Bidenbaka

There are beautiful math, there are ugly math, and there are ugly math, whose elengance can only be appreciated after suffering.

Zahl presenting one of the most important results in analysis over the past 20 years — The Kakeya Conjecture

Denote the Lower Box dimension as $\underline\dim_B$ and the packing dimension as $\dim_P$, is there a criterion to tell, for a given set $F$, when $\underline\dim_B(F)<\dim_P(F)$? Or, at least, are there examples for both sides of the inequality?

We know that the Hausdorff dimension is smaller equal to the Lower Box (LB) dimension and hence the Upper Box (UB) dimension. We also know that the Packing dimension is greater equal to the Hausdorff dimension but smaller equal to the UB dimension. Is there a criterion to tell when LB <= Packing dim

Ehsan Abedi: Processes on Wasserstein spaces and energy-minimizing particle representations in fractional Sobolev spaces https://arxiv.org/abs/2503.10859 https://arxiv.org/pdf/2503.10859 https://arxiv.org/html/2503.10859

Volume estimates for unions of convex sets, and the Kakeya set conjecture in three dimensions We study sets of $δ$ tubes in $\mathbb{R}^3$, with the property that not too many tubes can be contained inside a common convex set $V$. We show that the union of tubes from such a set must have almos...

I am happy to announce that the Kakeya set conjecture, one of the most sought after open problems in geometric measure theory, has now been proven (in three dimensions) by Hong Wang and Joshua Zahl! arxiv.org/abs/2502.17655 I discuss some ideas of the proof at terrytao.wordpress.com/2025/02/25/t...

Today I am reading Peres and Pörter's Brownian motion. In the part of deriving the Frostman's lemma, the original context states that $A\subset\mathbb{R}^d$ is a closed set while in other contexts $A$ is assumed to be compact. I reached to Deepseek and this prized goofy states this...

Update: note that $h\cdot M$ is a martingale and then the first moment vanishes, result follows.

But why is the middler term vanishing?

The Itô's isometry yields

To show this equality holds, we first square both sides and then take expectations on both sides, then we have

We wish to show that $\mathcal{H}$ is a complete subspace of $(\Omega,\mathcal{F}_\infty,\mathbb{P})$ and then by the Hilbert space structure it is closed. To show that we need the equaltiy

I am recently reading LeGall's Brownian motion. There is a step in proving Martingale Representation Theorem (Theorem 5.18) I don't understand (see Figure 1). The idea in proving assertion (i) is to use $\mathcal{H}$ to denote all such $Z$ in $L^2$, then prove it is a closed subspace (see Figure 2).