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I've edited, just skip the first attempt and go to the second one.

THE FRAMEWORK: let us consider a real topological vector space $V$.

We denote with $\mathscr C_k(V)$ the set of all continous functions $f:[0,T]^k\to V$ such that $f_{t_1\cdots t_k}=0$ whenever $t_i=t_{i+1}$ for some $0\le i\le k-1$.

We define the operator $\delta_k:\mathscr C_k(V)\to\mathscr C_{k+1}(V)$ as follows: $$ (\delta_kf)_{t_1\cdots t_{k+1}}:=\sum_{j=1}^{k+1}(-1)^jf_{t_1\cdots \widehat t_{j}\cdots t_{k+1}} $$ where the hat $\widehat{\cdot}$ means that argument is omitted.

So if $f\in\mathscr C_1(V)$ then $(\delta_1f)_{ts}=f_t-f_s$ and if $g\in\mathscr C_2(V)$ then $(\delta_2g)_{tus}=-g_{us}+g_{ts}-g_{tu}$.

Next we define the following norms: if $g\in\mathscr C_2(V)$ then, for $\mu>1$ we set $$ \|g\|_{\mu}:=\sup_{0\le s<t\le T}\frac{|g_{st}|}{|t-s|^{\mu}} $$ while for $h\in\mathscr C_3(V)$ we first set $$ \|h\|_{\mu,\rho}:=\sup_{0\le s<t\le T}\frac{|h_{tus}|}{|t-u|^{\rho}|u-s|^{\mu-\rho}} $$ (here $0<\rho<\mu$) and then define the norm: $$ \|h\|_{\mu}:=\inf\left\{\sum_j\|h_j\|_{\rho_j,\mu-\rho_j}:h=\sum_jh_j,\; h_j\in\mathscr C_3,0<\rho_j<\mu\right\}\;. $$

Finally let us denote for $k=2,3$ $$ \mathscr C_k^{\mu}:=\{h\in\mathscr C_k:\|h\|_{\mu}<+\infty\}. $$ and accept that $$ \ker\delta_k=\operatorname{Im}\delta_{k-1} $$ (this holds for every $k$).

THE PROBLEM: let us take $h\in\mathscr C_3^{\mu}$ such that $\delta_3h=0$. Then it is not difficult to prove that there exists $B\in\mathscr C_2$ such that $\delta_2B=h$. Now fix $0\le s<t\le T$ and consider on $[s,t]$ a sequence of partitions $\{\pi_n\}_n$ whose mash tends to zero.

To fix ideas we write $$ \pi_n=\{s=r_0^n<r_1^n<\cdots<r_{k_n}^n<r_{k_n+1}^n=t\} $$

Define then $$ M_{ts}^{\pi_n}:=B_{ts}-\sum_{l=0}^{k_n}B_{r_{l+1}^nr_l^n}\;. $$ Now accept this last one converges (up to passing to a subsequence): how can we show that the limit does not depend on the particular sequence of partitions chosen?

MY ATTEMPT: let us take another sequence of partitions whose mash tends to zero, say $$ \sigma_n=\{s=u_0^n<u_1^n<\cdots<u_{h_n}^n<u_{h_n+1}^n=t\} $$

and prove that $$ |M_{ts}^{\pi_n}-M_{ts}^{\sigma_n}|\to0\;\;\;n\to+\infty. $$

Now \begin{align*} M_{ts}^{\pi_n}-M_{ts}^{\sigma_n} &=\sum_{l=0}^{h_n}B_{u_{l+1}^nr_l^n} -\sum_{l=0}^{k_n}B_{r_{l+1}^nr_l^n}\\ &=\sum_{l=0}^{k_n} \left(B_{t_{4l+4}^nt_{4l+2}^n}-B_{t_{4l+3}^nt_{4l+2}^n}- B_{t_{4l+4}^nt_{4l+3}^n}\right)- \left(B_{t_{4l+3}^nt_{4l+3}^n}-B_{t_{4l+2}^nt_{4l+1}^n}- B_{t_{4l+3}^nt_{4l+2}^n}\right)\\ &=\sum_{l=0}^{k_n} (\delta_2B)_{t_{4l+4}^nt_{4l+3}^nt_{4l+2}^n}- (\delta_2B)_{t_{4l+3}^nt_{4l+2}^nt_{4l+1}^n}\\ &=\sum_{l=0}^{k_n} h_{t_{4l+4}^nt_{4l+3}^nt_{4l+2}^n}- h_{t_{4l+3}^nt_{4l+2}^nt_{4l+1}^n}\\ \end{align*} supposing wlog that $h_n\le k_n$ and setting, for $l\le h_n$ \begin{align*} t_{4l+1}^n&:=r_l^n\\ t_{4l+2}^n&:=u_l^n\\ t_{4l+3}^n&:=r_{l+1}^n\\ t_{4l+4}^n&:=u_{l+1}^n\\ \end{align*} while, for $l>h_n$ the definition of the odd terms stay the same, while $t_{4l+2}^n=t_{4l+4}^n:=t$.

Then we can write \begin{align*} |M_{ts}^{\pi_n}-M_{ts}^{\sigma_n}| &\le\sum_{l=0}^{k_n} |h_{t_{4l+4}^nt_{4l+3}^nt_{4l+2}^n}|+ |h_{t_{4l+3}^nt_{4l+2}^nt_{4l+1}^n}|\\ &=\sum_{l=0}^{k_n} \left|\sum_jh_{t_{4l+4}^nt_{4l+3}^nt_{4l+2}^n}^j\right|+ \left|\sum_ih_{t_{4l+3}^nt_{4l+2}^nt_{4l+1}^n}^i\right|\\ &\le\sum_{l=0}^{k_n}\left[ \sum_j\left|h_{t_{4l+4}^nt_{4l+3}^nt_{4l+2}^n}^j\right|+ \sum_i\left|h_{t_{4l+3}^nt_{4l+2}^nt_{4l+1}^n}^i\right|\right]\\ &=\sum_{l=0}^{k_n}\left[ \sum_j\left|\frac{h_{t_{4l+4}^nt_{4l+3}^nt_{4l+2}^n}^j}{|t_{4l+3}^n-t_{4l+2}^n|^{\rho_j}|t_{4l+4}^n-t_{4l+3}^n|^{\mu-\rho_j}}\right| |t_{4l+3}^n-t_{4l+2}^n|^{\rho_j}|t_{4l+4}^n-t_{4l+3}^n|^{\mu-\rho_j}\right]\\ +& \left[ \sum_i\left|\frac{h_{t_{4l+3}^nt_{4l+2}^nt_{4l+1}^n}^i}{|t_{4l+2}^n-t_{4l+1}^n|^{\rho_i}|t_{4l+2}^n-t_{4l+2}^n|^{\mu-\rho_i}}\right| |t_{4l+2}^n-t_{4l+1}^n|^{\rho_i}|t_{4l+3}^n-t_{4l+2}^n|^{\mu-\rho_i}\right]\\ \end{align*} and since this is true for every $\{h^j\}_j,\{h^i\}_i\subset\mathscr C_3(V)$ such that $\sum_jh^j=\sum_ih^i=h$ and $0<\rho_j,\rho_i<\mu$, passing to the $\inf$ on these parameters, we get \begin{align*} |M_{ts}^{\pi_n}-M_{ts}^{\sigma_n}| \le \|h\|_{\mu}\left[\sum_{l=0}^{k_n}\left(\max\{|t_{4l+3}^n-t_{4l+2}^n|,|t_{4l+4}^n-t_{4l+3}^n|\}\right)^{\mu}\\ +\left(\max\{|t_{4l+2}^n-t_{4l+1}^n|,|t_{4l+3}^n-t_{4l+2}^n|\}\right)^{\mu}\right]\\ \le\|h\|_{\mu}\left[\sum_{l=0}^{k_n}\left(|t_{4l+3}^n-t_{4l+2}^n|+|t_{4l+4}^n-t_{4l+3}^n|\right)^{\mu}\\ +\left(|t_{4l+2}^n-t_{4l+1}^n|+|t_{4l+3}^n-t_{4l+2}^n|\right)^{\mu}\right] \end{align*}

From this, I tried as follows: call $m_0=1$ and take $\pi_{m_0}$; then since the mash tends to zero, there exists $m_1>m_0$ such that $m_0$ elements of $\sigma_{m_1}$, call them $\widetilde{\sigma_{m_1}}:=\{u_{l_s}^{m_1}\}_{s=1}^{m_0}$ are closer as we want to the elements of $\pi_{m_0}$, say $|r_s^{m_0}-u_{j_s}^{m_1}|<\varepsilon_1$.

We have to work on $\sigma_{m_1}$: there exists $m_2>m_1$ such that considering suitable elements of $\pi_{m_2}$, say $\widetilde{\pi_{m_2}}:=\{r_{j_s}^{m_2}\}_{s=1}^{m_1}$, we can approximate the elements of $\sigma_{m_1}$, say $|u_s^{m_i}-r_{j_s}^{m_2}|<\varepsilon_2$, and so on.

Then I did the obvious: $$ M_{ts}^{\pi_{m_{2\beta}}}-M_{ts}^{\sigma_{m_{2\beta+1}}} =\left(M_{ts}^{\pi_{m_{2\beta}}}-M_{ts}^{\widetilde{\sigma_{m_{2\beta+1}}}}\right) +\left(M_{ts}^{\widetilde{\sigma_{m_{2\beta+1}}}}-M_{ts}^{\widetilde{\pi_{m_{2(\beta+1)}}}}\right) +\left(M_{ts}^{\widetilde{\pi_{m_{2(\beta+1)}}}}-M_{ts}^{\sigma_{m_{2(\beta+1)-1}}}\right)\;\;; $$ now the first and the last summand go to zero as $\beta\to+\infty$, but the second term doesn't a priori, since it is an approximation of the LHS.

I thought that maybe there is a way to modify the $\widetilde{\pi}$ and $\widetilde{\sigma}$ and/or exploiting the absolute continuity of $M$ in oder to compare the LHS and the second term of RHS, and maybe writing something like $$ \left|M_{ts}^{\widetilde{\sigma_{m_{2\beta+1}}}}-M_{ts}^{\widetilde{\pi_{m_{2(\beta+1)}}}}\right| =a_{\beta}|M_{ts}^{\pi_{m_{2\beta}}}-M_{ts}^{\sigma_{m_{2\beta+1}}}| $$ for some $0<a_{\beta}<1$ such that $\limsup_{\beta\to+\infty}a_{\beta}<1$, thing that allow me to conclude, but for the moment I'm stuck.

ATTEMPT 2:

Step 1: Let us fix $m_0\ge1$ and consider $\Pi_{m_0}$ (which has $k_{m_0}+2$ elements); then, since the meshes of both sequences of partitions go to $0$, there exists $m_1>m_0$, such that, looking at $\mathfrak S_{m_1}$ we can find in it $k_{m_0}+3$ elements we label as $\{u_{j_s}^{m_1}\}_{s=0}^{k_{m_0}+2}=:\widetilde{\mathfrak S}_{m_1}$ such that \begin{align*} \left|B_{r_{l+1}^{m_0}r_l^{m_0}}-B_{u_{j_{l+1}}^{m_1}u_{j_{l}}^{m_1}}\right|&\le\frac1{2(k_{m_0}+1)^3} \;\;,\;\;\;\;l=0,\dots,k_{m_0}\\ \left|B_{u_{j_{k_{m_0}+2}}^{m_1}u_{j_{k_{m_0}+1}}^{m_1}}\right|&\le\frac1{2(k_{m_0}+1)^3}\;. \end{align*} Thus we associate step $1$ to the pair $\left(\Pi_{m_0},\widetilde{\mathfrak S}_{m_1}\right)$.

Step 2: Let us consider $\mathfrak S_{m_1}$ (which has $h_{m_1}+2$ elements); then there exists $m_2>m_1$ such that looking at $\Pi_{m_2}$, we can find in it $h_{m_1}+3$ elements we label as $\{r_{j_s}^{m_2}\}_{j=0}^{h_{m_1}+3}=:\widetilde{\Pi}_{m_2}$ such that \begin{align*} \left|B_{u_{l+1}^{m_1}u_l^{m_1}}-B_{r_{j_{l+1}}^{m_2}r_{j_{l}}^{m_2}}\right|&\le\frac1{2(h_{m_1}+1)^3} \;\;,\;\;\;\;l=0,\dots,h_{m_1}\\ \left|B_{r_{j_{h_{m_1}+2}}^{m_2}r_{j_{h_{m_1}+1}}^{m_2}}\right|&\le\frac1{2(h_{m_1}+1)^3}\;. \end{align*} Thus we associate step $2$ to the pair $\left(\mathfrak S_{m_1},\widetilde{\Pi}_{m_2}\right)$.

Going on with this construction, we have \begin{align*} &{Step 2N}\leadsto \left(\mathfrak S_{m_{2N-1}},\widetilde{\Pi}_{m_{2N}}\right)\\ &{Step 2N+1}\leadsto \left(\Pi_{m_{2N}},\widetilde{\mathfrak S}_{m_{2N+1}}\right). \end{align*} Now it is well know that given a converging sequence (and it is known that $\{M_{ts}^{\Pi_n}\}_{n\ge1}$, up to passing to a subsequence, is such), every its subsequence converges to the same limit. Since we want to prove that $$ \left|M_{ts}^{\Pi_n}-M_{ts}^{\mathfrak S_n}\right|\stackrel{n\to+\infty}{\longrightarrow}0 $$ we can work with any subsequence of $\{M_{ts}^{\Pi_n}\}_{n\ge1}$ and $\{M_{ts}^{\mathfrak S_n}\}_{n\ge1}$. So, let us prove that $$ \left|M_{ts}^{\Pi_{m_{2N}}}-M_{ts}^{\mathfrak S_{m_{2N+1}}}\right|\stackrel{N\to+\infty}{\longrightarrow}0. $$

Now, observing that $k_m,h_m\ge m$ and $m_N\ge N$, we have \begin{align*} \left|M_{ts}^{\Pi_{m_{2N}}}-M_{ts}^{\mathfrak S_{m_{2N+1}}}\right| &\le\left|M_{ts}^{\Pi_{m_{2N}}}-M_{ts}^{\widetilde{\mathfrak S}_{m_{2N+1}}}\right| +\left|M_{ts}^{\widetilde{\mathfrak S}_{m_{2N+1}}}-M_{ts}^{\widetilde{\Pi}_{m_{2(N+1)}}}\right| +\left|M_{ts}^{\widetilde{\Pi}_{m_{2(N+1)}}}-M_{ts}^{\mathfrak S_{m_{2N+1}}}\right|\\ &\le\frac1{(k_{m_{2N}}+1)^2} +\left|M_{ts}^{\widetilde{\mathfrak S}_{m_{2N+1}}}-M_{ts}^{\widetilde{\Pi}_{m_{2(N+1)}}}\right| +\frac1{(h_{m_{2N+1}}+1)^2}\\ &\le\frac1{(2N+1)^2} +\frac1{(2N+2)^2} +\left|M_{ts}^{\widetilde{\mathfrak S}_{m_{2N+1}}}-M_{ts}^{\widetilde{\Pi}_{m_{2(N+1)}}}\right|. \end{align*}

but here I am stuck, because in order to repeat the argument for the last summand I should pass to a suitable subsequence wrt to $N$ in the whole inequality $$ \left|M_{ts}^{\Pi_{m_{2N}}}-M_{ts}^{\mathfrak S_{m_{2N+1}}}\right| \le\frac1{(2N+1)^2} +\frac1{(2N+2)^2} +\left|M_{ts}^{\widetilde{\mathfrak S}_{m_{2N+1}}}-M_{ts}^{\widetilde{\Pi}_{m_{2(N+1)}}}\right| $$ but I don't know how to do it in a rigorous way.

Moreover, even doing this, and getting something like $$ \left|M_{ts}^{\Pi_{m_{2N}}}-M_{ts}^{\mathfrak S_{m_{2N+1}}}\right| \le\sum_{k\ge N}a_k+\left|M_{ts}^{\Pi_{m_{X}}}-M_{ts}^{\mathfrak S_{m_{Y}}}\right| $$ where $\sum_ka_k$ is a convergent series, how can we control the last summand?

PS: this is taken from a proof contained in the 2010 paper by M.Gubinelli and S. Tindel ROUGH EVOLUTION EQUATIONS at the end of page 9 (they say to see another paper, but I didn't found nothing on it!)

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    $\begingroup$ I up-voted this for the sheer effort that's gone into it. $\endgroup$ Nov 28, 2017 at 16:38
  • $\begingroup$ This is very similar to the definition of Grower norms. $\endgroup$
    – Hu xiyu
    Nov 28, 2017 at 19:45

1 Answer 1

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Let us define then $\mathfrak {U}_n:=\Pi_n\cup\mathfrak S_n$ and $u_n+2$ as the cardinality of $\mathfrak U_n$; then, relabeling the elements of $\mathfrak U_n$, we can write $\mathfrak U_n=\{s=t_0^n<t_1^n<\cdots<t_{u_n}^n<t_{u_n+1}^n=t\}$

We recall that \begin{align*} M_{ts}^{\Pi_n}=B_{ts}-\sum_{i=0}^{k_n}B_{r_{i+1}^nr_i^n}\;\;\;\;\;\;\;\mbox{and}\;\;\;\;\;\;\;\;\;\; M_{ts}^{\mathfrak S_n}=B_{ts}-\sum_{i=0}^{h_n}B_{u_{i+1}^nu_i^n}\;. \end{align*}

We want to prove that $$ \left|M_{ts}^{\Pi_n}-M_{ts}^{\mathfrak S_n}\right|\stackrel{n\to+\infty}{\longrightarrow}0. $$ So, let us consider $$ \left|M_{ts}^{\Pi_n}-M_{ts}^{\mathfrak S_n}\right| \le \left|M_{ts}^{\Pi_n}-M_{ts}^{\mathfrak U_n}\right|+\left|M_{ts}^{\mathfrak U_n}-M_{ts}^{\mathfrak S_n}\right|. $$ Now \begin{align*} \left|M_{ts}^{\Pi_n}-M_{ts}^{\mathfrak U_n}\right| &=\left|\sum_{i=0}^{k_n}B_{r_{i+1}^nr_i^n}-\sum_{j=0}^{u_n}B_{t_{j+1}^nt_j^n}\right|\\ &=\left|\sum_{i=0}^{k_n}\left(B_{r_{i+1}^nr_i^n}-\sum_{\substack{t_j^n,t_{j+1}^n\in\mathfrak U_n\\r_i^n\le t_j^n<t_{j+1}^n\le r_{l+1}^n}}B_{t_{j+1}^nt_j^n}\right)\right|\\ &\le\sum_{i=0}^{k_n}\left|B_{r_{i+1}^nr_i^n}-\sum_{\substack{t_j^n,t_{j+1}^n\in\mathfrak U_n\\r_i^n\le t_j^n<t_{j+1}^n\le r_{l+1}^n}}B_{t_{j+1}^nt_j^n}\right|\\ &\le\sum_{i=0}^{k_n}\left|M_{r_{i+1}^nr_i^n}^{\mathfrak U_n\cap[r_i^n,r_{i+1}^n]}\right|\\ &\le c_{h,\lambda}\sum_{i=0}^{k_n}|r_{i+1}^n-r_i^n|^{\lambda}\\ &\le c_{h,\lambda}|\Pi_n|^{\lambda-1}\sum_{i=0}^{k_n}|r_{i+1}^n-r_i^n|\\ &= c_{h,\lambda}|\Pi_n|^{\lambda-1}|t-s|\stackrel{n\to+\infty}{\longrightarrow}0. \end{align*} The proof that even $\left|M_{ts}^{\mathfrak U_n}-M_{ts}^{\mathfrak S_n}\right|$ goes to 0 is the same because of the simmetry of $\Pi_n$ and $\mathfrak S_n$ with respect to $\mathfrak U_n$.

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