This exercise is from Higson's 'Analytic K - Homology'.
Let $H$ be the Hardy space $H^2(S^1)$. Show that the $C^*$-subalgebra of $\mathcal{B}(H)$ (bounded operators) generated by the Toeplitz operators $T_g$, $g\in C(S^1)$, contains all the compact operators on $H$.
I tried both the definitions of compact operators: 1) $T$ is compact if it maps unit ball to a precompact set. 2) $T$ is compact if for every bounded sequence, the image has a convergent subsequence
Any help is appreciated.
Various details raised by this question are scattered across the site, but the question itself doesn't appear to be an exact duplicate of anything - so I will collect various observations here as an answer that mostly points at other answers for details.
Different sources give different definitions of "compact operator" at various levels of generality, but for a bounded linear operator $T$ whose domain and codomain are the same Hilbert space $H$, most of these variations simplify considerably. One definition, also used in more general contexts, is that $T$ is compact iff $T$ maps the closed unit ball of $H$ to a subset of $H$ that is precompact (i.e. has compact closure) in the norm topology of $H$. (As you note in the question above, from the general equivalence of compactness and sequential compactness in metric spaces it follows that $T$ is compact iff for every bounded sequence $(\xi_n)_{n \geq 0}$ in $H$, the sequence $(T \xi_n)_{n \geq 0}$ has a norm convergent subsequence.)
With this definition in hand, one can deduce many elementary facts about compact operators, including that a finite rank operator is necessarily compact, that a norm limit of compact operators is necessarily compact, and that the adjoint of a compact operator is compact. (These facts generalize; e.g. the answers just linked include proofs that work for Banach spaces that might not be Hilbert spaces. Note that the meaning of "adjoint" is slightly different in the Banach space case but that the answers to that question discuss this issue.)
In the Hilbert space setting it is also possible to prove that every compact operator is a norm limit of finite rank operators. One way of seeing this is to choose an orthonormal basis $(e_n)_{n \geq 0}$ of the Hilbert space $H$, to let $P_n \in \mathcal{B}(H)$ denote the orthogonal projection onto the span of $\{e_j: 0 \leq j \leq n\}$, and to prove that if $A \in \mathcal{B}(H)$ is compact then $A$ is the norm limit of the specific sequence $(P_n A)_{n \geq 0}$ of finite rank operators. (This fact also generalizes somewhat, e.g. to whenever the identity on a Banach space is the strong limit of a uniformly bounded sequence of finite rank operators, although it does not generalize to every Banach space.) We will use a specific consequence of this below - that that any compact $A \in \mathcal{B}(H)$ is the norm limit of the sequence $(P_n A P_n)_{n \geq 0}$. Sketch of proof: note that \begin{align*} \|P_n A P_n - A\| & = \|(P_n A - A) P_n + A P_n - A\| \\ & \leq \|P_n A - A\| + \|A P_n - A\| \\ & = \|P_n A - A\| + \|(P_n A^* - A^*)^*\| \\ & = \|P_n A - A\| + \|P_n A^* - A^*\| \end{align*} and the right hand side goes to $0$ by applying the previously stated results to the compact operators $A$ and $A^*$. End of sketch.
Some intuition can perhaps be gained by using the orthonormal basis $(e_n)_{n \geq 0}$ to think of operators on $\mathcal{B}(H)$ as "infinite matrices." For any operator $A \in \mathcal{B}(H)$ there are scalars $(a_{jk})_{j,k \geq 0}$ with the property that for all $j, k \geq 0$ we have $A e_j = \sum_{k \geq 0} a_{kj} e_k$, and for arbitrary $\xi = \sum_{j \geq 0} \xi_j e_j$ in $H^2$ you can even compute $A \xi$ as an "infinite matrix product" of the "infinite matrix" $(a_{kj})_{j, k \geq 0}$ with the "infinite column vector" $(\xi_j)_{j \geq 0}$ in the usual way. In terms of matrices, the finite rank operator $P_n A$ (respectively, $A P_n$) corresponds to the matrix obtained from the matrix of $A$ by zeroing out all rows (respectively, columns) beyond the $n$th row. The matrix of $P_n A P_n$ is obtained by zeroing out all entries of $A$ in any row beyond the $n$th row or any column beyond the $n$th column. For general bounded $A$, the sequences $A P_n$, $P_n A$, and $P_n A P_n$ need not converge in the norm topology at all - but as shown above, each of these sequences converges to $A$ when $A$ is compact.
To put it more simply (if more vaguely), the intuition is that a compact operator on $H$ is one whose "matrix" (with respect to any orthonormal basis) is the norm limit of its various truncations to sequences of "matrices" of finite rank. If we specialize the above discussion to the Fourier orthonormal basis of $H^2$, we find via a short calculation that the finite rank truncations $P_n A P_n$ of any operator all lie in the Toeplitz algebra. It follows from the above general discussion that any compact operator on $H^2$ is also in the Toeplitz algebra.
More specifically, let $\mathcal{T}$ denote the $C^*$-subalgebra of $\mathcal{B}(H^2)$ generated by all Toeplitz operators with continuous symbol. Letting $\zeta: S^1 \to S^1$ denote the identity function on $S^1$, it follows that $\mathcal{T}$ contains the operator $T_{\zeta}$, which might be more easily recognized via its action on the usual Fourier orthogonal basis $(\zeta^j)_{j \geq 0}$ of $H^2$ as the unique bounded linear operator $S$ satisfying $S \zeta^j = \zeta^{j+1}$ for all $j \geq 0$). (I'm writing $S$ for $T_{\zeta}$ because I will not expressly consider Toeplitz operators with any other symbol, and the subscript $\zeta$ only complicates the notation.)
Because $\mathcal{T}$ is a $C^*$-algebra containing $S$, the $C^*$-algebra $\mathcal{T}$ also contains the operators $$ E_{mn} := S^m (I - S S^*) (S^*)^n, \qquad m, n \geq 0, $$ as well as any limit of any norm-convergent sequence of finite linear combinations of these operators.
A short calculation shows that for any $m, n \geq 0$, the operator $E_{mn}$ sends $\zeta^n$ to $\zeta^m$ and all other elements of the Fourier basis (that is, all $\zeta^j$ with $j \neq n$) to $0$. This calculation or others of its type can be used to show that for all $n \geq 0$ that the operator $E_{nn}$ is the orthogonal projection onto the span of $\zeta^n$, that for all $n \geq 0$ the orthogonal projection $P_n$ onto the span of $\{\zeta^j: 0 \leq j \leq n\}$ is $\sum_{j=0}^n E_{jj}$, and that for any operator $A \in \mathcal{B}(H)$ (whether compact or not) and any $n \geq 0$ the operator $P_n A P_n$ is given by the finite sum $$ P_n A P_n = \sum_{j, k \leq n} \langle A \zeta^j, \zeta^k\rangle E_{kj} $$ and in particular is an element of $\mathcal{T}$ (here I write $\langle \cdot, \cdot \rangle$ for the inner product of $H^2$, linear in the first entry). When $A$ is compact, the general fact that the sequence $P_n A P_n$ converges in norm to $A$ (noted above) then implies $A$ is in $\mathcal{T}$.
