AI 中文总结
该研究构造了非圆形曲线Γ₀，其所有单位切线迭代均为处处曲率为正的光滑嵌入闭曲线，解决了Tabachnikov的问题，且迭代曲线位于宽度与n无关的带形区域内。
AI 中文摘要
对于定向正则平面曲线γ，定义𝒯γ=γ+τ，其中τ为其单位切线。我们构造了一条非圆形曲线Γ₀，使得所有迭代𝒯ⁿΓ₀（n≥0）均为处处曲率为正的光滑嵌入闭曲线，这回答了Tabachnikov提出的问题。构造始于一条非紧凸曲线C，其两端渐近于平行线，满足𝒯C=C+(V,0)（V>0）。从C出发构造长凸闭曲线，匹配周长后，将每条曲线与其单位切线像为序列中下一条曲线的凸曲线比较，在合适的弧长坐标下，二者曲率函数的L¹差为指数小量。通过一致逆估计将这些近似关系转化为精确序列Γₙ₊₁=𝒯Γₙ，每个Γₙ位于宽度与n无关的带形区域内，而迭代圆的半径趋于无穷大。
英文摘要
For an oriented regular plane curve $γ$, define ${\cal T}γ=γ+τ$, where $τ$ is its unit tangent. We construct a noncircular curve $Γ_0$ such that every iterate ${\cal T}^nΓ_0$, $n\geq0$, is a smooth embedded closed curve with everywhere positive curvature. This answers a question of Tabachnikov. The construction starts with a noncompact convex curve $C$, asymptotic at its two ends to parallel lines, satisfying ${\cal T} C=C+(V,0)$ for some $V>0$. From $C$ we construct long closed convex curves. After matching perimeters, we compare each curve with a convex curve whose unit-tangent image is the next curve in the sequence. In suitable arclength coordinates, their curvature functions differ by an exponentially small amount in $L^1$. Uniform inverse estimates turn these approximate relations into an exact sequence $Γ_{n+1}={\cal T}Γ_n$. Each $Γ_n$ lies in a strip of width bounded independently of $n$, whereas the radii of iterated circles tend to infinity.
Comments14 pages
