# Show $C([-1;1])$ (inner product space) is the direct sum of $S_+$ and $S_-$ and the orthogonal complement of $S_+$ is $S_-$.

Let $C([-1;1])$ be a inner product space, consisting of continuous real functions defined on $[-1;1]$, with inner product given by $\langle f,g \rangle = \int^1_{-1} f(t)g(t)$.

$S_+ = \{f \in C([-1;1]) : f(t) = f(-t) \forall t \in [-1;1]\}$ $S_- = \{f \in C([-1;1]) : f(t) = -f(-t) \forall t \in [-1;1]\}$

I've shown that $S_+, S_-$ are subspaces of $C[-1;1]$.

Also, I've shown that $f \in C([-1;1])$ then $(f(t) + f(-t)) \in S_+$ and $(f(t) - f(-t)) \in S_-$.

However I cannot show $C([-1;1])$ is the direct sum of $S_+$ and $S_-$.

Also I'm having trouble showing that the orthogonal complement of $S_+$ is $S_-$ and vice versa.

• Your $S_+$ and $S_-$ as defined are the same. – Joe Johnson 126 Mar 10 '14 at 14:23
• It has been fixed. – Shuzheng Mar 10 '14 at 15:57

## 3 Answers

You can see that $$f(t)=\frac{1}{2}[(f(t)+f(-t))+(f(t)-f(-t))].$$ Thus $\mathcal{C}([-1,1])$ is the sum of the subspaces. You need only show that the intersection of the two subspaces is $\{0\}$.

To show they are orthogonal, integrate $g\in S_-$ against $f\in S_+$. Use $$g(-t)f(-t)=-g(t)f(t).$$

EDIT: To answer why $S_+=S_-^\perp$, let us examine a general case. Suppose that we have a direct sum $V=W\oplus U$ and $U\subseteq W^\perp$. Suppose that there is some vector $x=v+w$, with $v\neq 0$, such that $\langle x,v'\rangle=0$ for all $v'\in V$. Letting $v'=v$:

\begin{align*} 0 &= \langle x,v\rangle \\ &= \langle v,v\rangle + \langle w,v\rangle \\ &= \langle v,v \rangle \\ \end{align*}

But, the inner product on $\mathcal{C}([-1,1])$ does not allow for $\langle f,f\rangle=0$ if $f\neq 0$.

• If I show the intersection is $\{0\}$ (already done), can I conclude that the sum is unique ? Why ? – Shuzheng Mar 10 '14 at 16:12
• Think I got it, I can write $f = f_1 + f_2 = f_1^{'} + f_2^{'}$ and then conclude $f_1 - f_1^{'} = 0$ because it lies both in $S_+$ and $S_-$. – Shuzheng Mar 10 '14 at 16:15
• How do I show that two functions in $S_+$ does not have an inner product equal to zero ? – Shuzheng Mar 10 '14 at 16:35
• @NicolasLykkeIversen Does my edit help? – Joe Johnson 126 Mar 10 '14 at 19:28
• Thank you, Joe. I appreciate it. Your answer is complete. – Shuzheng Mar 12 '14 at 7:03

$f_+(t)=\frac{f(t)+f(-t)}{2}\in S_+$ and $f_-(t)=\frac{f(t)-f(-t)}{2}\in S_-$, then you have $f(t)=f_+(t)+f_-(t)$ so $C([-1;1])=S_++S_-$. If $f\in S_+\bigcap S_-\Rightarrow -f(t)=f(-t)=f(t)\quad\forall t\in[-1;1]\Rightarrow f(t)=0\quad\forall t\in[-1;1]$

Edit: For the part about that $S_+^\bot=S_-$, you take $f\in S_+$ and $g\in S_-$, so $$\langle f,g\rangle=\int_{-1}^{1}f(t)g(t)dt=\int_{-1}^{0}f(t)g(t)dt+\int_{0}^{1}f(t)g(t)dt=-\int_{1}^{0}f(-u)g(-u)du+\int_{0}^{1}f(t)g(t)dt=-\int_{0}^{1}f(u)g(u)du+\int_{0}^{1}f(t)g(t)dt=0$$

• How do I show two function in $S_+$ does not have an inner product equal to zero ? – Shuzheng Mar 10 '14 at 16:36
• You have shown that $S_-\subseteq S_+^\perp$. The OP doesn't know why $S_-\supseteq S_+^\perp$. – Joe Johnson 126 Mar 10 '14 at 19:36
• It's inmediate consequence of $C([-1,1])=S_++S_-$. – blues66 Mar 10 '14 at 19:57
• More generally, if I have $(V,\langle,\rangle)$ vectorial space with inner product, and $S_1$,$S_2$ subspaces such that $S_2\subseteq S_1^\bot$ and $V=S_1+S_2$, then $S_2=S_1^\bot$ – blues66 Mar 10 '14 at 20:07

You already have a map $\phi: C[-1, 1] \to S_+\oplus S_-$ defined as $$\phi(f)(t) = (f(t) + f(-t), f(t) - f(-t))$$ as you write yourself. Can you think of a nice inverse of this map? Also, if a function is part of both $S_+$ and $S_-$, what can you say about it?

• It must be zero if it lies in the intersection. However to prove $S_+$ has orthogonal complement $S_-$, I show that $f \in S_+$ and $g \in S_-$ satisfy $\langle f, g\rangle = 0$. However I must show that this is the "whole" orthogonal complement. So let $f \in C([-1;1])$ we can write $f = s_+ + s_-$ for unique functions in subspaces and $\langle f, s_+ + s_- \rangle = \langle f, s_+ \rangle$. How do I show that $\langle f, s_+ \rangle = 0$ if and only if $s_+ = 0$ so that I prove what I want ? – Shuzheng Mar 10 '14 at 16:42