Editing Openai/693e3ce6-229c-8008-97dc-ab720cb1f95a (section)

=== Assistant: Yes. ===
Yes. You can get γ=o(ρ)\gamma = o(\sqrt{\rho})γ=o(ρ) by taking a family of star graphs where ρ→0\rho\to 0ρ→0 (via the graph size) and choosing ρ\rhoρ extremely close to the support-change breakpoint so that the KKT inequality is almost tight on an inactive node.

I’ll spell it out cleanly.

==== Interpreting your statement in standard asymptotic terms: ====

: 

This is stronger than just “γ<ρ\gamma<\sqrt{\rho}γ<ρ” at a fixed instance.

==== Take the star Sm+1S_{m+1}Sm+1 (center ccc, leaves ℓ1,…,ℓm\ell_1,\dots,\ell_mℓ1,…,ℓm), seed s=ecs=e_cs=ec, and fix any constant α∈(0,1)\alpha\in(0,1)α∈(0,1). ====

For this instance (we already derived this earlier), there is a breakpoint

ρ0(m)=1−α2m\rho_0(m)=\frac{1-\alpha}{2m}ρ0(m)=2m1−α
such that for any ρ∈[ρ0(m), 1/m)\rho\in[\rho_0(m),\,1/m)ρ∈[ρ0(m),1/m) the optimizer has support S⋆={c}S^\star=\{c\}S⋆={c}, and the margin satisfies

γ  =  2α1+α (ρ−ρ0(m)).\gamma \;=\; \frac{2\alpha}{1+\alpha}\,(\rho-\rho_0(m)).γ=1+α2α(ρ−ρ0(m)).
So γ\gammaγ is exactly linear in how far ρ\rhoρ is above the breakpoint.

==== Let’s define a sequence ρm\rho_mρm by choosing a tiny offset above the breakpoint, for example: ====

ρm:=ρ0(m)+ρ0(m)2.\rho_m := \rho_0(m) + \rho_0(m)^2.ρm:=ρ0(m)+ρ0(m)2.
Then:
# ρm→0\rho_m\to 0ρm→0 as m→∞m\to\inftym→∞ since ρ0(m)=Θ(1/m)\rho_0(m)=\Theta(1/m)ρ0(m)=Θ(1/m).
# The margin is

γm=2α1+α (ρm−ρ0(m))=2α1+α ρ0(m)2=Θ ⁣(1m2).\gamma_m
= \frac{2\alpha}{1+\alpha}\,(\rho_m-\rho_0(m))
= \frac{2\alpha}{1+\alpha}\,\rho_0(m)^2
= \Theta\!\left(\frac{1}{m^2}\right).γm=1+α2α(ρm−ρ0(m))=1+α2αρ0(m)2=Θ(m21).
# Meanwhile,

ρm=Θ ⁣(1m).\sqrt{\rho_m} = \Theta\!\left(\frac{1}{\sqrt{m}}\right).ρm=Θ(m1).
# Therefore

γmρm=Θ ⁣(1/m21/m)=Θ ⁣(1m3/2)⟶0.\frac{\gamma_m}{\sqrt{\rho_m}}
= \Theta\!\left(\frac{1/m^2}{1/\sqrt{m}}\right)
= \Theta\!\left(\frac{1}{m^{3/2}}\right)
\longrightarrow 0.ρmγm=Θ(1/m1/m2)=Θ(m3/21)⟶0.
So indeed

  γm=o(ρm)  \boxed{\;\gamma_m = o(\sqrt{\rho_m})\;}γm=o(ρm)
for this explicit star-graph family.

==== Since ρm≈ρ0(m)\rho_m \approx \rho_0(m)ρm≈ρ0(m) for large mmm, and ρ0(m)=Θ(1/m)\rho_0(m)=\Theta(1/m)ρ0(m)=Θ(1/m), the choice ρm=ρ0+ρ02\rho_m=\rho_0+\rho_0^2ρm=ρ0+ρ02 gives ====

γm=Θ(ρ0(m)2)=Θ(ρm2).\gamma_m = \Theta(\rho_0(m)^2) = \Theta(\rho_m^2).γm=Θ(ρ0(m)2)=Θ(ρm2).
And ρ2=o(ρ)\rho^2 = o(\sqrt{\rho})ρ2=o(ρ) as ρ→0\rho\to 0ρ→0 because

ρ2ρ=ρ3/2→0.\frac{\rho^2}{\sqrt{\rho}} = \rho^{3/2}\to 0.ρρ2=ρ3/2→0.
So this construction actually gives the very clean scaling:

  γ=Θ(ρ2)⇒γ=o(ρ).  \boxed{\;\gamma = \Theta(\rho^2)\quad\Rightarrow\quad \gamma=o(\sqrt{\rho}).\;}γ=Θ(ρ2)⇒γ=o(ρ).

===== This shows you cannot hope for any worst-case lower bound like γ≳ρ\gamma \gtrsim \sqrt{\rho}γ≳ρ (or even γ≳ρ\gamma \gtrsim \rhoγ≳ρ) that depends only on (α,ρ)(\alpha,\rho)(α,ρ): by tuning ρ\rhoρ close to a breakpoint on a suitable graph family, γ\gammaγ can be polynomially smaller than ρ\sqrt{\rho}ρ, even when the true support is tiny (here ∣S⋆∣=1≤1/ρ|S^\star|=1\le 1/\rho∣S⋆∣=1≤1/ρ). =====

If you want, I can also give a variant where γ\gammaγ is super-polynomially small relative to ρ\sqrt{\rho}ρ (e.g., choose ρ−ρ0\rho-\rho_0ρ−ρ0 exponentially small in mmm), but the γ=Θ(ρ2)\gamma=\Theta(\rho^2)γ=Θ(ρ2) example already proves γ=o(ρ)\gamma=o(\sqrt{\rho})γ=o(ρ) cleanly.