Mathematics Asked on December 18, 2021
Let $h:mathbb R^m to mathbb R$ be a convex diifferentiable function and define $C_h := {x in mathbb R^m mid h(x) = 0}$, assumed to be non-empty.
Question. Given $x in mathbb R^m$ such that $nabla h(x) ne 0$, is there any geometric interpretation of the quantity $omega_h(x) := (h(x))_+/|nabla h(x)|$ vis à vis the set $C_h$ ?
Motivation. In the linear case where $h(x) = v^Tx$ for some unit-vector $v in mathbb R^m$, one has
Given a unit-vector $v in mathbb S_{m-1} := {v in mathbb R^m mid |v| = 1}$, let $mathscr L_v := {x + tv mid t ge 0} subseteq mathbb R^m$ be the ray generated by $v$ and starting at $x$. The distance of $x$ from $C_h$ can be written as $$ r_h(x) := d(x,C_h) := inf_{x' in C_h}|x'-x| = inf_{v in mathbb S_{m-1}}inf_{x' in mathscr L_v cap C_h}|x'-x|. $$
Now, by definition, $x' in mathscr L_v cap C_h$ if and only if there exists a time $t in [0,infty)$ such that $x' = x + tv$ and $h(x') = 0$. Thus, begin{eqnarray} r_h(x) = inf{t mid (v,t) in mathbb S_{m-1} times [0,infty),; h(x + tv) = 0}. tag{1} end{eqnarray}
If $h$ is convex and differentiable, then for all $(v,t) in mathbb S_{m-1} times [0,infty)$ one has begin{eqnarray*} begin{split} h(x+tv) &ge h(x) + tnabla h(x)^Tvge h(x) - t|nabla h(x)|,\ -h(x+tv) &le -h(x) - tnabla h(x)^Tv le -h(x) + t|nabla h(x)|, end{split} end{eqnarray*} where we have used Cauchy-Schwarz to reach the final term of each chain of inequalities. Combining the above inequalities and (1) one computes begin{eqnarray*} begin{split} r_h(x) &ge inf{t ge 0 mid h(x) - t|nabla h(x)|le 0} = frac{max(h(x),0)}{|nabla h(x)|},\ r_h(x) &ge inf{t ge 0 mid -h(x) + t|nabla h(x)|ge 0} = frac{max(-h(x), 0)}{|nabla h(x)|}, end{split} end{eqnarray*} We have thus proved
Theorem. If $h:mathbb R^m to mathbb R$ is convex differentiable, we have begin{eqnarray} r_h(x) ge frac{|h(x)|}{|nabla h(x)|},;forall x in mathbb R^m end{eqnarray}
If we instead assume $h$ Lipschitz continuous (but not necessarily convex), then $|h(x + tv) - h(x)| le Lt$ for all $(v,t) in S_{m-1} times [0,infty)$. Using (1), we may then compute begin{eqnarray} begin{split} r_h(x) &ge inf{t ge 0mid h(x) - tL ge 0} = frac{max(h(x),0)}{L},\ r_h(x) &ge inf{t ge 0mid h(x) + tL ge 0} = frac{max(-h(x),0)}{L}. end{split} end{eqnarray} Putting things together, we get
Theorem. If $h:mathbb R^m to mathbb R$ is $L$-Lipschitz, then we have begin{eqnarray} r_h(x) ge frac{|h(x)|}{L},;forall x in mathbb R^m. end{eqnarray}
Answered by dohmatob on December 18, 2021
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