Show subspace, and orthogonal subspace Let the real inner product space $V=C([-1,1],\mathbb R)$ with inner product $$<f,g>=\int_{-1}^{1}f(t)g(t)dt$$
Set $$S_+=\{f\in V:f(t)=f(-t) \forall t\in [-1,1]\}$$
$$S_-=\{f \in V:f(t)=-f(-t) \forall t\in[-1,1]\}$$
1) Show that $S_+,S_-$ is a subspace of $V$.
For those to be a subspace, I want to show the following, but I am not sure on how to do this. $$0 \in S_+$$$$ \forall u,v \in S_+ \Rightarrow u+v\in S_+$$ $$\forall u\in S_+, \alpha \in V\Rightarrow \alpha \cdot u\in S_+$$
2) Show that $S_+$ and $S_-$ is orthogonal subspace in $V$
For this I believe I have to use the inner product. I know the following, if $$<S_+,S_->=0$$
Then they are orthogonal. 
 A: Let $Z:[-1,1]\to\mathbb{R}$ be the zero function. Then $Z(t)=0$ for all $t\in[-1,1]$, so $Z(t)=Z(-t)=-Z(-t)$ for all $t\in[-1,1]$, so $Z\in S^+\cap S^-$.
For any $f,g\in S^+$, it holds that $(f+g)(t)=f(t)+g(t)=f(-t)+g(-t)=(f+g)(-t)$ for all $t\in[-1,1]$, so $f+g\in S^+$. By the exact same argument it follows that $f+g\in S^-$ for any $f,g\in S^-$.
The second condition is not that for any $v\in S^+$ and $\alpha\in V$ it holds that $\alpha\cdot v\in S^+$; the subspace condition is that the space be closed under multiplication with elements from the field over which the vector space is taken, which is $\mathbb{R}$ here. So let $f\in S^{\pm}$ and $\alpha\in\mathbb{R}$, then $(\alpha\cdot f)(t)=\alpha\cdot f(t)=\alpha\cdot f(-t)=(\alpha\cdot f)(-t)$, so $\alpha\cdot f\in S^{\pm}$ (where of course for $S^-$ you have the minuses, but that should be clear now).
Now let $f\in S^+$ and $g\in S^-$, and let $\phi:[0,1]\to[-1,0]$ be the continuous bijection sending $t$ to $-t$. Then
\begin{align}
\langle f,g\rangle&=\int_{-1}^1f(t)g(t)\,\mathrm{d}t \\
&=\int_{-1}^0f(t)g(t)\,\mathrm{d}t+\int_0^1f(t)g(t)\,\mathrm{d}t \\
&=\int_{\phi(1)}^{\phi(0)}f(t)g(t)\,\mathrm{d}t+\int_0^1f(t)g(t)\,\mathrm{d}t \\
&=\int_1^0f(\phi(t))g(\phi(t))\phi'(t)\,\mathrm{d}t+\int_0^1f(t)g(t)\,\mathrm{d}t \\
&=-\int_0^1f(-t)g(-t)\cdot-1\cdot\,\mathrm{d}t+\int_0^1f(t)g(t)\,\mathrm{d}t \\
&=\int_0^1f(t)\cdot-g(t)\,\mathrm{d}t+\int_0^1f(t)g(t)\,\mathrm{d}t \\
&=0,
\end{align}
which proves the orthogonality.
