Had a question from Katznelson recorded in my journal which is still bugging me; I believe I have solved the following exercise subject to a minor point: Let $B$ be a Banach spach on $\mathbb{T}$ with translation invariant norm $\|\cdot\|_B$. Denote by $B_c$ the closed subspace of $B$ such that the translation $\tau \mapsto f_{\tau}$ is continuous in the $B$-norm. Prove that $B_c$ is the closure of the set of trigonometric polynomials in $B$.
Let $A \subset B$ be the collection of trigonometric polynomials in $B$. Then we want to show that $\bar{A}=B_c$. First, suppose $f$ is an element of $B_c$. Then by a previous theorem, $\|K_n \star f - f\|_B \rightarrow 0$ as $n\rightarrow\infty$ where $K_n$ is the Fejér kernel. Observe that $K_n$ is in $L^1(\mathbb{T})$ and so by another exercise, $K_n \star f$ is contained in the homogeneous Banach space $B_c$ because $f$ is. Hence $f \in \bar{A}$.
Conversely, suppose $f \in \bar{A}$. Then there exists a sequence of trigonometric polynomials $\{f_n\}$ converging to $f$ in $B$-norm. We want to show that $\|f_{\tau}-f\|_B\rightarrow0$ as $\tau\rightarrow0$. Now, $ \begin{align*} \|f_{\tau}-f\|_B &\leq \|f_{\tau}-(f_n)_{\tau}\|_B + \|(f_n)_{\tau}-f_n\|_B + \|f_n-f\|_B \\ &\leq 2\|f_n-f\|_B + \|(f_n)_{\tau}-f_n\|_B, \end{align*} $ by translation invariance of the norm. Since $f_n \rightarrow f$, we can fix $n$ so that the first term in the last line is as small as desired. Here comes my question: is it then true that $f_n$ is an element of $B_c$, that is, do we know that the second term can be made small for small $\tau$?
As an alternate solution for the second direction of the inclusion, if each $f_n$ is contained in $B_c$, then it must be that $f \in B_c$ as $B_c$ is closed. But the question is still the same: given $g \in B$ a trigonometric polynomial, is it always the case that $\tau \mapsto g_{\tau}$ is continuous? Is $g \in B_c$? I'm unsure what fact about $\|\cdot\|_B$ or the complex exponentials would make this statement true.