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Let $M$ be the Banach algebra of measures on the circle with $L_1$ naturally sitting as a closed ideal of $M$. Then $M$ carries the strict topology implemented by the family of seminorms $\|\mu\|_f = \|f\ast \mu\|$ where $f\in L_1$.

Let $F\colon M\to C(X_M)$ be the Gelfand (Fourier-Stieltjes) transform, where $X_M$ is the maximal ideal space of $M$. Is $F$ strictly-to-norm continous?

Edit: Of course, the answer is negative as explained in the comments. I would be then interested in the following follow-up question.

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    $\begingroup$ So are you asking if $\| f*\mu_n\| \to 0$ for all $f\in L^1$ implies that $\|\widehat{\mu_n}\|_{\infty} \to 0$, with $\widehat{\mu}$ denoting the classical Fourier transform? $\endgroup$
    – Christian Remling
    Commented Jul 25, 2018 at 20:15
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    $\begingroup$ As @ChristianRemling's comment suggests, the answer is negative (if I have not made an error, just take a bounded net in $L^1$ that forms an approximate identity, e.g. Fejer kernels). I don't follow the "intuition" you mention in the last sentence; what is going on is that you have the Gelfand transform $L_1\to C_0$ and it is natural to guess that this would extend to a strictly-to-strictly continuous map $M(L_1) \to M(C_0) = C_b$, but this extension is unlikely to be strictly-to-norm continuous $\endgroup$
    – Yemon Choi
    Commented Jul 26, 2018 at 3:10
  • $\begingroup$ @Jan_Ch: it's considered bad form to change the title to a different question after the original question has been answered. Next time please ask the new question separately. $\endgroup$
    – Nik Weaver
    Commented Jul 27, 2018 at 14:13

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