Prove that the orthogonal space of E is the zero vector

Question

Let $H$ be $C[-1,1]$, let $G=\{f \in H \mid f(1) = 0\}$ and let $\displaystyle E= \left\{f \in H \mid \int_{-1}^{1}f(t) dt = 0\right\}$ and let it have inner product:
$$\langle f,g \rangle \mapsto \int_{-1}^{1}f(t)\overline{g(t)} dt$$ Prove that the ortogonal of $E$ in $G$ ($E^{\perp G}$) is ${0}$.

Answer 1

If we let $e(t) = 1$ for all $t$, then $E = \{ f \in H | \langle e, f \rangle = 0 \} = \{ e \}^\bot = (\operatorname{sp} \{ e \})^\bot$.

We will show that $E^\bot = \operatorname{sp} \{ e \}$. If $x \in \operatorname{sp} \{ e \}$, and $f \in E$, then $\langle x, f \rangle = 0$, hence $x \in E^\bot$, so we have $\operatorname{sp} \{ e \} \subset E^\bot$. Now suppose that $x \in E^\bot$. Then I claim that $x$ is constant (ie, $x \in \operatorname{sp} \{ e \}$).

Let $\tau_{n}(x) = n\int_{-1}^x (1_{[-\frac{1}{n},0]}(t)-1_{[0,\frac{1}{n}]}(t))dt$. If $t \in (-1,1)$, then it is straightforward to show that $\lim_n \langle \tau_n, x \rangle = x(0)$. Now select $t_0,t_1 \in (-1,1)$ and let $\phi_n(t) = \tau_n(t-t_0) - \tau_n(t-t_1)$. Then we have $\lim_n \langle \phi_n, x \rangle = x(t_0)-x(t_1)$. Furthermore, for $n$ sufficiently large, we have $\phi_n \in E$. Hence if $x \in E^\bot$, then for $n$ sufficiently large, we have $\langle \phi_n, x \rangle = 0$, from which we obtain $x(t_0) = x(t_1)$. Since $t_0,t_1 \in (-1,1)$ were arbitrary, and $x$ is continuous, we see that $x$ is constant. Hence $E^\bot = \operatorname{sp} \{ e \}$.

If $x \in E^{\bot}$, then $x$ is a constant function. If $x$ is constant, and $x\in G $, then $x(1) = 0$, and hence $x=0$.

Answer 2

You should prove that if $f$ is orthogonal to $E$, then it is constant. Argue by contradiction: if $f(a)\ne f(b)$ for some $a,b\in (-1,1)$ then there is a function $g\in E$ such that $\int fg\ne 0$.

For example: the graph of $g$ is what you get by connecting the dots $(-1,0)$, $(a-\epsilon,0)$, $(a,1)$, $(a+\epsilon,0)$, $(b-\epsilon,0)$, $(b,-1)$, $(b+\epsilon,0)$, $(1,0)$. Choose $\epsilon>0$ small enough so you can show $\int fg\ne 0$.