We know that every function can be written as the sum of an odd and even function. That is:
$$f(x)={\rm E}+ {\rm O}$$
where ${\rm E}$ is the even function which is no other than ${\rm E}=\frac{f(x)+f(-x)}{2}$ and the odd function is no other than ${\rm O} = \frac{f(x)-f(-x)}{2}$. Also this decomposition is unique (left as an exercise to the reader). Prove that the sum of the odd and the even function is a direct sum.
Solution
Indeed we have seen that it is a sum so all we have to prove now is that the intersection of these two functions is the zero one. However, intersecting odd and even functions there is no other option for a function being both at the same time rather than the zero. Hence, this sum is a direct one, meaning that:
$$f(x)={\rm E} \oplus {\rm O}$$
$$f(x)={\rm E}+ {\rm O}$$
where ${\rm E}$ is the even function which is no other than ${\rm E}=\frac{f(x)+f(-x)}{2}$ and the odd function is no other than ${\rm O} = \frac{f(x)-f(-x)}{2}$. Also this decomposition is unique (left as an exercise to the reader). Prove that the sum of the odd and the even function is a direct sum.
Solution
Indeed we have seen that it is a sum so all we have to prove now is that the intersection of these two functions is the zero one. However, intersecting odd and even functions there is no other option for a function being both at the same time rather than the zero. Hence, this sum is a direct one, meaning that:
$$f(x)={\rm E} \oplus {\rm O}$$
More analytically :
ReplyDeleteLet \(\displaystyle{A}\) be a non-empty subset of \(\displaystyle{\mathbb{R}\) such that
\(\displaystyle{-x\in A\,\,,\forall\,x\in A}\) . Consider the abelian group
\(\displaystyle{\left(G,+\right)}\)
where :
\(\displaystyle{G=\left\{f:A\longrightarrow \mathbb{R}\right\}}\) and \(\displaystyle{+}\) is the
usual operation of addition of real functions.
The zero-element of this group is the function \(\displaystyle{\mathbb{O}:A\longrightarrow \mathbb{R}\,\,,\mathbb{O}(x)=0}\) .
Let \(\displaystyle{H=\left\{f\in S: f(-x)=f(x)\,,\forall\,x\in A\right\}\subseteq S}\) be the set of all the
even functions and
\(\displaystyle{K=\left\{f\in S: f(-x)=-f(x)\,,\forall\,x\in A\right\}\subseteq S}\) be the set of all the
odd functions.
Obviously, \(\displaystyle{\mathbb{O}\in H\,,K}\). Let \(\displaystyle{f_1\,,g_1\in H\,\,,,f_2\,,g_2\in K}\) .
Then,
\(\displaystyle{\forall\,x\in A: (f_1-g_1)(-x)=f_1(-x)-g_1(-x)=f_1(x)-g_1(x)=(f_1-g_1)(x)}\)
\(\displaystyle{\forall\,x\in A: (f_2-g_2)(-x)=f_2(-x)-g_2(-x)=-f_2(x)+g_2(x)=-(f_2-g_2)(x)}\)
which means that \(\displaystyle{f_1-g_1\in H}\) and \(\displaystyle{f_2-g_2\in K}\) .
So, \(\displaystyle{H\,,K\leq \left(G,\cdot\right)}\) and since the group \(\displaystyle{\left(G,+\right)}\)
is an abelian group, we have that \(\displaystyle{H\,,K\trianglelefteq \left(G,+\right)}\) .
Also, \(\displaystyle{H\cap K=\left\{\mathbb{O}\right\}}\) .
Let \(\displaystyle{f\in S}\) . As Tolaso proved, the function
\(\displaystyle{f_1:A\longrightarrow \mathbb{R}\,,f_1(x)=\dfrac{f(x)+f(-x)}{2}}\) is an even function
and the function
\(\displaystyle{f_2:A\longrightarrow \mathbb{R}\,,f_2(x)=\dfrac{f(x)-f(-x)}{2}}\) is an odd function,
that is \(\displaystyle{f_1\in H\,,f_2\in K}\) and also :
\(\displaystyle{\forall\,x\in A: (f_1+f_2)(x)=\dfrac{f(x)+f(-x)+f(x)-f(-x)}{2}=f(x)}\) and thus :
\(\displaystyle{f=f_1+f_2}\) . Therefore, \(\displaystyle{G=H\oplus K}\) .