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

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.

• I think there is an error in defining $\mathcal{M}_n$ this way. The condition that the finite sum is convergent seems trivially true. Did you mean for the finite sum to equal $0$? Commented Jan 25, 2022 at 7:32
• @TheoBendit You're right i'm sorry, I edited it. Commented Jan 25, 2022 at 7:35
• Can you find a Cauchy sequence in $K$ which does not converge? Commented Jan 25, 2022 at 7:56

$$|\sum\limits_{n=1}^{N}a_n|\leq \sqrt N \sqrt {\sum\limits_{n=1}^{N}|a_n|^{2}}$$ 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].
• The claim that $(a_n) \in \ell_2$ implies that $(a_1, \dots a_n, 0, 0, \dots)$ belongs to $K$ is not true. Commented Jan 25, 2022 at 7:41
• @KaviRamaMurthy I understood your solution very well, thanks for your help! Now I'm trying to find the ortogonal space of each $\mathcal M_N^\perp$. I got $$\mathcal M_N^\perp=\left[ \{1,1,...,1,0,0,...\} \right]$$ (the $N$ first terms are $1$ and the rest are all $0$). I hope it's correct, tough my reasoning to reach that result may not be completely rigurous. Commented Jan 25, 2022 at 8:47