= Solution
Use the standard <bounded linear operator> setting for $K:U\to V$, so the <adjoint operator> $K^*:V\to U^*$ in the updates is defined on every datum. Put $q=K^*Ku_\lambda$. The <generalized singular vector> relation is $\lambda q\in\partial J(u_\lambda)$ with $\|Ku_\lambda\|=1$ and $\lambda=J(u_\lambda)\geq0$. For an <absolutely one-homogeneous functional>, a <subgradient> $p$ at $u$ is characterized by $\langle p,u\rangle=J(u)$ and $\langle p,v\rangle\leq J(v)$ for all $v$. Consequently $\lambda q$ is also a <subgradient> at every positive multiple of $u_\lambda$, and $bq\in\partial J(0)$ for every $0\leq b\leq\lambda$.
Use the initialization $u^0=p^0=0$ and start the first update at $k=0$; the PDF lists $k=1,2,\ldots$ despite specifying only index-zero initial data. Suppose $p^k=b_kq$ with $0\leq b_k\leq\lambda$. Define
$$
c_{k+1}=[\gamma-\alpha(\lambda-b_k)]_+.
$$
If it is positive, $u^{k+1}=c_{k+1}u_\lambda$ and $p^{k+1}=\lambda q$ satisfy the optimality equation and the dual update. If it is zero, take $u^{k+1}=0$ and $p^{k+1}=(b_k+\gamma/\alpha)q\in\partial J(0)$. Thus this compatible branch has
$$
b_{k+1}=\min\{\lambda,b_k+\gamma/\alpha\}.
$$
Induction proves the <finite termination of Bregman iteration on a singular vector> formulas
$$
\boxed{p^k=\min\{k\gamma/\alpha,\lambda\}q,\qquad u^k=\min\{\gamma,(k\gamma-\alpha\lambda)_+\}u_\lambda\quad(k\geq1).}
$$
Once the dual coefficient reaches $\lambda$, the next reconstruction is $\gamma u_\lambda$. Therefore a guaranteed stopping index for this branch is
$$
\boxed{k_*=\left\lceil\frac{\alpha\lambda}{\gamma}\right\rceil+1.}
$$
The formulas also cover $\lambda=0$, where exact reconstruction occurs at the first update. The <subgradient> conditions are sufficient for global minimization because each objective is convex.
There is a genuine uniqueness issue in the printed claim about arbitrary iterates. Strict convexity of data fidelity in $Ku$ ensures that any two minimizers at a given step have the same image under $K$. Induction therefore makes every minimizer branch have the same dual variables and attain exact data at the stated finite index. It does not make their states equal to the specified $\gamma u_\lambda$ if $K$ has a <null space>.
For a counterexample take $U=\mathbb R^2$, $V=\mathbb R$, $K(x,y)=x$, $J(x,y)=|x|$, $u_\lambda=(1,0)$, and $\lambda=1$. Every step leaves $y$ completely unconstrained. Choosing the minimizing state's second coordinate to be $k$ gives valid updates and <subgradients> but never reaches $(\gamma,0)$ and does not converge as a state <sequence>. Thus \b[finite exact-data recovery holds for all minimizer branches; exact recovery of the stated vector needs uniqueness or the aligned branch]. If $K$ is injective, equal reconstructed data identify the state, and the literal vector conclusion follows.
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