Question about orthogonal complement and euclidean space

Question

Consider a set $L=\{x\in L_2[-1,1]: x(t)=0 \ \forall t\le0 \}$ It is easy to see that the orthogonal complement is a set $L^⊥=\{x\in L_2[-1,1]: x(t)=0 \ \forall t\ge0 \}$. But how to prove that on a segment $[0,1]$ in $L^⊥$ consists only such functions? It's enough to show that $(L^⊥)^⊥=L$? And one more question, I have to show that normed space is euclidean space iff for any $x$ and $y$ perform parallelogram equation, i.e $||x+y||^2+||x-y||^2=2(||x||^2+||y||^2)$. I proved this one way. in the opposite direction, I take $(x,y)=\frac{1}{4}(||x+y||^2-||x-y||^2)$ and I proved the linearity, but I have some problem with proving that $(\lambda x,y)=\lambda(x,y)$. Firsly we can easly prove that for $-1$ and by induction that this is done $\forall \lambda\in\mathbb Z$ and therefore $\forall \lambda\in\mathbb N$.(This follows from linearity). I'm also already proof that $\forall \lambda\in\mathbb Q$, but I have some problems to prove this for irrational $\lambda$

Answer 1

For the first part, suppose $x\in L^{\perp}$. Then $$ x = \chi_{[-1,0]}x + \chi_{[0,1]}x $$ and $\chi_{[-1,0]}x\in L^{\perp}$, $\chi_{[0,1]}x\in L$. So $$ x-\chi_{[-1,0]}x = \chi_{[0,1]}x \in L \cap L^{\perp}= \{0\}. $$ That gives $x\in L^{\perp}$ iff $x=\chi_{[-1,0]}x$.

For the second part, $$ |(x,y)| \le \|x+y\|^2+\|x-y\|^2=2(\|x\|^2+\|y\|^2). $$ Assume that $x\ne 0$ and $y\ne 0$. Choose rational numbers $r$, $s$ so that $0 < \|rx\|^2=|r|^2\|x\|^2 \le 1$ and $0 < \|sy\|^2 =|s|^2\|x\|^2\le 1$. Then $$ |rs||(x,y)|=|(rx,ry)| \le 2(\|rx\|^2+\|sy\|^2) \le 4 \\ |(x,y)| \le 1/|rs|, \\ |(x,y)| \le \|x\|\|y\|. $$ Using this Cauchy-Schwarz inequality then gives continuity of the inner product, and allows you to replace scalars with rational real and imaginary components with real components.