Prove that a certain subspace $M$ of space $ (C_{00},\|\cdot\|_2)$ is closed and determine $M^{\perp}$ Could someone help me with this exercise?
Let $$M=\left\{x\in C_{00}: \sum_{k=1}^\infty{\frac{x(k)}{k}=0}\right\}$$ which is a subspace of the prehilbertian space $(C_{00},\|\cdot\|_2)$ . Prove that $M$ is closed and calculate $M^ \perp$.
Clarification
$C_{00}$ is the space of sequences with a finite number of terms different from $0.$
 A: You can see $M$ as the kernel of the linear functional $$\phi:x\longmapsto \sum_k\frac{x(k)}k.$$
So if you show that $\phi$ is bounded, then $M$ is closed. And $\phi$ is bounded because by Cauchy-Schwarz
$$
|\phi(x)|\leq\sum_k\frac{|x(k)|}k\leq \frac{\pi}{\sqrt 6}\,\left(\sum_k|x(k)|^2\right)^{1/2}=\frac{\pi}{\sqrt6}\,\|x\|.
$$
As for $M^\perp$, it will be given by those functionals that have the same kernel as $\phi$. 
Edit: The answer by Fred provides a glimpse into looking at $M^\perp$ from the abstract point of view (and it is what I hinted in my answer). But one can also deal with $M^\perp$ directly. In fact, one can show very easily that $M^\perp=\{0\}$. Indeed, suppose that $y\in M^\perp$, this means that $\sum_k x(k)\overline{y(k)}=0$ for all $x\in M$. Let $m$ be the maximum index such that $y(m)\ne0$. Fix $j\leq m$, and construct an $x\in M$ in the following manner: let $x(j)=1$, $x(m+1)=-(m+1)/j$, and $x(k)=0$ for $k\in\mathbb N\setminus \{j,m+1\}$. 
This way $x\in M$ and $$0=\langle x,y\rangle=\overline{y(j)}.$$
As we can do this for every $j=1,\ldots,m$, it follows that $y=0$. So $M^\perp=\{0\}$. 
A: $\def\czero{c_{00}}$
$\def\ltwo{\mathscr l_2}$
As already presented in the answer of Martin Argerami, we know that $M$ is a closed hyperplane in $\czero$.  I consider the interesting part of the question to be that concerning 
 the orthogonal complement of $M$ in $\czero$.   I want to make this answer pedagogically helpful, which means I don't even want to explicitly tell you the slightly surprising, and entirely explicit,  conclusion, but lead you to discover it. 
Let's establish notation:  $\czero \subset \ltwo$ is a norm--dense subspace.  Let $y_0 = (1/k)_{k \ge 1} \in \ltwo$.  Let $\phi \in \ltwo^*$ be defined by
 $\phi(x) = \langle x, y_0\rangle$.  Then $M = \ker(\phi) \cap \czero =    \{y_0\}^\perp \cap \czero$.  You want to compute the orthogonal complement of $M$ in $\czero$, but it could be helpful to compare what is happening in $\czero$ to what is happening in $\ltwo$.  Therefore, I propose a change of notation as follows:  for any set $S$, whether contained in $\czero$ or not, I propose that $S^\perp$ shall mean its orthogonal complement in $\ltwo$. Thus what you want to compute is 
$M^\perp \cap \czero$, in my proposed notation.
Well then, what is $M^\perp$?  A priori, since $M \subseteq \ker(\phi)$, we have $M^\perp \supseteq \ker(\phi)^\perp = \mathbb C y_0$.  Is this a proper containment or an equality?  The only reason why it would be an equality would be that $M$ is dense in $\ker(\phi)$, either in norm or in a weak topology. What can you say about the density of $M$ in $\ker(\phi)$?
Addendum:  Generalizing, one has the following proposition, which has the same proof.  
Proposition. Let $H$ be a Hilbert space and $H_0$ a proper dense subspace.  Let $y_0 \in H \setminus H_0$.  Let $K = {y_0}^\perp = \ker(\phi)$,  where $\phi(x) = \langle x, y_0 \rangle$ for $x \in H$, and let $K_0 = K \cap H_0$.  Then $K_0^\perp \cap H_0 = (0)$ .
This gives a negative answer to this question.
