The space $C_{00}$ and its kernel with a continuous function

Question

Let $C_{00}$ be the vector space of all complex sequences having finitely many non-zero terms equipped with the inner product $\displaystyle \langle x,y \rangle=\sum_{n=1}^{\infty}x_n \overline{y_n}$ for all $x=(x_n)$ and $y=(y_n)$ in $C_{00}$.

Define, $f:C_{00} \to \Bbb C$ by $\displaystyle f(x)=\sum_{n=1}^{\infty}\frac{x_n}{n}$. Let , $N$ be the kernel of $f$. Then which of the following is TRUE ?

(A) $C_{00} \not=N$.

(B) $N$ is closed.

(C) $C_{00}$ is not a complete inner product space.

(D) $C_{00}=N \oplus N^{\perp}$.

$x=(1,1,0,0,...)\in C_{00}$ but $x\not \in N$. So (A) is TRUE.

Since $f$ is continuous, $N$ is closed. So (B) is TRUE.

Please check whether I'm right or wrong.

I can easily check that $C_{00}$ is NOT complete. So (C) is TRUE.

Since $N$ is a closed subspace of $C_{00}$, (D) is TRUE.

Answer 1

You are right about (A), (B), and (C), but wrong about (D). In fact, $N^\perp=\{0\}$ and therefore $N\oplus N^\perp=N\neq C_{00}$.

Answer 2

Let $x = (x_n)_{n=1}^\infty \in N^\perp$ and assume $x_n = 0$, $\forall n > m$.

For $j \in \{1, \ldots, m\}$ define $$y = \overline{x_j}\left(e_j - \frac{m+1}{j}e_{m+1}\right) = \left(0, \ldots, 0, \underbrace{\overline{x_j}}_{j}, 0, \ldots, 0, \underbrace{-\frac{m+1}{j}\cdot \overline{x_j}}_{m+1}, 0, \ldots\right) \in c_{00}$$

and notice that $y \in N$.

We have:

$$0 = \langle x, y\rangle = |x_j|^2 \implies x_j = 0$$

This implies $x = 0$.

Thus, $N^\perp = \{0\}$ so it cannot be $c_{00} = N \oplus N^\perp = N$.

To be able to conclude $M \oplus M^\perp = H$ for any $M$ closed subspace of $H$, the space $H$ has to be a Hilbert space. As you checked, $c_{00}$ is not.