Example of direct decomposition of a Banach space such as $C[0,1]$ into closed linear subspaces While reading a theorem on estimates in Banach spaces I came across the following:

Let $X$ be a Banach space and let $X_1,X_2 \subset X$ be two closed
  linear subspaces such that $X = X_1 \oplus X_2$, i.e. $X_1 \cap X_2 =
 \{0\}$ and every vector $x\in X$ can be written as $x=x_1 +
 x_2$ with $x_1 \in X_1$ and $x_2 \in X$.

Say $X$ is the space of continuous functions on $[0,1]$and we want a direct decomposition of this space into two closed linear subspaces $X_1$ and $X_2$.
Can we do so by splitting the domain $[0,1]$ into say $[0,0.5]$ and $[0.5,1]$? If so, I'm really not sure how to correctly interpret $X_1 \cap X_2 = \{0\}$, or the expression $x = x_1 + x_2$. Does $\{0\}$ refer to the zero function or the empty set? If its the zero function,  we would now have three zero functions (on $X$, $X_1$, and $X_2$) so I don't see how they can be compared.
Does someone know a simple example of a direct decomposition of this space into two closed linear subspaces $X_1$ and $X_2$ so that I can clear up the issues I have described above?
 A: $\{0\}$ means the subset of $X$ which consists of one element, namely the zero element of the space. In the context of functions, $0$ is the function that is identically zero. 

we would now have three zero functions 

No, it's the same zero function. Since $X_1\subset X$, the elements of $X_1$ are functions defined on all of $[0,1]$. We don't actually change their domain. "Splitting the domain" isn't really what is happening, it's more like "splitting the support set of functions", making them zero or not on some part of their domain.
Construction
If we define
$$X_1=\{f\in X:f\equiv 0 \text{ on } [1/2,1]\}$$
$$X_2=\{f\in X:f\equiv 0 \text{ on } [0, 1/2]\}$$
then it's true that $X_1\cap X_2 = \{0\}$. But the property $X=X_1\oplus X_2$ fails: every element $f\in X_1\oplus X_2$ satisfies $f(1/2)=0$, so we don't get, for example, the constant function $1$. 
To get a direct sum, try
$$X_1=\{f\in X:f\equiv 0 \text{ on } [1/2,1]\}$$
$$X_2=\{f\in X:f\equiv c \text{ on } [0, 1/2] \text{ for some constant } c\}$$
This still satisfies $X_1\cap X_2 = \{0\}$. But now every $f\in X$ can be written as the sum according to the above: 
$$f_1(x)=\begin{cases}
f(x) - f(1/2), & x\in [0,1/2] \\
0, & x\in [1/2,1]
\end{cases}$$
and 
$$f_2(x)=\begin{cases}
f(1/2), & x\in [0,1/2] \\
f(x), & x\in [1/2,1]
\end{cases}$$
A: In analogy with even and odd functions, you can consider
$$
X_1 = \{ f \in X \mid f(1-t) = f(t)\ \forall t \in [0,1] \}; \\
X_2 = \{ f \in X \mid f(1-t) = -f(t)\ \forall t \in [0,1] \}.
$$
Then $X_1 \cap X_2 = \{0\}$, and for each $f \in C[0,1]$, you can write $f = f_1 + f_2$ where
$$
f_1(t) := \frac{f(t)+f(1-t)}{2} \in X_1; \quad f_2(t) := \frac{f(t)-f(1-t)}{2} \in X_2.
$$
That $X_1$ and $X_2$ are closed also follows from the fact that uniform convergence implies pointwise convergence, so that any limit point of $X_1$ resp. $X_2$ must satisfy the same equality.
