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Let $S^1=\mathbb{R}/\mathbb{Z}$ and define the Fourier projection operator $P_N$ for each $N \in \mathbb{N}$ as \begin{equation} P_N(f)=\sum_{n=-N}^N \langle f, e_n \rangle_{L^2} e_n \end{equation} where $e_n(x)=e^{inx}$, $f : S^1 \to \mathbb{C}$ is a smooth function and $\langle , \rangle_{L^2}$ is the $L_2$ inner product.

Let us denote the Frechet space of smooth functions on $S^1$ as $\mathcal{E}(S^1)$. Then, I know that $\{P_N\}_{N=1}^\infty$ form a equicontinuous family of linear maps on $\mathcal{E}(S^1)$.

Next, consider the space of periodic distiributions $\mathcal{E}'(S^1)$ equipped with strong dual topology. Then, the domain $P_N$ may be extended to $\mathcal{E}'(S^1)$ via dual pairing.

My question is whether or not $\{P_N \}_{N=1}^\infty$ form a equicontinuious family of linear maps on $\mathcal{E}'(S^1)$ as well. The generalized definition of equicontinuity is as follows:

https://en.wikipedia.org/wiki/Equicontinuity#Equicontinuous_linear_maps

I strongly believe that the projection still remain as equicontinuous on $\mathcal{E}'(S^1)$, but cannot justify this myself. Could anyone please help me or recommend any references?

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    $\begingroup$ This question has been answered but it might be of interest to note that the result is a consequence of the fact that if we have a Schauder basis for a Frechet space $E$, then its dual sequence is a basis for the dual with the weak topology $\sigma(E´,E)$. The fact that in your case $E´$ is a nuclear $DF$-space allows one to replace weak by strong, since in such spaces weak and strong convergence of sequences coincide. We also use that fact that in barrelled spaces, the kind of equicontinuity condition you impose is equivalent to the fact that the sequence is a basis. $\endgroup$
    – quarta
    Commented Oct 17, 2023 at 9:56
  • $\begingroup$ @quarta By "strong" topology or "strong" convergence on $E'$, what exactly do you mean? My understanding is that $\sigma(E',E)$ means that $T_{\alpha} \to T$ iff $T_{\alpha}(f) \to T(f)$ for all $f \in E$ and all nets $\{T_{\alpha}\} \subset E'$. $\endgroup$
    – Isaac
    Commented Oct 17, 2023 at 11:15
  • $\begingroup$ Also, in my case, can we just work with sequences instead of general "nets"? I vaguely remember relevant theorems, but cannot recall in detail. $\endgroup$
    – Isaac
    Commented Oct 17, 2023 at 11:16

1 Answer 1

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By "extended to $\mathcal E'(S^1)$ via dual pairing", I assume you mean that the extension of $P_N:\mathcal E(S^1)\to\mathcal E(S^1)$ to $\mathcal E'(S^1)$ is given by its transpose ${}^t\mkern-2mu P_N:\mathcal E'(S^1)\to\mathcal E'(S^1)$.

There are several ways of proving the equicontinuity of $\lbrace\mkern1mu{}^t\mkern-2mu P_N\rbrace_{N=1}^\infty$, some more high-tech, some more elementary. The way shown below is rather elementary.


There are mainly two ingredients:

  • from the equicontinuity of $\lbrace P_N\rbrace_{N=1}^{\infty}$ on $\mathcal E(S^1)$, it is easy to check that if $B$ is a bounded subset of $\mathcal E(S^1)$, then $\bigcup_{N=1}^\infty P_N(B)$ is also a bounded subset of $\mathcal E(S^1)$,
  • from the definition of the strong dual topology, the polars in $\mathcal E'(S^1)$ of bounded subsets of $\mathcal E(S^1)$ form a basis of the filter of neighbourhoods of the origin in $\mathcal E'(S^1)$.

Then, it is easy to check that if $V$ is the polar in $\mathcal E'(S^1)$ of a bounded subset $B$ of $\mathcal E(S^1)$ and $U$ is the polar in $\mathcal E'(S^1)$ of $\bigcup_{N=1}^\infty P_N(B)$, then $(\mkern2mu{}^t\mkern-2mu P_N)(U)\subseteq V$ for every $N\geq 1$: indeed, $$|\langle\mkern2mu{}^t\mkern-2mu P_N(T),f\rangle|=|\langle T,P_N(f)\rangle|\leq 1$$ for every $N\geq 1$, $T\in U$ and $f\in B$.

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  • $\begingroup$ Thank you so much for your detailed answer! $\endgroup$
    – Isaac
    Commented Oct 17, 2023 at 11:16
  • $\begingroup$ Perhaps, could you answer the following question too? mathoverflow.net/questions/457008/… $\endgroup$
    – Isaac
    Commented Oct 23, 2023 at 23:13

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