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.
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.