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

==== Take a star graph with: ====
* one center node ccc connected to mmm leaf nodes ℓ1,…,ℓm\ell_1,\dots,\ell_mℓ1,…,ℓm,
* seed s=ecs=e_cs=ec,
* teleportation α∈(0,1)\alpha\in(0,1)α∈(0,1),
* regularization parameter ρ>0\rho>0ρ>0.

Degrees:
* dc=md_c=mdc=m,
* dℓ=1d_{\ell}=1dℓ=1 for each leaf.

For the ℓ1\ell_1ℓ1-regularized PPR objective (your equation (3)), the weighted ℓ1\ell_1ℓ1 penalty is
* λc=ραm\lambda_c = \rho\alpha\sqrt{m}λc=ραm at the center,
* λℓ=ρα\lambda_{\ell} = \rho\alphaλℓ=ρα at each leaf. open-problem-fountoulakis22a

Because of symmetry, in a certain parameter regime the optimizer has the form:
* xc⋆>0x_c^\star > 0xc⋆>0,
* xℓ⋆=0x_{\ell}^\star = 0xℓ⋆=0 for every leaf.

===== In this regime (all leaves are zero), the center coordinate satisfies =====

(Qx⋆−αD−1/2s)c+λc=0,(Qx^\star - \alpha D^{-1/2}s)_c + \lambda_c = 0,(Qx⋆−αD−1/2s)c+λc=0,
which gives

xc⋆  =  2α(1−ρm)(1+α)m.x_c^\star \;=\; \frac{2\alpha(1-\rho m)}{(1+\alpha)\sqrt{m}}.xc⋆=(1+α)m2α(1−ρm).
This is positive as long as ρ<1/m\rho < 1/mρ<1/m.

===== A leaf ℓ\ellℓ stays at zero iff its KKT inequality holds: =====

∣(Qx⋆−αD−1/2s)ℓ∣≤λℓ=ρα.|(Qx^\star - \alpha D^{-1/2}s)_\ell| \le \lambda_\ell = \rho\alpha.∣(Qx⋆−αD−1/2s)ℓ∣≤λℓ=ρα.
For the star, this inequality reduces to a very clean threshold:

ρ  ≥  ρ0:=1−α2m.\rho \;\ge\; \rho_0 := \frac{1-\alpha}{2m}.ρ≥ρ0:=2m1−α.
So for

  ρ∈[1−α2m,  1m)  \boxed{\;\rho \in \Big[\frac{1-\alpha}{2m},\;\frac{1}{m}\Big)\;}ρ∈[2m1−α,m1)
the unique optimum has support S⋆={c}S^\star=\{c\}S⋆={c} (so ∣S⋆∣=1≤1/ρ|S^\star|=1\le 1/\rho∣S⋆∣=1≤1/ρ automatically).

===== All inactive coordinates are leaves and are symmetric, so the margin is attained on any leaf: =====

γ=λℓ−∣(Qx⋆−αD−1/2s)ℓ∣.\gamma = \lambda_\ell - |(Qx^\star - \alpha D^{-1/2}s)_\ell|.γ=λℓ−∣(Qx⋆−αD−1/2s)ℓ∣.
A direct substitution yields the closed form:

  γ  =  α1+α(2ρ−1−αm)  =  2α1+α(ρ−ρ0).  \boxed{\;
\gamma \;=\; \frac{\alpha}{1+\alpha}\Big(2\rho - \frac{1-\alpha}{m}\Big)
\;=\; \frac{2\alpha}{1+\alpha}\big(\rho-\rho_0\big).
\;}γ=1+αα(2ρ−m1−α)=1+α2α(ρ−ρ0).
Two key takeaways:
* At the breakpoint ρ=ρ0\rho=\rho_0ρ=ρ0: γ=0⇒γ<ρ (trivially).\gamma = 0 \quad\Rightarrow\quad \gamma < \sqrt{\rho}\ \text{(trivially)}.γ=0⇒γ<ρ (trivially).
* If you choose ρ=ρ0+δ\rho=\rho_0+\deltaρ=ρ0+δ with δ>0\delta>0δ>0 tiny, then γ=2α1+αδ,\gamma = \frac{2\alpha}{1+\alpha}\delta,γ=1+α2αδ, so you can make γ\gammaγ as small as you want, hence certainly <ρ<\sqrt{\rho}<ρ.