Prove $\mathcal M_n$ is closed subspace of $\ell^2$

Question

I've been solving problems from my Functional Analysis course and don't know how to approach this one:

Given $\ell^2=\left\{\{a_n\}_{n\in\mathbb N}\subset \mathbb C : \sum_{n=1}^\infty|a_n|^2<\infty\right\}$, $N\in\mathbb N$ and $$\mathcal M_N=\left\{\{a_n\}\subset \ell^2 : \sum_{n=1}^N a_n=0\right\},\phantom{aa} \mathcal K=\left\{\{a_n\}\subset \ell^2 : \sum_{n=1}^\infty a_n=0\right\}.$$ Prove $\mathcal M_N$ is a closed subspace of $\ell^2$. Is $\mathcal K$ also closed?

I considered defining the linear mappings $T_N:\ell^2\to \mathbb C$ defined as $$T_N(\{a_n\})=\sum_{n=1}^N a_n.$$ If I managed to prove it's continuous it would be done for $\mathcal M_N$. To prove it I think I must use the equivalence between continuity and boundedness for linear mappings between normed spaces. Minkowski's inequality gives me that $$\left\| \sum_{n=1}^N a_n\right\|\leq\sum_{n=1}^N\| a_n \|,$$ but I'm not sure how can I relate this to the hypothesis that the sum of the squares of the $a_n$ is finite (and how to use that to conclude each $T_N$ is bounded). For $\mathcal K$, I got no clue where to start.

How can I solve this? Any help or hint will be appreciated.

Answer 1

$|\sum\limits_{n=1}^{N}a_n|\leq \sqrt N \sqrt {\sum\limits_{n=1}^{N}|a_n|^{2}}$ by Cauchy_Schwarz inequality. This proves that $T$ is continuous (with $\|T\| \leq \sqrt N$) and hence $\mathcal M_N$ is closed. $\mathcal K$ is not closed. In fact, it is dense. In fact the kernel of any dis-continuous linear functional is dense. See relation between linear functional and kernel on nls

[If $(a_n) \to \sum a_n$ is continuous then there exists a constant $C$ such that $|\sum a_n | \leq C \sqrt {\sum |a_n|^{2}}$ for all $(a_n) \in \ell^{2}$. Look at $(1,\frac 1 2, \frac 1 3,...)$ to get a contradiction].