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I uploaded a photo of my problem because latex isn't working very well here.

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Note that it's a bit of awkward notation to use the index $n$ for both the sequence, and the definition of $d_{\mathcal{S}}$. I'll use $k$ for the latter

  1. If $p_{\alpha, \beta}(f_n-f)\to 0$ for all $\alpha,\beta$. Fix $\varepsilon>0$ and pick first $K$ such that $\sum_{k=K+1}^\infty c_k <\varepsilon/2$, then pick $N=N(K,\varepsilon)$ such that $\sum_{k=1}^K p_{\alpha_k,\beta_k}(f_n-f)<\varepsilon/2$.

  2. Use the "trivial" bound $\| \mathcal{F} F\|_\infty \leq \| F\|_1$, so that $$ p_{\alpha,\beta}(\hat{f}_n-\hat{f})\lesssim \| \partial^{\alpha}(x^\beta(f_n-f)(x))\|_1, $$ and now use the product rule to bound the right hand side in terms of a finite sum of terms of the form $$ \| x^{\beta'}\partial^{\alpha}(f_n-f)(x)\|_{1}, $$ with $\beta'\leq \beta$. Now use that $f_n-f$ is Schwartz again to get, for instance, that $$ |x^{\beta'}\partial^\alpha(f_n-f)(x)|\lesssim p_{\alpha'', \beta''}(f_n-f)(1+|x|)^{-2}, \qquad x\in \mathbb{R}^n, $$ for some $\alpha'', \beta''$. This gives the integrability and continuity.

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