= Solution
The <Schumacher compression> theorem concerns a <memoryless quantum information source> with average <density matrix> $\rho$ on a finite-dimensional <Hilbert space>. For $n$ uses the average state is $\rho_n=\rho^{\otimes n}$. Compression consists of an encoding <quantum channel> into a space of dimension $D_n$ and a decoding channel back to the original block space, with asymptotic rate $n^{-1}\log_2D_n$ qubits per source use. \b[The optimal reliable rate is $S(\rho)$: every larger rate is achievable, and no smaller rate is reliable.] This holds for average pure-source fidelity; the direct construction also preserves a purification reference, and the converse for that stronger <entanglement fidelity> criterion holds with fidelity tending to zero below the threshold. The encoding and decoding are deterministic channels; no free classical transmission of the source label is allowed.
Diagonalize $\rho=\sum_j\lambda_j|j\rangle\langle j|$ and omit zero-probability eigenvalues when sampling. Define the <quantum typical subspace> projector $P_{n,\delta}$ by retaining product eigenvectors with
$$
2^{-n(S+\delta)}\leq\lambda_{j_1}\cdots\lambda_{j_n}\leq2^{-n(S-\delta)}.
$$
The sampled random variable $-\log_2\lambda_j$ has expectation $S=S(\rho)$. The <weak law of large numbers> gives $w_n=\operatorname{Tr}(\rho_nP_{n,\delta})\to1$. Every retained eigenvalue is at least $2^{-n(S+\delta)}$, and the total probability is at most one, so
$$
\operatorname{rank}P_{n,\delta}\leq2^{n(S+\delta)}.
$$
This proves the two <typical subspace theorem> estimates used in the direct part.
Write $P=P_{n,\delta}$. Encode its range isometrically into a compressed space and reserve one orthogonal failure flag. Decode the range back isometrically and the failure flag to a fixed pure state $|\varphi\rangle$. This <typical-subspace compression with a failure flag> is a physical channel whose composite action is
$$
\mathcal N_n(X)=PXP+\operatorname{Tr}[(I-P)X]|\varphi\rangle\langle\varphi|.
$$
Indeed its Kraus operators are $P$ and $|\varphi\rangle\langle e_a|$, where the $e_a$ form an orthonormal basis of $\ker P$, and their adjoint products sum to $I$. The <Kraus formula for entanglement fidelity> gives
$$
F_e(\rho_n,\mathcal N_n)=\sum_\alpha|\operatorname{Tr}(\rho_nK_\alpha)|^2\geq w_n^2\longrightarrow1.
$$
For any pure-state ensemble of $\rho_n$, the average squared <quantum fidelity> is at least this number: apply convexity of $|z|^2$ separately to the amplitudes $\langle\psi_x|K_\alpha|\psi_x\rangle$ with their source probabilities. Thus average fidelity also tends to one. The compressed dimension is at most $2^{n(S+\delta)}+1$, and its rate is at most $S+\delta+o(1)$. Choose $\delta<R-S$ for any prescribed $R>S$. One may also take $\delta_n=n^{-1/4}$: the finite spectral information variance and <Chebyshev's inequality> still give $w_n\to1$, with rate tending to $S$. A pure source, $S=0$, can simply be re-prepared exactly without sending any qubits.
For the coherent converse, we prove the <finite-dimensional quantum compression converse>. If the compressed space has dimension $D$, the state of reference plus compressed system has a pure-state decomposition with <Schmidt rank> at most $D$. Applying decoder Kraus operators locally cannot increase the Schmidt rank of those vectors, so the final reference–output state has <Schmidt number> at most $D$. Let $|\Psi_{\rho_n}\rangle$ purify the source, with squared Schmidt coefficients equal to the decreasing eigenvalues $\mu_1\geq\mu_2\geq\cdots$ of $\rho_n$. For any normalized vector $|v\rangle$ of Schmidt rank at most $D$, its reference support lies in a rank-at-most-$D$ projector $Q$. Hence
$$
|\langle\Psi_{\rho_n}|v\rangle|^2
\leq\|(Q\otimes I)|\Psi_{\rho_n}\rangle\|^2
=\operatorname{Tr}(Q\rho_{n,R})\leq\sum_{j=1}^D\mu_j.
$$
The last bound follows by maximizing $\sum_j\mu_j\langle j|Q|j\rangle$ subject to $0\leq\langle j|Q|j\rangle\leq1$ and their sum at most $D$. Average over a Schmidt-rank-bounded decomposition of the final state to obtain $F_e\leq\sum_{j=1}^D\mu_j$.
At rate $R<S$, choose $0<\delta<S-R$. The total source probability outside the <quantum typical subspace> tends to zero, while every eigenvalue inside it is at most $2^{-n(S-\delta)}$. Therefore for $D_n\leq2^{nR}$,
$$
F_e\leq\sum_{j=1}^{D_n}\mu_j\leq(1-w_n)+2^{nR}2^{-n(S-\delta)}\longrightarrow0.
$$
This argument allows arbitrary encoding and decoding channels, not only subspace projection.
For completeness, average pure-source reliability also forces the same rate, even though it is a weaker criterion than entanglement fidelity. Here is a separate converse for that formulation. Let the emitted block pure states be $\psi_x$, with probabilities $p_x$, and let $\sigma_x$ be the decoded states. If their mean squared fidelity tends to one, then their mean <trace distance> $\bar\delta_n$ tends to zero, since $\tfrac12\|\sigma_x-\psi_x\|_1\leq\sqrt{1-\langle\psi_x|\sigma_x|\psi_x\rangle}$. This inequality follows by choosing a purification with that overlap and using contraction of trace distance under a partial trace.
We need only an elementary <entropy continuity from Jordan decomposition> bound, derivable from Question 4. If $\delta=\tfrac12\|\sigma-\tau\|_1$, write the positive and negative parts of $\sigma-\tau$ as $\delta\omega_+,\delta\omega_-$. The common density matrix
$$
\frac{\sigma+\delta\omega_-}{1+\delta}=\frac{\tau+\delta\omega_+}{1+\delta}
$$
and the two mixture bounds give
$$
|S(\sigma)-S(\tau)|\leq\delta\log_2d+g(\delta),\qquad
g(\delta)=(1+\delta)h_2\left(\frac\delta{1+\delta}\right).
$$
Here $g$ is increasing and concave, $g(0)=0$. Apply this in block dimension $d^n$, both to the average states and to individual outputs, using convexity of trace distance and concavity of $g$. Since the original ensemble is pure, its <Holevo quantity> is $S(\rho_n)=nS$. The decoded ensemble has
$$
\chi_{\mathrm{out}}=S\left(\sum_xp_x\sigma_x\right)-\sum_xp_xS(\sigma_x)
\geq nS-2\bar\delta_n n\log_2d-2g(\bar\delta_n)=nS-o(n).
$$
The <Holevo quantity under a quantum channel> cannot increase. To specify the entropy ingredient, <Strong subadditivity of Von Neumann entropy> states $S(XB)+S(BE)\geq S(B)+S(XBE)$. Attach the classical source label $X$, dilate the decoder isometrically to $BE$, and trace out $E$. The displayed inequality says $I(X:B)\leq I(X:BE)$; the isometry preserves the input mutual information. Since $I(X:B)$ of a labelled ensemble is its Holevo quantity, this proves the needed decoder monotonicity. Before decoding, the system has dimension $D_n$, so its Holevo quantity is at most $\log_2D_n$. Thus $\log_2D_n\geq nS-o(n)$ and the rate cannot be below $S$. \b[Both operational fidelity formulations therefore have the claimed noiseless coding threshold.]
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