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small fixes
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Parcly-Taxel committed Sep 21, 2024
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2 changes: 1 addition & 1 deletion Carleson/ForestOperator.lean
Original file line number Diff line number Diff line change
Expand Up @@ -201,7 +201,7 @@ lemma second_tree_pointwise (hu : u ∈ t) (hL : L ∈ 𝓛 (t u)) (hx : x ∈ L
· unfold defaultD; positivity
_ < _ := by rw [mul_comm]; gcongr
have d1 : dist_{x, D ^ (s₂ - 1)} (𝒬 u) (Q x) < 1 := by
have := le_cdist_iterate (x := x) (r := D ^ (s₂ - 1)) (by sorry) (𝒬 u) (Q x) (100 * a)
have := le_cdist_iterate (x := x) (r := D ^ (s₂ - 1)) (by positivity) (𝒬 u) (Q x) (100 * a)
calc
_ ≤ dist_{x, D ^ s₂} (𝒬 u) (Q x) * 2 ^ (-100 * a : ℤ) := by
rw [neg_mul, zpow_neg, le_mul_inv_iff₀ (by positivity), mul_comm]
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8 changes: 4 additions & 4 deletions blueprint/src/chapter/main.tex
Original file line number Diff line number Diff line change
Expand Up @@ -524,7 +524,7 @@ \chapter{Proof of Metric Space Carleson, overview}
For each $\fu\in \fU$ and each $\fp\in \fT(\fu)$
we have $\scI(\fp) \ne \scI(\fu)$ and
\begin{equation}\label{forest1}
4\fp\lesssim 1\fu.
4\fp\lesssim \fu.
\end{equation}
For each $\fu \in \fU$ and each $\fp,\fp''\in \fT(\fu)$ and $\fp'\in \fP$
we have
Expand Down Expand Up @@ -2897,7 +2897,7 @@ \section{Proof of the Forest Union Lemma}
$$
B_{\fu}(\fcc(\fu), 1) \subset B_{\fu'}(\fcc(\fu'), 500) \subset B_{\fp}(\fcc(\fp), 4)\,.
$$
Together with $\scI(\fp) \subset \scI(\fu') = \scI(\fu)$, this gives $4\fp \lesssim 1\fu$, which is \eqref{forest1}.
Together with $\scI(\fp) \subset \scI(\fu') = \scI(\fu)$, this gives $4\fp \lesssim \fu$, which is \eqref{forest1}.
\end{proof}

\begin{lemma}[forest convex]
Expand Down Expand Up @@ -4387,7 +4387,7 @@ \section{The pointwise tree estimate}

\begin{proof}
Let $s_1 = \underline{\sigma}(\fu, x)$. By definition, there exists a tile $\fp \in \fT(\fu)$ with $\ps(\fp) = s_1$ and $x \in E(\fp)$. Then $x \in \scI(\fp) \cap L$. By \eqref{dyadicproperty} and the definition of $\mathcal{L}(\fT(\fu))$, it follows that $L \subset \scI(\fp)$, in particular $x' \in \scI(\fp)$, so $x \in I_{s_1}(x')$.
Next, let $s_2 = \overline{\sigma}(\fu, x)$ and let $\fp' \in \fT(\fu)$ with $\ps(\fp') = s_2$ and $x \in E(\fp')$. Since $\fp' \in \fT(\fu)$, we have $4\fp' \lesssim 1\fu$. Since $\tQ(x) \in \fc(\fp')$, it follows that
Next, let $s_2 = \overline{\sigma}(\fu, x)$ and let $\fp' \in \fT(\fu)$ with $\ps(\fp') = s_2$ and $x \in E(\fp')$. Since $\fp' \in \fT(\fu)$, we have $4\fp' \lesssim \fu$. Since $\tQ(x) \in \fc(\fp')$, it follows that
$$
d_{\fp}(\fcc(\fu), \tQ(x)) \le 5\,.
$$
Expand All @@ -4399,7 +4399,7 @@ \section{The pointwise tree estimate}
$$
d_{B(x, D^{s_2})}(\fcc(\fu), \tQ(x)) \le 5 \cdot 2^{5a}\,.
$$
Finally, by applying \eqref{seconddb} $2^{100a}$ times, we obtain
Finally, by applying \eqref{seconddb} $100a$ times, we obtain
$$
d_{B(x, D^{s_2-1})}(\fcc(\fu), \tQ(x)) \le 5 \cdot 2^{-95a} < 1\,.
$$
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