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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.

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The fact that you are missing is that translation is a diagonal operator in the basis of exponentials. This simple fact is fundamental to the whole Fourier business, since from translations we get derivatives etc. Details follow.

In order to prove that $B_c$ contains all trigonometric polynomials, it suffices to show that it contains monomials $t\mapsto \exp(int) $. The latter follows from $\exp(in(t+\tau))-\exp(int) =(\exp(in\tau)-1)\exp(int)$. Indeed, the norm of this difference is $|\exp(in\tau)-1|\|\exp(int)\|\to0$ as $\tau\to0$.

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Duh, that was silly! Thanks! – dls Jul 25 '12 at 17:18

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