Is a homogeneous Banach space on $\mathbb T$ always well defined? A homogeneous Banach space $B$ on group $\mathbb T=\mathbb R/2\pi\mathbb Z$ is a linear subspace $B$ of $L^1(\mathbb T)$ having a norm $\|\ \|_B\ge\|\ \|_{L^1}$ under which it is a Banach space, and having the following properties:


*

*If $f\in B$ and $\tau\in\mathbb T$, then $f_\tau\in B$ and $\|f_\tau\|_B=\|f\|_B$ (where $f_\tau=f(t-\tau)$).

*For all $f\in B\quad $, $\tau, \tau_0\in\mathbb T,\quad $$\lim_{\tau\to\tau_0}\|f_\tau-f_{\tau_0}\|=0.$
I am wondering whether this is well-defined. For instance, if we have $f_1,f_2\in B\subseteq L^1(\mathbb T)$ and $f_1=f_2$ in $L^1(\mathbb T)$, then the values of $f_1$ and $f_2$ differ at most at a zero measure set. But it is also possible that $f_1\ne f_2$ in $B$ and in which case, it does not make sense to say the $B$-norm of $f_1$ or $f_2$ since they are considered to be one element in $L^1(\mathbb T)$ and also in the subspace $B$ of $L^1(\mathbb T)$. Whereas, $\|f_1\|_B\ne\|f_2\|_B$ is possible.
 A: Let's be extremely explicit about all the objects involved. We have the vector space $\mathcal{L}^1(\mathbb{T})$ of all measurable functions $f \colon \mathbb{T} \rightarrow \mathbb{C}$ such that $\int_{\mathbb{T}} |f| \, dt < \infty$ and on this space we have a map $\| \cdot \|_{L^1} \colon \mathcal{L}^1(\mathbb{T}) \rightarrow \mathbb{R}$ given by
$$ \| f \|_{L^1} = \int_{\mathbb{T}} |f| \, dt. $$
The map $\| \cdot \|_{L^1}$ is not a norm on $\mathcal{L}^1(\mathbb{T})$, only a semi-norm so we quotient out and replace $\mathcal{L}^1(\mathbb{T})$ with the space $L^1(\mathbb{T})$ of equivalence classes (equal a.e). Elements in $L^1(\mathbb{T})$ should be denoted by $[f]$ (but they aren't) and the $L^1$ norm descends to $L^1(\mathbb{T})$ and defines an actual norm
$$ \| [f] \|_{L^1} = \| f \|_{L^1}. $$
Now, a homogeneous Banach subspace $B$ is a subspace of $L^1(\mathbb{T})$, not of $\mathcal{L}^1(\mathbb{T})$ so the elements of $B$ are themselves equivalence classes. Going to your scenario, when you write $f_1,f_2 \in B \subseteq L^1(\mathbb{T})$, the only way to interpret it with what I wrote above is that $[f_1],[f_2] \in B \subseteq L^1(\mathbb{T})$. Then indeed it is possible that $f_1,f_2$ differ on a zero measure set but it is not possible that "$f_1 \neq f_2$ in $B$" because $B$ itself is a subspace of equivalence classes and this doesn't even make any sense.
The point is that there is some abuse of notation going on with all of this. When people say that $B = C(\mathbb{T})$ with the $\| \cdot \|_{\infty}$ norm is a homogeneous Banach space on $\mathbb{T}$, they actually mean the space 
$$ B = \{ [f] \, | \, f \in C(\mathbb{T}) \} $$
of equivalence classes (so that it becomes a subspace of $L^1(\mathbb{T})$). In each equivalence class $[f]$ one has the continuous function $f$ but also other functions $g \in [f]$ which are not necessarily continuous but equal to $f$ a.e. However, if $f \in C(\mathbb{T})$ then in each equivalence class $[f]$, there is only one continuous function which is $f$ itself. Hence, one can identify $[f]$ with $f$ and define $\| [f] \|_{\infty} = \| f \|_{\infty}$.
Now, it is possible that $[f_1] = [f_2] \in B \subseteq L^1(\mathbb{T})$ and that $f_1$ is continuous and $f_2$ is not but $f_1 = f_2$ a.e. Note that this doesn't mean that $f_1 \neq f_2$ in $C(\mathbb{T})$ as $f_2$ doesn't even belong there. However, if $[f_1] = [f_2] \in B \subseteq L^1(\mathbb{T})$ and both $f_1,f_2$ are continuous then $f_1 = f_2$ (not only a.e) and so $f_1,f_2$ are indeed the same function in $C(\mathbb{T})$.
A: I rephrase myself. 
Suppose $A$ is a veckor space and that $B$ is a subspace, then any two vectors  of $B$ are equal if and only if they are equal in $A$. 
