ScalingStacks

Proof. [04U0]

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Proof.

By translating and subtracting a linear function assume that x=p=0x=p=0. Assume by way of contradiction that we can find hk→0h_{k}\rightarrow 0 and some δ>0\delta>0 such that

dn−1​(hk)>δ​hk1/2d_{n-1}(h_{k})>\delta h_{k}^{1/2}

for all kk. We first show that vv is trapped by two tangent planes at 00.

Let x1,kx_{1,k} and x2,kx_{2,k} be the points on ∂Shk,0v​(0)\partial S_{h_{k},0}^{v}(0) where the hyperplanes perpendicular to the shortest axis of the John ellipsoid become tangent to ∂Shk,0v​(0)\partial S_{h_{k},0}^{v}(0), and let p1,kp_{1,k} and p2,kp_{2,k} denote subgradients at these points. Since

d1​(hk)​d2​(hk)​…​dn​(hk)<C​hkn+12,d_{1}(h_{k})d_{2}(h_{k})...d_{n}(h_{k})<Ch_{k}^{\frac{n+1}{2}},

we have that dn​(hk)<Cδn−1​hkd_{n}(h_{k})<\frac{C}{\delta^{n-1}}h_{k} for all kk. By this observation and convexity we can rotate and pass to a subsequence such that

p1,k→c1​(δ)​en,p2​(k)→−c2​(δ)​en.p_{1,k}\rightarrow c_{1}(\delta)e_{n},\quad p_{2}(k)\rightarrow-c_{2}(\delta)e_{n}.

Then vv is trapped by the planes ±c⁡(δ)​xn\pm c(\delta)x_{n}. We conclude that

Shk,0v(0)⊂{|xn|<C(δ)hk}.S_{h_{k},0}^{v}(0)\subset\{|x_{n}|<C(\delta)h_{k}\}.

To complete the proof, we show that the volumes of sections obtained with tilted supporting planes are too large. Take the largest aa such that v≥a​xnv\geq ax_{n} and consider the sections

Sk=S(1+a​C​(δ))​hk,av​(0).S_{k}=S_{(1+aC(\delta))h_{k},a}^{v}(0).

Then SkS_{k} engulf Shk,0v​(0)S_{h_{k},0}^{v}(0). Furthermore,

sup{|xn|:x∈Sk}=Rk​hk,\sup\{|x_{n}|:x\in S_{k}\}=R_{k}h_{k},

where Rk→∞R_{k}\rightarrow\infty as k→∞k\rightarrow\infty. Indeed, if not, then for some small ϵ\epsilon and a sequence bi→0b_{i}\rightarrow 0 we would have v⁡(x′,bi)>(a+ϵ)​biv(x^{\prime},b_{i})>(a+\epsilon)b_{i} for all x′x^{\prime}. Convexity and v⁡(0)=0v(0)=0 imply that v>(a+ϵ)​xnv>(a+\epsilon)x_{n} for all xn>bix_{n}>b_{i}, which in turn implies that

v>(a+ϵ)​xn,v>(a+\epsilon)x_{n},

contradicting the definition of aa.

Finally, let (xk′,Rk​hk)∈Sk(x^{\prime}_{k},R_{k}h_{k})\in S_{k} be the point in SkS_{k} furthest in the ene_{n} direction. Since vv grows at least quadratically, we have

|xk′|<C⁡(δ,a)​hk1/2.|x^{\prime}_{k}|<C(\delta,a)h_{k}^{1/2}.

Recall that Shk,0v(0)⊂{|xn|<C(δ)hk}S_{h_{k},0}^{v}(0)\subset\{|x_{n}|<C(\delta)h_{k}\}. Since di​(hk)>δ​hk1/2d_{i}(h_{k})>\delta h_{k}^{1/2} for all i≤n−1i\leq n-1, SkS_{k} contains the cone with vertex (xk′,Rk​hk)(x^{\prime}_{k},R_{k}h_{k}) and base given by a ball of radius hk1/2​(δ−C⁡(a,δ)/Rk)h_{k}^{1/2}(\delta-C(a,\delta)/R_{k}) on the hyperplane {xn=C(δ)hk}\{x_{n}=C(\delta)h_{k}\}. We conclude that

|Sk|≥c⁡(δ,a)​Rk​hkn+12,|S_{k}|\geq c(\delta,a)R_{k}h_{k}^{\frac{n+1}{2}},

contradicting our definition of Σv\Sigma_{v} for kk large. ∎

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