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Let $\mathbb{T}^d$ be the $d$-dimensional torus, $\mathscr{S}:=C^\infty(\mathbb{T}^d)$ the Schwartz space, $\mathscr{S}'$ the space of tempered distributions.

We consider a dyadic partition of unity $(\rho_{-1},\rho_{0})$ and define $$ \rho_{j}(x):=\rho_0(x2^{-j}),x\in \mathbb{T}^d,j \in \mathbb{N}. $$ In most of the papers, publications, Besov-Holder space are defined by $$ \mathscr{C}^\alpha:=\Big\{f \in \mathscr{S}',\left\Vert f\right\Vert_\alpha:=\sup_{j \geq -1}\left\Vert \mathscr{F}^{-1}_{\mathbb{T}^d}\rho_j*f\right\Vert_{L^{\infty}(\mathbb{T}^d)}<\infty\Big\}, $$ where $\mathscr{F}^{-1}_{\mathbb{T}^d}$ is the inverse Fourier transform on $\mathbb{T}^d$.
Consider the heat kernel $$ p(r,x)=\sum_{k \in \mathbb{Z}^d}e^{2\pi\mathrm{i}\langle x,k\rangle}e^{-r|k|^2}\quad x \in \mathbb{T}^d,\alpha\in]0,2[ $$ and define $P_0:=\mathrm{Id}_{\mathscr{S}'},P_rf:=p(r,\cdot)*f,f \in \mathscr{S}',r>0.$

For $g \in \mathscr{C}^\alpha,$ can we claim that $r\longmapsto P_rg$ is continuous at $0$?

We have, for $r \in [0,1]$, $j \in \mathbb{N}$, $$ \left\Vert P_rg-g\right\Vert_{\alpha}\lesssim 2^{j\alpha}\min(1,h^{\alpha/2})\left\Vert g\right\Vert_{\alpha}+\sup_{m \in \mathbb{N}\cap[j,\infty[}2^{m\alpha}\left\Vert\mathscr{F}^{-1}_{\mathbb{T}^d}\rho_j*g\right\Vert_{L^\infty(\mathbb{T}^d)},$$ and the first term clearly goes to $0$ when $h\to 0.$

How to treat $\sup_{m \in \mathbb{N}\cap[j,\infty[}2^{m\alpha}\left\Vert\mathscr{F}^{-1}_{\mathbb{T}^d}\rho_j*g\right\Vert_{L^\infty(\mathbb{T}^d)},$ when $j\to\infty$ from the definition that $g \in \mathscr{C}^\alpha$?

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    $\begingroup$ $P_rg$ is smooth, but smooth functions aren't dense in $\mathcal{C}^\alpha$, so it cannot hold in general. $\endgroup$ Commented Jun 2 at 21:33
  • $\begingroup$ @MartinHairer, is there an alternative way to prove continuity at $0$? What must be done in this case to ensure having continuity? This is crucially needed to prove Schauder estimates for Holder-Besov spaces $\endgroup$
    – mathex
    Commented Jun 2 at 23:55
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    $\begingroup$ Define $\mathcal{C}^\alpha$ as the closure of $\mathcal{C}^\infty$ under the norm you describe. $\endgroup$ Commented Jun 3 at 7:36

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