Small-action bridge to a relaxed equilibrium profile

Since $\rho$ is smooth and periodic, repeated integration by parts gives $(1+4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{)}^m\widehat{\rho }(k)=\widehat{(1-\mathrm{\Delta }{)}^m\rho }(k)$, so $\mathrm{\mid }\widehat{\rho }(k)\mathrm{\mid }\le C_m(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-m}$ for every integer $m\ge 1$. The number of lattice points with $\mathrm{\parallel }k{\mathrm{\parallel }}_{\mathrm{\infty }}=r$ is at most $(2r+1{)}^d-(2r-1{)}^d\le C_d{r}^{d-1}$. Thus for every multi-index $\alpha$, choosing $m>(d+\mathrm{\mid }\alpha \mathrm{\mid })\mathrm{/}2$ makes ${\sum }_k(2\pi \mathrm{\mid }k\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }\widehat{\rho }(k)\mathrm{\mid }$ finite, so the Fourier series and its differentiated series converge uniformly on compact time intervals including $s=0$. For $s\ge 1$, the same shell bound and the Fourier formula in [F2] give $\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }({\rho }_s-\bar{\rho }){\mathrm{\parallel }}_{\mathrm{\infty }}\le e^{-4{\pi }^{2}s}{\sum }_{k\mathrm{\ne }0}(2\pi \mathrm{\mid }k\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }\widehat{\rho }(k)\mathrm{\mid }{e}^{-4{\pi }^2(\mathrm{\mid }k{\mathrm{\mid }}^2-1)}\le C_{\alpha ,\rho }{e}^{-4{\pi }^{2}s}$. The zero Fourier coefficient is $\widehat{\rho }(0)=\bar{\rho }$, proving the asserted smoothness and $C^{\mathrm{\infty }}$ convergence.

The heat profile stays in $[\epsilon ,1-\epsilon ]$: the affine rescaling $v_0=(\rho -\epsilon )\mathrm{/}(1-2\epsilon )$ lies in $[0,1]$, and its Fourier heat profile $v_s$ preserves $[0,1]$ by [F2]. The multiplier is linear and fixes constants, so ${\rho }_s=\epsilon +(1-2\epsilon )v_s$; its mean is $\bar{\rho }$ by the zero-mode calculation in step 1.1.

Define $a(t)=0$ for $t\le 0$ and $a(t)=e^{-1\mathrm{/}t}$ for $t>0$, and set $\zeta (t)=a(t)\mathrm{/}(a(t)+a(1-t))$ on $[0,1]$. For each derivative order, the right derivatives of $a$ at zero are sums of terms $t^{-m}{e}^{-1\mathrm{/}t}$, which tend to zero; hence $a$ is smooth and flat at zero. The denominator is positive on $[0,1]$, $\zeta (0)=0$, $\zeta (1)=1$, and $\zeta$ is flat at both endpoints in the sense that all derivatives of $\zeta$ vanish at zero and all derivatives of $\zeta -1$ vanish at one. Define ${\sigma }_t^S=\bar{\rho }+\zeta (t)({\rho }_{S+1-t}-\bar{\rho })$. This is smooth by [F2], has mean $\bar{\rho }$, has values in $[\epsilon ,1-\epsilon ]$ by step 2.1, starts at $\bar{\rho }$, and ends at ${\rho }_S$.

Put $g_t={\mathrm{\partial }}_t{\sigma }_t^S$, which has zero mean. Define the smooth periodic potential ${\varphi }_t$ by its Fourier coefficients ${\widehat{\varphi }}_t(0)=0$ and ${\widehat{\varphi }}_t(k)={\widehat{g}}_t(k)\mathrm{/}(4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2)$ for $k\mathrm{\ne }0$, and set ${\kappa }_t^S=\mathrm{\nabla }{\varphi }_t$. Repeated integration by parts applied to the smooth $g_t$ gives rapid coefficient decay, so the series for ${\varphi }_t$ and all its space and time derivatives converge uniformly. Termwise differentiation gives $\mathrm{\Delta }{\varphi }_t=-g_t$, hence ${\mathrm{\partial }}_t{\sigma }^S+\mathrm{\nabla }\cdot {\kappa }^S=0$. At $t=1$, $g_1=-\mathrm{\Delta }{\rho }_S$, so ${\varphi }_1={\rho }_S-\bar{\rho }$ and ${\kappa }_1^S=\mathrm{\nabla }{\rho }_S$. Flatness of $\zeta -1$ at $1$ makes every derivative of ${\sigma }^S$ agree there with the reversed heat segment; the fixed linear Fourier inverse gives the same matching for ${\kappa }^S$.

For $k\mathrm{\ne }0$, the Fourier coefficient of $\mathrm{\nabla }{\varphi }_t$ has magnitude $\mathrm{\mid }{\widehat{g}}_t(k)\mathrm{\mid }\mathrm{/}(2\pi \mathrm{\mid }k\mathrm{\mid })$. Repeated integration by parts gives $\mathrm{\mid }{\widehat{g}}_t(k)\mathrm{\mid }\le C_m\mathrm{\parallel }{g}_t{\mathrm{\parallel }}_{C^{2m}}(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-m}$; choosing $m$ with $2m>d-1$ makes ${\sum }_{k\mathrm{\ne }0}(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-m}\mathrm{/}\mathrm{\mid }k\mathrm{\mid }$ finite by the lattice-shell bound in step 1.1, hence $\mathrm{\parallel }{\kappa }_t^S{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_m^{\mathrm{\prime }}\mathrm{\parallel }{g}_t{\mathrm{\parallel }}_{C^{2m}}$. The formula for $g_t$ and the $C^{\mathrm{\infty }}$ decay in step 1.1, together with ${\mathrm{\partial }}_s{\rho }_s=\mathrm{\Delta }{\rho }_s$ from [F2], imply ${\sup }_{0\le t\le 1}\mathrm{\parallel }{g}_t{\mathrm{\parallel }}_{C^{2m}}\to 0$ and ${\sup }_{0\le t\le 1}\mathrm{\parallel }\mathrm{\nabla }{\sigma }_t^S{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$ as $S\to \mathrm{\infty }$. By step 2.1, $\chi ({\sigma }_t^S)\ge \epsilon (1-\epsilon )$; since the space-time domain has measure one, [F1] bounds the connector action by $\frac{1}{4\epsilon (1-\epsilon )}{\sup }_{t,x}\mathrm{\mid }{\kappa }_t^S+\mathrm{\nabla }{\sigma }_t^S{\mathrm{\mid }}^2$, which tends to zero.