The forward subgradient step and backward subgradient step are the set-valued maps
Thus means . They are respectively the explicit and implicit time discretizations of subgradient flow . At a differentiable point the forward update is , while the backward update evaluates the gradient at the new point. For proper closed convex data the latter is the proximal operator
To prove at most one output, suppose . Then and . The two subgradient inequalities imply monotonicity of a convex subdifferential, giving
Since , . Convexity supplies this monotonicity; lower semicontinuity is not needed for this at-most-one argument.
Under the full printed assumptions the output actually exists at every . Properness ensures a finite point and excludes . Proper lower-semicontinuous convex has an affine minorant, so the quadratic proximal objective is coercive. Lower semicontinuity makes its minimum attained on a compact sublevel set. Convexity and the positive quadratic curvature make it strongly convex, hence its minimizer is unique. The subgradient optimality condition, with the differentiable quadratic term, is exactly . Thus
This also identifies precisely which assumptions provide existence, uniqueness and the update's optimality interpretation.
For a step size , define the set-valued maps
The forward subgradient step maps to the set . If is differentiable this is the explicit gradient step . The backward subgradient step consists of the satisfying , an implicit step for the subgradient flow. It is the resolvent of a monotone operator associated with .
Suppose . Then . The two defining subgradient inequalities are
Their sum proves monotonicity of a convex subdifferential, . But , so
Convexity supplies the subgradient inequalities and monotonicity; membership in the subdifferential ensures the two function values are finite, so subtraction is legitimate. Positivity of supplies the decisive sign. Properness rules out the identically infinite and negative-infinity pathologies in the overall setting, but lower semicontinuity is not needed for this at-most-one argument. Its role is in existence, proved next. The backward step cannot have two values, though uniqueness alone has not yet shown its domain is all of .