claimroutineinferred_dep

Choosing gamma and K yields the barrier inequalities

Chooses

\gamma

small so

\Delta_z\mathcal F\le V(s)\mathcal F

with

V=Cs^{-2}e^{-cs/h}

finishes the barrier proof by combining both bounds

Claim·§5 Estimating the energy on a fixed fiber·inside propEnergy barrier on a fixed fiber

Statement (paper)

With

\gamma

chosen so small that

C_3\gamma^2<c_0/2

and

K

fixed to meet the ball inclusions and so that

h<1/(4K)

lies in the range of , one has for

2a\le s=|\tau|\le3

that

\Delta_z\mathcal F/\mathcal F\le C_\gamma s^{-2}\sigma^{2m}e^{-(c_0-C_3\gamma^2)\sigma}\le V(s)=Cs^{-2}e^{-cs/h}

, which is . Moreover, since the scaled radii

4a/h=4K

and

2a/h=2K

are fixed, the upper bound at

|\tau|=4a

gives for

f

, and the lower bound at

|\tau|=2a

gives for

g_T

In plain English

With

\gamma

small enough that

C_3\gamma^2<c_0/2

and

K

fixed suitably, dividing the wall-decay bound on

\Delta_z\mathcal F

by the interior lower bound on

\mathcal F

gives

\Delta_z\mathcal F\le V(s)\mathcal F

with

V(s)=Cs^{-2}e^{-cs/h}

for

2a\le s\le 3

. The fixed scaled radii

4K

and

2K

also turn the upper and lower energy bounds into the input and output inequalities.

Why it's here

This claim finishes the proof of the proposition by fixing the constants and combining the two preceding bounds. The upper bound on

\Delta_z\mathcal F

decays like

e^{-c_0\sigma}

while the lower bound on

\mathcal F

loses only

e^{-C_3\gamma^2\sigma}

and a power of

\sigma

. Taking

\gamma

small leaves net exponential decay

e^{-(c_0-C_3\gamma^2)\sigma}

, and the power of

\sigma

is absorbed into it, which gives the potential

V(s)

. At the fixed radii

|\tau|=4a

and

2a

the same bounds give the input and output inequalities directly.

Classified routine:

Direct consequence of the preceding claims (energy/wall-flux upper bound and interior lower bound): dividing the two bounds and fixing the constants

\gamma

and

K

, with the power of

\sigma

absorbed into the exponential decay, is pure bookkeeping.

Uses

claimEnergy and wall flux boundsclaimInterior energy lower bound via ball quadratic

Used by

Nothing in the paper

Figures

Panel (a) draws the fiber in modulus coordinates: the horizontal axis is $|W|$ with tick $w_0$, and the vertical axis is $|Q|=|\Phi|/a$ with ticks $1/20$ and $1/10$. A quarter-ellipse arc labeled $r=0$ runs from $(0,1/10)$ to $(w_0,0)$ and bounds the shaded region $r<0$. A dotted line at height $1/20$ meets the arc, and the part of the arc below it is drawn heavy and labeled "flux wall". A small dashed box near the origin, containing a smaller solid box, is marked (b). Panel (b) enlarges that corner: a dashed rectangle $|W|\le2\gamma$, $|Q|\le1/100$, labeled "Gaussian inversion, $g(Q)=1$", contains a darker solid rectangle $|W|\le\gamma$, $|Q|\le1/400$, labeled "centers". The figure shows that the region where the wall flux lives is separated from the central region where the Gaussian multiplier is inverted. — Figure 1schematicp. 19
The fixed fiber in modulus coordinates. The wave flux through $r=0$ is supported on the heavy arc, where $|Q|\le1/20$ and hence $|W|\ge\sqrt{3}\,w_0/2$ . This separation gives Gaussian decay $e^{-cs/h}$ even for $|v|\le w_0/3$ . In the enlargement (b), once $K\gamma$ is large, the radius- $h$ balls about the marked centers project into the dashed patch. There $|W-v|\le4\gamma$ on the enlarged cap $|v|\le2\gamma$ , so Gaussian inversion costs at most $e^{C\gamma^2s/h}$ . The aperture $\gamma$ is fixed independently of $h$ ; its size relative to $w_0$ is schematic.
What it shows
Panel (a) draws the fiber in modulus coordinates: the horizontal axis is $|W|$ with tick $w_0$ , and the vertical axis is $|Q|=|\Phi|/a$ with ticks $1/20$ and $1/10$ . A quarter-ellipse arc labeled $r=0$ runs from $(0,1/10)$ to $(w_0,0)$ and bounds the shaded region $r<0$ . A dotted line at height $1/20$ meets the arc, and the part of the arc below it is drawn heavy and labeled "flux wall". A small dashed box near the origin, containing a smaller solid box, is marked (b). Panel (b) enlarges that corner: a dashed rectangle $|W|\le2\gamma$ , $|Q|\le1/100$ , labeled "Gaussian inversion, $g(Q)=1$ ", contains a darker solid rectangle $|W|\le\gamma$ , $|Q|\le1/400$ , labeled "centers". The figure shows that the region where the wall flux lives is separated from the central region where the Gaussian multiplier is inverted.
Read from the image by claude-opus-5.5 · confidence 85%
SupportspropEnergy barrier on a fixed fiberclaimCurrent and density bounds with wall decayclaimInterior density lower bound by Gaussian inversionclaimInterior energy lower bound via ball quadraticclaimChoosing gamma and K yields the barrier inequalities

Sources

PaperOpen

Section: §5 Estimating the energy on a fixed fiber
Page: —
Label: —
Line: sections/barrier.tex:233

Leannot aligned

Phase 2 aligns this node to a Lean declaration.

Friction

info
inferred_dep
outline: "Choose gamma, K; divide to get potential" rests on claim-prop-fixed-barrier-5
info
inferred_dep
outline: "Choose gamma, K; divide to get potential" rests on claim-prop-fixed-barrier-7