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Prove that if $x$ is a fixed point for $g$ then its projection to $S$ minimizes $f$ over $S$.

Discussion in 'Mathematics' started by ex.nihil, Oct 8, 2018.

  1. ex.nihil

    ex.nihil Guest

    Let $f:\mathbb{R}^n \longrightarrow \mathbb{R}$ be a convex and differentiable function and let $S$ be a nonempty, closed, and convex subset of $\mathbb{R}^n$.

    The function $g:\mathbb{R}^n \longrightarrow \mathbb{R}^n$ is defined by: $$ g(x) := -\big[ \nabla f\big(P_S(x)\big) - P_S(x)\big], \;\; \forall x \in \mathbb{R^n} \tag{1} $$ Where $P_S(x)$ is the projection of $x$ onto the set $S$.

    Prove that if $x$ is a fixed point for $g$ then its projection to $S$ minimizes $f$ over $S$.


    ATTEMPT

    Since $x$ is a fixed point for $g$ then we have $g(x)=x$.

    Therefore rearrange $(1)$ to obtain: $$ \nabla f\big(P_S(x)\big) = P_S(x) - x \tag{2} $$ Case 1: $(x \in S)$
    We have that $P_S(x) = x$ and thus $\nabla f(x)=x-x=0$.

    Since $f$ is convex then $x=\text{argmin}_{x \in S} f(x)$, as desired.

    Case 2: $(x \notin S)$
    It is possible to write $(2)$ as: $$ \nabla f\big(P_S(x)\big) = P_S(x) - x = d_S(x) = \inf_{s\in S} \Vert x-s\Vert \tag{3} $$ Where $d_S$ is the distance relative to $S$.


    I am frankly stuck, and unsure what to do. Moreover, it is not very clear to me how to use the closedness and convexity of $S$.

    Any help is greatly appreciated.

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