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The squared distance to a convex set is the infimal convolution of the set's indicator functional of a constraint set and the half-squared Hilbert space norm. Its convex conjugate is the sum of the half-squared norm and the support function of the set.

The infimal convolution is $(E\square F)(u)={\inf }_v\{E(v)+F(u-v)\}$. It optimizes a decomposition of $u$ between two costs. It is convex when both costs are convex functions, and its convex conjugate is $E^{\ast }+F^{\ast }$ whenever the extended-real operations are well defined.

Let $G=E\square F$. To prove convexity, take $u_1,u_2$ with finite $G$ values, $0<t<1$, and $\epsilon >0$. By the definition of infimum, choose $v_i$ withPut $u_t=t{u}_1+(1-t)u_2$ and $v_t=t{v}_1+(1-t)v_2$. Convexity of $E$ and $F$ givesLet $\epsilon \downarrow 0$. If either endpoint value is infinite, the convexity inequality is automatic, and $t=0,1$ gives equality. Thus the infimal convolution of two convex functions is convex under the stated properness assumption. The proof uses approximate minimizing splits, so no attainment of the inner infimum is assumed.

For $p\in {\mathcal{U}}^{\ast }$, the definition of the convex conjugate giveswhere the substitution $w=u-v$ is a bijection of the independent pairs. ThereforeThis infimal convolution identity does not need convexity of $E,F$ or attainment of the inner infimum. The assumptions ensure that both functions have nonempty effective domains; their conjugates never take $-\infty$, so the separated sum is well defined, allowing $+\infty$.

Write ${\iota }_C$ for the indicator functional of a constraint set $C$ and $q(u)=\Vert u{\Vert }^2/2$. ThenBoth terms are convex, hence their infimal convolution is the convex squared distance to a convex set. The closest point theorem in a Hilbert space supplies the unique metric projection onto a closed convex set $P_{C}u$ and gives $d_C(u)=\Vert u-P_{C}u{\Vert }^2/2$.

Identify the Hilbert space with its dual space using the Riesz representation theorem. Completing the square yields $q^{\ast }(p)=\Vert p{\Vert }^2/2$, while ${\iota }_C^{\ast }(p)={\sup }_{v\in C}⟨p,v⟩={\sigma }_C(p)$ is the support function. The conjugate of the squared distance to a convex set is thereforeFor the closed unit ball, the metric projection onto a closed convex set is $P_{C}u=u$ if $\Vert u\Vert \le 1$, and $P_{C}u=u/\Vert u\Vert$ otherwise. ThereforeThe last equality uses the Cauchy-Schwarz inequality to compute the unit ball's support function, attained in the direction of $p$ when $p\ne 0$.