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On a Banach space one can define the set of minimizers of $E(x)+\Vert x-q{\Vert }^2/2$, but existence and uniqueness need additional geometry or coercivity hypotheses. The optimality condition is $0\in \partial E(x)+\mathcal{J}(x-q)$, with the duality mapping $\mathcal{J}=\partial (\Vert \cdot {\Vert }^2/2)$. The Hilbert completion-of-square shift by a dual vector does not generalize unchanged. On $({\mathbb{R}}^2,\Vert \cdot {\Vert }_1)$ with $E(x)=x_1+x_2$ and $q=0$, every $x_1,x_2\le 0$ with $x_1+x_2=-1$ is a minimizer; the vector $(-1,-1)$ is not one.

The duality mapping takes a vector to its supporting dual functionals for half its squared norm. Equivalently $p\in \mathcal{J}(x)$ when $\Vert p{\Vert }_{\ast }=\Vert x\Vert$ and $⟨x,p⟩=\Vert x{\Vert }^2$. This follows from the dual-norm inequality and equality in the conjugate relation for squared norms. In a Hilbert space, the Riesz representation theorem identifies it with the identity; in a general Banach space it can be set-valued.

On a real Hilbert space, for a proper lower-semicontinuous convex function $E$, the proximal operator isThe quadratic makes this objective strongly convex and coercive; the affine lower bound available for a proper closed convex function and weak lower semicontinuity give existence. The minimizer is unique. The subdifferential optimality condition isFor a positive parameter $t$, ${\mathrm{prox}}_{tE}$ uses $tE(x)$ in the same formula. On a general Banach space, the analogous norm-squared minimization is Banach-space proximal minimization; it can be set-valued and the identity-valued Hilbert derivative must be replaced by the duality mapping.

First use the usual Hilbert space convention for the proximal operator, identifying the dual element $z$ with its representing vector. For $c\ne 0$, completing the square givesSet $v=cx-y$ and multiply the objective by $c^2$. Its variable part becomes $\Vert v-[c(q-z)-y]{\Vert }^2/2+\alpha c^{2}J(v)$. Thus the proximal operator under affine rescaling isNegative $c$ is allowed. If $c=0$ and $J(-y)<\infty$, the penalty is constant and the answer is $q-z$. If $c=0$ and $J(-y)=\infty$, $E$ is identically infinite, so there is no proper proximal problem.

The PDF actually assumes an arbitrary Banach space, so the Hilbert input shift cannot be asserted literally under its hypotheses. Define ${\mathrm{Prox}}_F(s)=\arg {\min }_v\{F(v)+\Vert v-s{\Vert }^2/2\}$ as a possibly empty or set-valued minimizer set. Absolute homogeneity of the norm still gives the valid change of variablesThis retains the linear term, rather than shifting by a dual vector. Equivalently its optimality condition uses the duality mapping $\mathcal{J}=\partial (\Vert \cdot {\Vert }^2/2)$:For a concrete failure of the Hilbert shortcut, take $({\mathbb{R}}^2,\Vert \cdot {\Vert }_1)$, $J=0$, $q=y=0$, $c=1$ and $z=(1,1)$. Every $x_1,x_2\le 0$ with $x_1+x_2=-1$ minimizes $\Vert x{\Vert }_1^2/2+x_1+x_2$ with value $-1/2$. The naive shifted answer $(-1,-1)$ has value zero and is not a minimizer. Hence the simple shifted formula requires Hilbert geometry; general Banach proximal minimizers need not be unique. For $c=0$ with finite constant penalty the Banach condition is $-z\in \mathcal{J}(x-q)$, not generally $x=q-z$.