Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A\subset[0,1]^n\subset\mathbb{R}^n$ be a closed semi-algebraic set.

Let $f_k: [0,1]\rightarrow\mathbb{R}$, $f_k(x_i)=\mu^{n-1}(A_{|_{x_i}}+B^{n-1}_{1/k})$ where $A_{|_{x_i}}=A\cap H^i_{x_i}$ with $H^i_{x_i}$ the hyperplane orthogonal to the $i$-axis through $x_i$, $\mu^{n-1}$ the $n-1$-dimensional Lebesgue measure and $B^{n-1}_\varepsilon$ the $n-1$-dimensional $\varepsilon$-ball around zero. I.e. for any $x_i \in [0,1]$ it measures the cross section of $A$ that is widened by an $\varepsilon$-ball.

Show that $f_k$ converges uniformly to $f(x_i)=\mu^{n-1}(A_{|_{x_i}})$ for $x_i\in [0,1]$.

$A_{|_{x_i}}$ is closed, so $A_{|_{x_i}}$ is the limit of $A_{|_{x_i}}+B^{n-1}_{1/k}$. From the properties of the Lebesgue measure we get that $f_k$ converges pointwise. But I have no idea how to prove uniform convergence.

EDIT:

Some more thoughts of mine: The series above converges uniformly, if

$f_k'(x_i)=\mu^{n-1}(A_{|_{x_i}}+B_{1/k}) -\mu^{n-1}(A_{|_{x_i}})$

converges uniformly to zero. Since $f_k''(x_i)=\mu^{n-1}(\delta(A_{|_{x_i}})+B_{1/k})\ge f_k'(x_i)$ for all $x_i$ it would suffice that $f_k''$ converges uniformly to zero.

Here, discontinuities of $f_k''$ are a problem. It seems that $f_k''$ is discontiuous in $x\in[0,1]$ iff the set-valued map $M^A:[0,1]\subset\mathbb{R}\rightrightarrows\mathbb{R}^{n-1}; M^A(x_i)=A_{|_{x_i}}$ is Hausdorff discontiuous in $x_i$.

Uniform convergence should follow if $M^A$ is Hausdorff discontiuous in finitely many points only. $A$ is semi-algebraic. For a set defined by a single polynomial (in-) equality the number of discontinuities is finite. Does it stay finite under unions / intersections (complement seems obvious)?

Any help is greatly appreciated! I have the feeling that I'm missing the easy path and taking a detour through the djungle...

Thank you

J.P.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.