Show that $\mathbb{R} ^ {\mathbb{R}} = M \oplus L.$ 
Let $L = \{ f \in \mathbb{R}^{\mathbb{R}} : f(-t) = f(t) \} $ and $M = \{ f \in \mathbb{R}^{\mathbb{R}} : f(-t) = -f(t) \}$. Show that $\mathbb{R}^{\mathbb{R}} = M \oplus L.$

I am having trouble with this. It is easy to show that $ M \cap L $ equals null vector. The problem starts when I have to prove that each function in $\mathbb{R}^{\mathbb{R}}$ can be written as sum of some function in $M$ and $L$. 
First function that came to my mind was constant funtion. How can I write it as sum of even and odd function? 
 A: The comments suggest simple formulas, but suppose no formulas came to your mind and you wanted to somehow "brute force" this problem.
Consider an arbitrary $f \in \mathbb R^{\mathbb R}$. For each $t$, consider $f(t)=A$ and $f(-t)=B$. Let's write down everything that we need to be true:
\begin{align*}
L(t) + M(t) &= A \\
L(-t) + M(-t) &= B \\
L(t) - L(-t) &= 0 \\
M(t) + M(-t) &= 0
\end{align*}
Let's suppose for the moment that $t \ne 0$, and so we can think of $L(t)$, $M(t)$, $L(-t)$, and $M(-t)$ as four independent unknown quantities that we need to solve for. So this is a system four linear equations in four unknowns, and we can solve it. 
The case $t=0$ is a special case, we must have $M(0)=0$ and we can just take $L(0)=f(0)$.
A: For
$f(t) \in \Bbb R^{\Bbb R} \tag 1$
define
$f_+(t) = \dfrac{f(t) + f(-t)}{2}, \tag 2$
and
$f_-(t) = \dfrac{f(t) - f(-t)}{2}; \tag 3$
then
$f_+(-t) = \dfrac{f(-t) + f(-(-t))}{2} =  \dfrac{f(t) + f(-t)}{2} = f_+(t) \tag 4$
and
$f_-(-t) = \dfrac{f(-t) - f(-(-t))}{2} = \dfrac{f(-t) - f(t)}{2} = -f_-(t); \tag 5$
evidently
$f_+(t) \in L \tag 6$
and
$f_-(t) \in M; \tag 7$ 
also, it is easy to see that
$f(t) = f_+(t) + f_-(t); \tag 8$
thus
$\Bbb R^{\Bbb R} = L + M; \tag 9$
now if
$g(t) \in L \cap M, \tag{10}$
we have
$g(t) = g(-t) \tag{11}$
and
$g(-t) = -g(-t); \tag{12}$
these two together yield
$g(t) = -g(t) \Longrightarrow g(t) = 0; \tag{13}$
thus
$L \cap M = \{0\}, \tag{14}$
and thus we may write
$\Bbb R^{\Bbb R} = L \oplus M. \tag{15}$
Note the non-zero constant functions are all in $L$; the only constant in $M$ is $0$.
