How to show: $\exists$ L such that $X\neq L\oplus L^{\perp}$

Question

$X$ : inner product space

$L$ : closed subspace of $X$

How to show: $\exists$ L such that $X\neq L\oplus L^\perp$

let $X= (\text{the set of all finite sequences}, \|\cdot\|_2 )$

$L=\lbrace x\in X: \sum^\infty_{n=1}x_n/n=0\rbrace$

Then: How to show that $L^\perp =\lbrace 0\rbrace$?

Answer 1

Let be $y\in L^\perp$. Suppose that $y\neq 0$. Then, there is some $j\in\Bbb N$ such that $y_j\neq 0$. Since $y$ is a finite sequence, there is also $k\in \Bbb N$ such that $y_k=0$.

Now define $x$ this way: $x_j=1$, $x_k=-\frac kj$, and the rest of terms $0$.

$x$ is in $L$ since $$\sum_{n=1}^\infty\frac{x_n}n=\frac 1j-\frac k{jk}=0$$

The inner product $\langle x,y\rangle$ is $$\sum_{n=1}^{\infty} x_n\bar y_n$$ but since $y_k=0$ and $x_n$ is $0$ for each $n$ other than $j$ and $k$, the sum has only one nonzero term, and it is at index $n=j$: $x_j\bar y_j=\bar y_j$, so that $\langle x,y\rangle=\bar{y}_j\neq 0$, a contradiction with that $y\in L^\perp$.