I have been just reading "An introduction to harmonic analysis" by Yitzhak Katznelson, and try to finish the problem in it. There is a problem in Chapter 1:
Let $B$ be a Banach space on $\mathbb{T}$( $\mathbb{T}$ denotes torus), satisfying $\|f_\tau\|=\|f\|$, where $f_\tau=f(t+\tau)$. Define $B_c$ the set of all $f\in B$ such that $\tau\mapsto f_\tau$ is a continuous $B$ valued function. Then $B_c$ is the closure of the set of trigonometric polynomials in B.
To think about this s problem, I have tried to think of why trigonometric polynomial is so important in Banach space on $\mathbb T$ with translation invariant property, is it possible to find out a Banach space with translation invariant that even not include the space of trigonometric polynomials?
Best Answer
$\newcommand{\T}{\mathbb T}\newcommand{\Z}{\mathbb Z}$Given any $f$ in $B_c$, and any $n\in {\Z}$, define $$ \hat f(n) = \frac1{2\pi }\int_0^{2\pi } e^{-int} f_t\, dt. $$ Observe that $\hat f(n)$ is a well defined element of $B$ because the integrand is a continuous, hence Riemann integrable $B$-valued function. One should therefore notice that $\hat f(n)$ is not a scalar and hence should not be confused with the Fourier coefficient of $f$ (more to follow).
For $s$ in $\T$ denote by $T_s$ the map $$ T_s:f\in B\mapsto f_s\in B, $$ and notice that $T_s$ is an isometric linear map on $B$ by hypothesis, hence $$ T_s\big (\hat f(n)\big ) = \frac1{2\pi }\int_0^{2\pi } e^{-int} T_s(f_t)\, dt = $$$$ = \frac1{2\pi }\int_0^{2\pi } e^{-int} f_{s+t}\, dt = \frac1{2\pi }\int_0^{2\pi } e^{-in(t'-s)} f_{t'}\, dt' = $$$$ = \frac{e^{ins}}{2\pi }\int_0^{2\pi } e^{-int} f_{t}\, dt = e^{ins} \hat f(n). $$ It follows that, for every $s$ in $\T$, $$ \hat f(n)(s) = \hat f(n)(0+s) = T_s\big (\hat f(n)\big )(0) = e^{ins} \hat f(n)(0) = ce^{ins}, $$ where the constant $c$ is given by $c= \hat f(n)(0)$.
Therefore we see that, although $\hat f(n)$ is not a scalar, as noted above, it turns out to be a scalar multiple of the trigonometric polynomial $p(s)=e^{ins}$. This also shows that $B_c$ does indeed contain at least some trigonometric polynomials!
To prove that the trigonometric polynomials are dense in $B_c$, it is enough to show that any continuous linear functional $\varphi $ on $B_c$ vanishing on all trigonometric polynomials must itself vanish. So let us fix such a $\varphi $.
Given any $f$ in $B_c$, consider the continuous, scalar valued function $g$ given by $$ g(t) = \varphi (f_t),\quad\forall t\in \T. $$ Then the $n^{\text{th}}$ Fourier coefficient (sic) of $g$ is given by $$ \hat g(n) = \frac1{2\pi }\int_0^{2\pi } e^{-int} \varphi (f_t)\, dt = $$$$ = \varphi \left(\frac1{2\pi }\int_0^{2\pi } e^{-int} f_t\, dt\right) = \varphi \big (\hat f(n)\big ) = 0, $$ because $\varphi $ vanishes on the trigonometric polynomial $\hat f(n)$ by our assumption.
Any continuous function whose Fourier coefficients vanish must coincide with the zero function, so we deduce that $$ 0=g(0) = \varphi (f_0) = \varphi (f), $$ whence $\varphi =0$, as desired.
Remark: I am following the OP's convention according to which $f_\tau (t)$ is given by $f(t+\tau )$ (rather than $f(t-\tau )$, as in Katznelson's book).