= Solution
The small-parameter limit depends on <data compatibility with the regularizer domain>. Set
$$
m=\inf_{v\in\operatorname{dom}J}D(v),\qquad 0\leq m\leq D(0).
$$
For every such $v$, optimality gives
$$
m\leq D(u_\alpha)\leq\Phi_{\alpha,f}(u_\alpha)
\leq D(v)+\alpha J(v).
$$
Taking the upper limit as $\alpha\downarrow0$ and then the <infimum> over $v$ proves $\Phi_{\alpha,f}(u_\alpha)\to m$. Its two nonnegative contributions above $m$ must vanish, yielding the general result
$$
\boxed{D(u_\alpha)\longrightarrow m,\qquad
\alpha J(u_\alpha)\longrightarrow0.}
$$
In particular, the requested zero-misfit limit holds when $f\in\overline{K(\operatorname{dom}J)}$. A sufficient condition is an exact solution $v$ with $Kv=f$ and $J(v)<\infty$; then
$$
0\leq D(u_\alpha)\leq\alpha J(v),\qquad
0\leq\alpha J(u_\alpha)\leq\alpha J(v)\longrightarrow0.
$$
\b[Membership of $f$ in $\mathcal R(K)$ alone is insufficient.] Take $\mathcal U=\mathcal V=\mathbb R$, $K=I$, $f=1$, and $J$ the <indicator functional of a constraint set> $\{0\}$. All positive-parameter objectives are <proper extended-real functions> with <coercivity> and <sequential lower semicontinuity>, with the unique minimizer $u_\alpha=0$, yet $D(u_\alpha)=1/2$ for every $\alpha$. Thus the first printed limit needs compatibility with the regulariser's <effective domain>; the second limit remains true under the given assumptions.
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