work ?

Answer 1

"Ignoring issues" is never good advice.

An obvious difference you have with the finite-dimensional case is that in finite dimension you can take any coefficients $c_1,\ldots,c_n$ and form $\sum_jc_je_j$. In an infinite-dimensional Hilbert space you need to make sure that $\sum_j|c_j|^2<\infty$ to be able to consider $\sum_jc_je_j$.

Regarding bounded operators, a drastic example is given by the trace. Given an orthonormal basis $\{e_n\}$ one can write, as in the finite-dimensional case, $$\tag1 \operatorname{Tr}(A)=\sum_n\langle Ae_n,e_n\rangle. $$ But this fails to make sense for most operators. The operators $A$ for which this makes sense are called the trace-class operators and they are a subclass of the compact operators. Here's an example of how things can go wrong when $A$ is not trace-class. Let $A$ be the selfadjoint compact operator $$ A=\bigoplus_n\begin{bmatrix} 0& 1/n\\ 1/n&0\end{bmatrix}. $$ This is the same as saying that, for some fixed orthonormal basis $\{e_n\}$, $$ Ae_{2n}=\frac{1}n\,e_{2n-1},\qquad Ae_{2n-1}=\frac 1n\,e_{2n}. $$ The naive computation in $(1)$ would give $\operatorname{Tr}(A)=0$. But the property that makes the trace interesting is that $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$, which is equivalent to say that $(1)$ does not depend on the choice of the orthonormal basis. If we use the orthonormal basis $$ \frac{e_1+e_2}{\sqrt2},\frac{e_1-e_2}{\sqrt2}, \frac{e_3+e_4}{\sqrt2},\frac{e_3-e_4}{\sqrt2},\ldots $$ now the diagonal of $A$ is $$\tag2 1,-1,\frac12,-\frac12,\ldots $$ This still converges to $0$, so it seems like all is good. Until we realize that any permutation of an orthonormal basis is an orthonormal basis, and that the series converges conditionally. So the diagonal of $A$ can be made any permutation of the sequence in $(2)$, which means that for each $r\in\mathbb R\cup\{\infty,-\infty\}$ there exists an orthonormal basis such that $(1)$ gives $\operatorname{Tr}(A)=r$.