Background: I'm reading about Hilbert spaces that require a complete metric space using inner product, where every Cauchy sequence of points $x_m$,$x_n$ on the metric space, $X$, has a limit, also in $X$. One difficulty I have is what motivates Cauchy sequences in Hilbert spaces, in the first place.
It is said the points in the sequences $x_n$, $x_m$ converge to $||x_n-x_m|| < r$, $r\in \mathbb{R}$, $r>0$ with $n,m>N$ for some $N \in \mathbb{N}$.
This reminds me of how compact operators $T:X\to Y$ between Hilbert spaces $X,Y$ are the limit of finite-rank operators: If one orders the eigenvalues of that compact operator, $T$, from largest to smallest, $|\lambda_1|\ge|\lambda_2|\ge|\lambda_3|...$, then the partial sums $T_n=\sum_{i=1}^n \lambda_i \phi_i$, (using basis functions, $\phi_n$,) form a Cauchy sequence with practical value, as the sequence approaches $T$, or at least $|T-T_n|<r$ for some $n>N$.
Next, in analogous fashion, we consider how expressing a function, $f(x)\in L^2$ in an orthonormal basis, $\{\psi(x)\}$, results in a sequence of generalized Fourier coefficients $\{c_i \} \in \ell^2$, where $c_i= <f(x),\psi_i(x)>$. Again, ordering the Fourier coefficients from largest to smallest, $|c_1|\ge|c_2|\ge|c_3|...$, the partial sums $f_n=\sum_{i=1}^n c_i \psi_i$, form a Cauchy sequence with practical value, as the sequence approaches $f(x)$, or at least $|f(x)-f_n(x)|<r$ for some $n>N$.*
Idea: Since the first partial sum $T_n$, a Cauchy sequence, and its limit, $T$, express compact operators, the second partial sum, $f_n$, and its limit, $f$, must be some sort of operator. Clearly it's a multiplication operator in the "eigenvalue domain", as $c_n=<\psi_n,\sum_{i=1}^\infty c_i \psi_i>=<\psi_n, c_n \psi_n>$.
Question: Expressing a function, $f(x)\in L^2$ in an orthogonal basis, is equivalent to a multiplication operator. Is it a compact operator? If not, what is its classification?
*Bonus: The above is is long enough, already, but given two projection operators $P,Q$ with $P$ being an orthogonal projection into a separable Hilbert space, that do not have commuting eigenfunctions, the operation, $PQP$ (not trivially zero,) is self-adjoint, with contractive eigenvalues that can be ordered, $1\ge|\lambda_1|\ge|\lambda_2|\ge|\lambda_3|...$ from largest to smallest. Is this $PQP$ compact?
Concerning the last part let $\{e_n\}_{n=1}^\infty$ denote an orthonormal basis in $\mathcal{H}.$ Consider the projection $P$ onto the closed subspace spanned by $\{e_{2n}\}_{n=1}^\infty$ and the projection $Q$ onto the closed subspace spanned by $\{\cos\theta_n e_{2n}+\sin\theta_n e_{2n-1}\}_{n=1}^\infty$ for a sequence $0<\theta_n<\pi/2.$ Then $P$ and $ Q$ do not admit common eigenvectors. We have $$ PQPe_{2n} =PQe_{2n}=P(\cos^2\theta_ne_{2n}+\cos\theta_n\sin\theta_ne_{2n-1})=\cos^2\theta_ne_{2n}$$The sequence $e_{2n}$ tends to $0$ weakly, but for a constant sequence $\theta_n$ the elements $PQPe_{2n}$ do not tend to $0.$ Hence $PQP$ is not compact. However if $\cos^2\theta_n={1\over 2^n}$ the operator $PQP$ is compact, even Hilbert-Schmidt,as $$\sum\|PQPe_n\|^2=\sum\|PQPe_{2n}\|^2=\sum{1\over 4^n}<\infty$$