Function space, Harmonic analysis 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?
 A: $\newcommand{\T}{\mathbb T}\newcommand{\Z}{\mathbb Z}$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).
