= Solution
Comparison with zero and nonnegativity of the two terms gives
$$
0\leq D(u_\alpha)+\alpha J(u_\alpha)
\leq D(0)+\alpha J(0)=\frac12\|f\|^2.
$$
Therefore
$$
\boxed{0\leq J(u_\alpha)\leq\frac{\|f\|^2}{2\alpha},\qquad
J(u_\alpha)\longrightarrow0\quad(\alpha\to\infty).}
$$
For a concrete <variational regularization> example, take a <Hilbert space> and $J(u)=\|u\|^2$. The bound forces $u_\alpha\to0$ in <norm>. Since $K$ is a <bounded linear operator>, $Ku_\alpha\to0$, and continuity of the squared <norm> gives
$$
\boxed{D(u_\alpha)\longrightarrow\tfrac12\|f\|^2.}
$$
The same conclusion is not forced by $J(u_\alpha)\to0$ for a regulariser with a nontrivial zero set.
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