1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.