A direct sum of two function spaces Let 
$$W=\{f\colon\mathbb{R}^2\to\mathbb{R}\},$$ $$U=\{f\in W\colon f(x,y)=f(y,x)\},$$ $$V=\{f\in W\colon f(x,y)=-f(y,x)\}.$$ How to prove that $W=U\bigoplus V$?
I started with the direct sum condition. So, let $f_1\in U$, $f_2\in V$. Then 
$$f_1+f_2=\theta \implies f_1=f_2=\theta.$$
This combined with the definitions of $U$ and $V$ leads to 
$$f_1(x,y)+f_2(x,y)=0  \Leftrightarrow f_1(y,x)=f_2(y,x)=0.$$
 A: The target
$$
W=U\oplus V
$$
means that for any $f\in W$, it observes a unique decomposition
$$
f=f_1+f_2,
$$
such that $f_1\in U$ and $f_2\in V$.
Thus let us take an arbitrary $f=f(x,y)\in W$. Note that
$$
f(x,y)=\frac{f(x,y)+f(y,x)}{2}+\frac{f(x,y)-f(y,x)}{2}.
$$
Define
\begin{align}
f_1(x,y)&=\frac{f(x,y)+f(y,x)}{2},\\
f_2(x,y)&=\frac{f(x,y)-f(y,x)}{2}.
\end{align}
Then obviously, $f_1\in U$ and $f_2\in V$.
The uniqueness of this decomposition could be proven as follows. Suppose
$$
f_1+f_2=f=g_1+g_2,
$$
where $f_1,g_1\in U$ and $f_2,g_2\in V$. It suffices to show that $f_1=g_1$ and $f_2=g_2$, meaning that if we have two decompositions, they must be identical.
Note that
$$
f_1+f_2=g_1+g_2\iff f_1-g_1=g_2-f_2.
$$
Define
$$
h=f_1-g_1=g_2-f_2.
$$
Since $f_1,g_1\in U$, we have $h\in U$. Similarly, as $f_2,g_2\in V$, we also have $h\in V$. These facts force $h\in U\cap V$, i.e.,
\begin{align}
h\in U&\Longrightarrow h(x,y)=h(y,x),\\
h\in V&\Longrightarrow h(x,y)=-h(y,x).
\end{align}
Hence
$$
h(y,x)=h(x,y)=-h(y,x)\Longrightarrow h(y,x)=0\Longrightarrow h=0.
$$
Consequently,
$$
f_1-g_1=g_2-f_2=h=0,
$$
i.e.,
\begin{align}
f_1&=g_1,\\
f_2&=g_2.
\end{align}
To sum up, we have proven that $\forall\,f\in W$, it observes a unique decomposition
$$
f=f_1+f_2,
$$
with $f_1\in U$ and $f_2\in V$. Therefore, we may conclude that
$$
W=U\oplus V.
$$
