Hölder bound for filtered subfilter energy transfer

Put ${\delta }_{s}u(x,t)=u(x-\mathrm{\ell }s,t)-u(x,t)$ and $m_{\mathrm{\ell }}(x,t)=\int K(s){\delta }_{s}u(x,t)ds$. By unit mass and the Euclidean representation in [F2], $m_{\mathrm{\ell }}={\bar{u}}_{\mathrm{\ell }}-u$. Expanding componentwise and using linearity [F5] gives $\int K(s){\delta }_{s,i}u{\delta }_{s,j}uds={\overline{u_i{u}_j}}_{\mathrm{\ell }}-{\bar{u}}_{\mathrm{\ell },i}{u}_j-u_i{\bar{u}}_{\mathrm{\ell },j}+u_i{u}_j$. Subtracting $m_{\mathrm{\ell },i}{m}_{\mathrm{\ell },j}=({\bar{u}}_{\mathrm{\ell },i}-u_i)({\bar{u}}_{\mathrm{\ell },j}-u_j)$ yields ${\tau }_{\mathrm{\ell },ij}$. Thus the covariance identity holds at each fixed time under the stated spatial regularity, without a time-regularity hypothesis on its spatial derivatives.

By [F1], each stress component is ${\tau }_{\mathrm{\ell },ij}=\int K(s){\delta }_{s,i}u{\delta }_{s,j}uds-m_{\mathrm{\ell },i}{m}_{\mathrm{\ell },j}$. From [F5], nonnegativity of $K$, and [A1], $\mathrm{\mid }\int K(s){\delta }_{s,i}u{\delta }_{s,j}uds\mathrm{\mid }\le A(t{)}^2{\mathrm{\ell }}^{2\alpha }{M}_{2\alpha }$ and $\mathrm{\mid }{m}_{\mathrm{\ell },i}\mathrm{\mid }\le A(t){\mathrm{\ell }}^{\alpha }{M}_{\alpha }$. Therefore $\mathrm{\mid }{\tau }_{\mathrm{\ell },ij}\mathrm{\mid }\le A(t{)}^2{\mathrm{\ell }}^{2\alpha }(M_{2\alpha }+M_{\alpha }^2)$.

By [F2], spatial differentiation at fixed $t$ passes through the convolution. Since ${\mathrm{\partial }}_{s_j}{u}_i(x-\mathrm{\ell }s,t)=-\mathrm{\ell }{\mathrm{\partial }}_j{u}_i(x-\mathrm{\ell }s,t)$, integration by parts in $s_j$ over a box containing $\mathrm{supp}K$ gives ${\mathrm{\partial }}_j{\bar{u}}_{\mathrm{\ell },i}={\mathrm{\ell }}^{-1}\int {\mathrm{\partial }}_{j}K(s)u_i(x-\mathrm{\ell }s,t)ds$. Compact support gives $\int {\mathrm{\partial }}_{j}K(s)ds=0$, so this equals ${\mathrm{\ell }}^{-1}\int {\mathrm{\partial }}_{j}K(s){\delta }_{s,i}u(x,t)ds$. The product-rectangle integrals are justified by [F4], and integration by parts by [F3]. Applying [F5] and [A1] gives $\mathrm{\mid }{\mathrm{\partial }}_j{\bar{u}}_{\mathrm{\ell },i}(x,t)\mathrm{\mid }\le A(t){\mathrm{\ell }}^{\alpha -1}\int \mathrm{\mid }{\mathrm{\partial }}_{j}K(s)\mathrm{\mid }\mathrm{\mid }s{\mathrm{\mid }}^{\alpha }ds\le A(t){\mathrm{\ell }}^{\alpha -1}{N}_{\alpha }$.

Summing the $d^2$ products in ${\mathrm{\Pi }}_{\mathrm{\ell }}=-{\sum }_{i,j}{\tau }_{\mathrm{\ell },ij}{\mathrm{\partial }}_j{\bar{u}}_{\mathrm{\ell },i}$ and using steps 1.3 and 1.2 gives $\mathrm{\mid }{\mathrm{\Pi }}_{\mathrm{\ell }}(x,t)\mathrm{\mid }\le d^2(M_{2\alpha }+M_{\alpha }^2)N_{\alpha }A(t{)}^3{\mathrm{\ell }}^{3\alpha -1}=C_{d,K,\alpha }A(t{)}^3{\mathrm{\ell }}^{3\alpha -1}$.

Fix a compact interval $J\subset I$. By [A2] and [F9], $J$ is Lebesgue measurable, including when it is a singleton. Set ${\mathcal{L}}_J:=\{J\cap E:E\in \mathcal{L}(\mathbb{R})\}$ and ${\lambda }_J(B):={\lambda }_1(B)$ for $B\in {\mathcal{L}}_J$; this is a measure on $J$, since its empty-set value and countable additivity are inherited from ${\lambda }_1$. Here $dt$ means $d{\lambda }_J$. If $J$ is empty, both displayed time integrals are zero; otherwise lift the fields periodically to a period cell $Q=[0,L{]}^d$. Choose closed boxes $S\subset {\mathbb{R}}^d$ and $Q^{\mathrm{\prime }}\subset {\mathbb{R}}^d$ with $\mathrm{supp}K\subset S$, $Q\subset Q^{\mathrm{\prime }}$, and $x-\mathrm{\ell }s\in Q^{\mathrm{\prime }}$ whenever $x\in Q$ and $s\in S$. By [F8], $Q^{\mathrm{\prime }}\times J$ and $Q\times J\times S$ are compact. The lifts of $u$ and ${\mathrm{\partial }}_{j}u$ are uniformly continuous on $Q^{\mathrm{\prime }}\times J$, so the maps $(x,t,s)\mapsto u(x-\mathrm{\ell }s,t)$ and $(x,t,s)\mapsto {\mathrm{\partial }}_{j}u(x-\mathrm{\ell }s,t)$ are uniformly continuous on $Q\times J\times S$. The convolution formulas and the spatial derivative formula from [F2] therefore make ${\bar{u}}_{\mathrm{\ell }}$, ${\overline{u\otimes u}}_{\mathrm{\ell }}$, $\mathrm{\nabla }{\bar{u}}_{\mathrm{\ell }}$, and ${\mathrm{\Pi }}_{\mathrm{\ell }}$ continuous on $Q\times J$. Thus $F_{\mathrm{\ell }}(t):={\sup }_{x\in {\mathbb{T}}_L^d}\mathrm{\mid }{\mathrm{\Pi }}_{\mathrm{\ell }}(x,t)\mathrm{\mid }$ is finite and continuous: $\mathrm{\mid }{F}_{\mathrm{\ell }}(t)-F_{\mathrm{\ell }}(s)\mathrm{\mid }\le {\sup }_x\mathrm{\mid }{\mathrm{\Pi }}_{\mathrm{\ell }}(x,t)-{\mathrm{\Pi }}_{\mathrm{\ell }}(x,s)\mathrm{\mid }\to 0$ as $s\to t$. For each open set $V\subseteq \mathbb{R}$, continuity makes $F_{\mathrm{\ell }}^{-1}(V)$ relatively open in $J$, hence equal to $J\cap O$ for some open $O\subseteq \mathbb{R}$; [F10] gives $O\in \mathcal{L}(\mathbb{R})$, so $F_{\mathrm{\ell }}^{-1}(V)\in {\mathcal{L}}_J$. Thus $F_{\mathrm{\ell }}:J\to [0,\mathrm{\infty })$ is ${\mathcal{L}}_J$-measurable. By step 2.1, $0\le F_{\mathrm{\ell }}(t)\le C_{d,K,\alpha }{\mathrm{\ell }}^{3\alpha -1}A(t{)}^3$ on $J$. The given $A^3$ is nonnegative and ${\mathcal{L}}_J$-measurable with finite ${\lambda }_J$ integral, while $C_{d,K,\alpha }{\mathrm{\ell }}^{3\alpha -1}$ is a finite nonnegative scalar by [F2] and $\mathrm{\ell }>0$. Apply [F7] on the measure space $(J,{\mathcal{L}}_J,{\lambda }_J)$ and use [F6] to obtain the claimed finite integral inequality, including the zero-scalar case without forming $0\cdot \mathrm{\infty }$. When $\alpha >1\mathrm{/}3$, $3\alpha -1>0$, so its right-hand side tends to zero as $\mathrm{\ell }\downarrow 0$; at $\alpha =1\mathrm{/}3$ its scale factor is $1$, which gives no decay conclusion.