ScalingStacks

Verified tagged author-source HTML · 1904.03696v1 · cited publication edition alignment unverified.

00L5

Proof. The assertion is clear if e1​(x)=0e_{1}(x)=0. For e1​(x)≠0e_{1}(x)\neq 0, let en​(x)=e1⊗n​(x)e_{n}(x)=e_{1}^{\otimes n}(x), note that

|e1∨​(x)|1n​FS​(∥⋅∥n)∨=(|en∨​(x)|FS​(∥⋅∥n)∨)1n.\lvert e_{1}^{\vee}(x)\rvert_{\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}}=(\lvert e_{n}^{\vee}(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}})^{\frac{1}{n}}.

By Corollary 3.15, one has

|en∨​(x)|FS​(∥⋅∥n)∨=max⁡{|en∨​(x)​(sn,j)|κ^​(x)⋅∥sn,j∥n−1}\lvert e_{n}^{\vee}(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}}=\max\Big\{\lvert e_{n}^{\vee}(x)(s_{n,j})\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert_{n}^{-1}\Big\}

so

|e1∨​(x)|1n​FS​(∥⋅∥n)∨=max⁡{|en∨​(x)​(sn,j)|κ^​(x)⋅∥sn,j∥n−1}1n.\lvert e_{1}^{\vee}(x)\rvert_{\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}}=\max\Big\{\lvert e_{n}^{\vee}(x)(s_{n,j})\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert_{n}^{-1}\Big\}^{\frac{1}{n}}.

Tautologically, one has

sn,j​(z)=(e1⊗n)∨​(x)​(sn,j)=en∨​(x)​(sn,j),s_{n,j}(z)=(e_{1}^{\otimes n})^{\vee}(x)(s_{n,j})=e_{n}^{\vee}(x)(s_{n,j}),

so the criterion holds. By these defining equations, it is easy to see that the image of the open dual unit disc bundle is an open set. ∎

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