# The KLS conjecture, slicing, thin shells, and concentration (Mon, Jan 30)

## Localization

• Consider a log-concave density $F : \R^n \to \R_+$ with corresponding measure $\mu$, and suppose that $D \seteq \mathrm{diam}(\mathrm{supp}(F)) < \infty$ (where $\mathrm{supp}(F) \seteq \{ x \in \R^n : F(x) > 0 \}$).

Let $S_1,S_2,S_3$ be three measurable sets partitioning $\R^n$ and denote $d(S_1,S_2) = \min \{ \|x-y\|_2 : x \in S_1, y \in S_2 \}$.

KK Lemma (Karzanov and Khachiyan): It holds that $\mu(S_3) \geq \frac{d(S_1,S_2)}{D} \min \{ \mu(S_1), \mu(S_2) \}.$

Our goal is to reduce this to the $1$-dimensional case. Define $C \seteq \frac{d(S_1,S_2)}{D}$. The idea is to define two functions $g_1,g_2 : \R^n \to \R$ by

$g_1(x) \seteq \begin{cases} C F(x) & x \in S_1 \\ 0 & x \in S_2 \\ -F(x) & x \in S_3 \end{cases} \qquad g_2(x) \seteq \begin{cases} C F(x) & x \in S_2 \\ 0 & x \in S_1 \\ -F(x) & x \in S_3\,. \end{cases}$

Suppose, for the sake of contradiction, that $C \mu(S_1), C \mu(S_2) \geq \mu(S_3)$. This gives $\int g_1(x)\,dx = C \mu(S_1) - \mu(S_3) > 0$ and $\int g_2(x)\,dx = C \mu(S_2) - \mu(S_3) > 0$.

Now we are going to define a sequence $K_0 \subseteq K_1 \subseteq K_2 \subseteq \cdots$ of convex bodies so that $K \seteq \bigcap_{i=1}^{\infty} K_i$ is either a point or a segment and such that $\int_{K_i} g_1(x)\,dx > 0$ and $\int_{K_i} g_2(x)\,dx > 0$ for each $i=1,2,\ldots$.

• For $K_0$, simply choose a ball large enough to contain $\mathrm{supp}(F)$.

Consider any halfspace $H$ such that

$\int_{K_i \cap H} g_1(x)\,dx = \frac12 \int_{K_i} g_1(x)\,dx\,.$

Call such a halfspace $H$ bisecting.

Then either $H$ (or its complement $\R^n \setminus H$) will satisfy

$\int_{K_i \cap H} g_2(x)\,dx > 0\,.$

We will choose $H$ and take $K_{i+1} \seteq K_i \cap H$. Let’s see how to choose $H$.

• Claim: For any $(n-2)$-dimensional affine subspace $A$, there is at least one bisecting hyperplane that contains $A$.

Note that the boundary of a halfspace $H$ is an $(n-1)$-dimensional hyperplane. So consider a halfspace $H$ that contains $A$ in its boundary (think of a plane containing a line). We can rotate $H$ about $A$ so that $H$ becomes its complenetary halfspace $H'$.

If $\int_H g_1 - \int_{\R^n \setminus H} g_1 = 0$ initially, we are done. If not, then $\int_H g_1 - \int_{\R^n \setminus H} g_1$ changes sign as we rotate from $H$ to $H'$. By continuity, we find a bisecting halfspace whose bounding hyperplane contains $A$.

• Let $A_0,A_1,\ldots,$ be a sequence of all $(n-2)$-dimensional affine spaces of the form $v_0 + \mathrm{span}(v_1,\ldots,v_{n-2})$ where the vectors $\{v_j\}$ have rational coordinates (there is a countable number of such subspaces). Let $K_{i+1} \seteq K_i \cap H_i$ where $H_i$ is a bisecting halfspace whose boundary contains $A_i$.

Claim: The intersection $K \seteq \bigcap_{i\geq 1} K_i$ is a point or an interval.

Assume to the contrary that $K$ is at least $2$-dimensional. Then its projection onto some pair of coordinate axes is still $2$-dimensional. Without loss, we can assume it’s the first two axes. Then there must be a pair of rational numbers $r_1$ and $r_2$ such that $(r_1,r_2,x_3, \ldots,x_n)$ is in the interior of $K$. But now the $(n-2)$-dimensional subspace $\mathrm{span}(r_1 e_1, r_2 e_2)$ occurs as some $A_i$.

Recall that the halfspace $H_i$ has its boundary containing $A_i$. Since $(r_1,r_2,x_3,\cdots,x_n)$ is in the interior of $K$, it must also be in the interior of $K_{i+1}$. But that means that $H_i$ properly cuts $K_{i+1}$. But this is a contradiction, as $H_i$ should fully contain $K_{i+1} = K_i \cap H_i$.

• We conclude that $\int_K g_1 \geq 0, \int_K g_2 \geq 0$ on the intersection $K$.

Note that we can easily get $\int_K g_1 > 0, \int_K g_2 > 0$ for the original functions by starting with functions $\hat{g}_1 = g_1 - \delta \mathbf{1}_{K_0}$ and $\hat{g}_2 = g_2 - \delta \mathbf{1}_{K_0}$ for $\delta > 0$ sufficiently small and then finding $K$ where $\int_K \hat{g}_1 \geq 0, \int_K \hat{g}_2 \geq 0$.

Since the restriction of a log-concave function to an interval is still log-concave, this is indeed a reduction of the KK Lemma to the $1$-dimensional case.

## Finding a better needle

For other applications, it helps to replace $g_1$ and $g_2$ by nicer functions at every step. One can consult Lovasz-Simonovits Lemma 2.5 for a (rather technical) argument proving the following.

• Localization Lemma: Given $\int_{\R^n} g_1 > 0$ and $\int_{\R^n} g_2 > 0$, there are two points $a,b \in \R^n$ and a linear function $\ell : [0,1] \to \R_+$ such that

\begin{align*} \int_0^1 \ell(t)^{n-1} g_1((1-t) a + tb)\,dt &> 0 \\ \int_0^1 \ell(t)^{n-1} g_2((1-t) a + tb)\,dt &> 0 \end{align*}
• One can instead derive the conclusion that there is a number $\gamma \in \R$ such that

\begin{align*} \int_0^1 e^{\gamma t} g_1((1-t) a + tb)\,dt &> 0 \\ \int_0^1 e^{\gamma t} g_2((1-t) a + tb)\,dt &> 0\,, \end{align*}

which I find a bit more natural.