 Solution

Fix $x$ and minimize $J_x(y)=f(y)+\Vert y-x{\Vert }^2/(2\tau )$. A proper lower semicontinuous convex function has an affine minorant $f(y)\ge a\cdot y+b$, as follows by separating a point below its closed epigraph. HenceThe quadratic dominates the linear term, proving coercivity. Properness supplies at least one finite trial value, lower semicontinuity passes to limits, and finite-dimensional compactness makes a bounded minimizing sequence converge along a subsequence to a minimizer. Thus the proximal operator exists at every $x$.

The subdifferential sum rule applies because the quadratic is finite and continuous everywhere. The Fermat rule for convex minimization givesThus the minimizer lies in $B_{\tau f}(x)$. The uniqueness proved in part (a), or strict convexity of the quadratic sum, now yieldsThis proof displays the separate roles of properness, lower semicontinuity, convexity and finite dimension. In particular, compactness here is not inferred merely from strict convexity.