= Solution
The assumptions prove convergence to the set of <J-minimizing solutions>, and subsequential convergence to a member of that set. Convergence to a single fixed member requires uniqueness or a suitable selection rule.
Let $u^\dagger$ be any <J-minimizing solution>, whose existence is granted. It is feasible for every noisy-data constraint because $Au^\dagger=f$ and $\|f-f_\delta\|\leq\delta$. Minimality therefore gives $J(u_\delta)\leq J(u^\dagger)$. All reconstructions lie in this strongly <sequentially compact space> sublevel set. For any sequence $\delta_k\downarrow0$, pass to a subsequence with $u_{\delta_k}\to u_*$. Boundedness of $A$ and the <triangle inequality> imply
$$
\|Au_{\delta_k}-f\|\leq\|Au_{\delta_k}-f_{\delta_k}\|+\|f_{\delta_k}-f\|\leq2\delta_k,
$$
so $Au_*=f$. By <sequential lower semicontinuity>,
$$
J(u_*)\leq\liminf_k J(u_{\delta_k})\leq J(u^\dagger).
$$
Thus $u_*$ is a <J-minimizing solution>. The same compactness argument shows the <distance to a set> of exact minimizers tends to zero: otherwise a subsequence at a fixed positive distance would converge to an exact minimizer, a contradiction. If that minimizer is unique, every convergent subsequence has the same limit, giving the asserted full <strong convergence>.
For a counterexample to full convergence under the printed hypotheses alone, take $\mathcal U=\mathcal V=\mathbb R$, $A=0$, $f=f_\delta=0$, and $J(u)=\max\{|u|-1,0\}$. This is nonnegative and continuous, with compact sublevel sets $[-1-C,1+C]$ for $C\geq0$. Every point of $[-1,1]$ is both an exact and a residual-method minimizer. Choosing $u_{1/k}=(-1)^k$ gives a valid sequence of minimizers that does not converge. The reusable corrected statement is <compact-sublevel convergence of the residual method>.
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