= Solution
Assume the displayed variational minimizer exists and $K:U\to V$ is bounded and linear. The smooth data-fidelity term and the <subdifferential sum rule> give
$$
p_\alpha=\frac1\alpha K^*(f^\delta-Ku_\alpha)\in\partial J(u_\alpha).
$$
Use the supplied <source condition in variational regularization> $p^\dagger=K^*w\in\partial J(u^\dagger)$. Put $e=f^\delta-f$ and $q=K(u_\alpha-u^\dagger)$, so $\|e\|\leq\delta$. By the <symmetric Bregman distance> identity,
$$
\alpha D_J^{\mathrm{sym}}(u_\alpha,u^\dagger)=\langle e-q-\alpha w,q\rangle=\frac14\|e-\alpha w\|^2-\left\|q-\frac{e-\alpha w}{2}\right\|^2.
$$
Dropping the last square and using $\|e-\alpha w\|^2\leq2\|e\|^2+2\alpha^2\|w\|^2$ proves the <source-condition estimate for symmetric Bregman distance>:
$$
\boxed{D_J^{\mathrm{sym}}(u_\alpha,u^\dagger)\leq\frac{\|e-\alpha w\|^2}{4\alpha}\leq\frac{\delta^2}{2\alpha}+\frac\alpha2\|w\|_V^2.}
$$
The source condition also makes $u^\dagger$ a $J$-minimizing exact solution: for $Ku=f$, its <subgradient> inequality has zero pairing with $u-u^\dagger$. Existence of $u_\alpha$ is not implied by properness and lower semicontinuity alone; it is a hypothesis when no coercivity/compactness condition is provided.
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