Research proof index

There is a measurable exceptional set $G' \subseteq G$ with $2\mu(G') \le \mu(G)$ such that, for every measurable $f$ bounded by $\mathbf{1}_F$, the integral over $G \setminus G'$ of the finitary oscillatory singular integral, summed only over the scales from $\sigma_1(x)$ to $\sigma_2(x)$, is at most $C(a,q)\mu(G)^{1-1/q}\mu(F)^{1/q}$.

$\lim_{t \to 0^+} F(it) / G(it) = 18 / \pi^2$. Proof outline (following blueprint Lemma 8.8): 1. Change of variables: lim_{t→0⁺} F(it)/G(it) = lim_{s→∞} F(i/s)/G(i/s) 2. Apply functional equations: - F(i/s) = s^12F(is) - 12s^11/πF₁(is)E₄(is) + 36s^10/π²E₄(is)² - G(i/s) = s^10H₄(is)³(2H₄(is)² + 5H₄(is)*H₂(is) + 5H₂(is)²) 3. Divide to get: F(i/s)/G(i/...

FmodGReal is differentiable on (0, ∞).

Let $\mathfrak F$ be a forest at level $n$, let $A\subseteq G$ be measurable, and let $f$ be measurable with $\lVert f(x)\rVert\le\mathbf 1_F(x)$. The integral over $A$ of the norm of the total forest Carleson sum is bounded by

$C(a,q,n)\,\mathrm{dens}_2(\mathfrak F)^{1/q-1/2}\,\lVert f\rVert_2\,\mu(A)^{1/2}.$

Let $u$ and $u'$ be distinct forest tops at level $(k,n,j)$. If a tile $p$ belongs to the tree rooted at $u'$ and its spatial cube lies below the spatial cube of $u$, then its phase center is quantitatively far from that of $u$ at the scale of $p$:

$2^{Z(n+1)}<d_p(\mathcal Q(p),\mathcal Q(u)).$

Two almost-everywhere strongly measurable functions on the circle that agree almost everywhere have the same Fourier coefficient at every fixed integer frequency $n$.