How to prove L is a Hilbert space?

Question

Let us consider the space $\ell^2$ with a function $[\cdot,\cdot]:\ell^2\rightarrow\mathbb{C}$ defined by $[x,y]=-x_1\overline{y_1}+\frac{1}{2}x_2\overline{y_2}+\frac{1}{4}x_3\overline{y_3}+\ldots$, where $x=(x_1,x_2,\ldots),~y=(y_1,y_2,\ldots)\in\ell^2$. Now let us consider the subspace $L$ defined by $L=\{x\in\ell^2:x_1=\frac{1}{2}x_2+\frac{1}{4}x_3+\ldots\}$. How can I prove that $(L,[\cdot,\cdot]|_L)$ is a Hilbert space?

Answer 1

By Cauchy's inequality applied to the vectors $u=(2^{-1/2},2^{-2/2},2^{-3/2},...)$ and $v=(2^{-1/2}x_2,2^{-2/2}x_3,2^{-3/2}x_4,...)$ we get

$$\require{cancel}\begin{align}\left(\sum_{n=2}^{\infty}2^{-n+1}|x_n|^2\right)&=\cancelto{{}=\,1}{\left(\sum_{n=2}^{\infty}2^{-(n-1)}\right)}\left(\sum_{n=2}^{\infty}2^{-n+1}|x_n|^2\right)\\&=\|u\|^2\|v\|^2\geq (u\cdot v)^2\\&=\left|\sum_{n=2}^{\infty}\frac{1}{2^{n-1}}x_n\right|^2\\&=|x_1|^2\end{align}$$

Therefore $$[x,x]=-|x_1|^2+\sum_{n=2}^{\infty}2^{-n+1}|x_n|^2\geq 0$$ for $x=(x_1,x_2,x_3,...)\in L$.

The equality can only hold when the equality holds in the step where we used Cauchy's inequality. Therefore, it can only hold when $u,v$ are proportional. This means that $x_2=x_3=...$. But, in $L$, this implies that $x_1=x_2\sum_{n=1}^{\infty}2^{-n}=x_2$. So, $x$ would have to be constant and in $\ell^2$. Hence $x=0$.