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For every $m\in\mathbb{N}_+$, $k=0,\dots,m-1$, denote $I_{m,k}:=(\frac{k}{m},\frac{k+1}{m})$.

Denote $W = H^1(S^1,\mathbb R^{2n})$ = The Hilbert space of $C^1$ maps maps $v(t)$ from the circle to $\mathbb R^{2n}$ with $v$ and $\dot v$ in $L^2$. We will think of those maps as maps $v\colon[0,1]\to\mathbb R^{2n}$ with $v(0)=v(1)$.

Consider the spaces:

\begin{eqnarray*} W_m &:=& W\cap\{v\in H^1([0,1],\mathbb R^{2n}):\text{$v|_{I_{m,k}}$ is linear}\} \end{eqnarray*}

Consider the projection $P$ from $W$ to $W_m$. Does $P$ have a kernel in the sense of convolution operator or Hilbert-Schmidt operator? Is there a closed formula?

(This is reminiscent of the situation in Fourier series, where the projection onto the space of truncated Fourier sums is a convolution with Dirichlet Kernel)

Thanks

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  • $\begingroup$ You don't tell us what the inner product on $W$ is, but if it's the obvious one (namely, $\langle v,w\rangle = \sum_1^{2n} \int v_iw_i + \sum_1^{2n}\int \dot{v}_i\dot{w}_i$) then you can reduce to maps into $\mathbb{R}$, since the coordinates are independent. Is that the inner product you had in mind? $\endgroup$
    – Nik Weaver
    Commented Jan 28, 2016 at 2:18
  • $\begingroup$ Yes. It's the obvious inner product. After reducing to $\mathbb R$ what is next? Is there a kernel for the projection in the one dimensional case? $\endgroup$
    – Yaniv Ganor
    Commented Jan 28, 2016 at 15:07

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