# Closed set in $\ell^p$

I am struggling to solve the following problem.

Let $$C := \{x = (x_n)_n \in \ell^p : \sum_{n=1}^{\infty} x_n = 0\}$$ with $$1 \le p < \infty$$. Prove that $$C$$ is closed in $$\ell^p$$ if and only if $$p = 1$$.

I started from the implication: $$p = 1 \Rightarrow C$$ is closed in $$\ell^1$$. Thus, I have to demonstrate that $$\forall x^k = (x^k_n)_k \subset C$$ such that $$x^k \rightarrow y$$ in $$\ell^1$$ for $$k \rightarrow \infty$$, then $$y \in C$$. This means that $$\forall \epsilon > 0, \exists \overline{k}$$ such that $$\forall k \ge \overline{k}$$, $$||x^k-y||_{\ell^1} = \sum_{i=1}^n |x^k_n -y_n| < \epsilon$$. Therefore, the convergence of the sequence in $$\ell^1$$ implies that $$|x^k_n -y_n| < \epsilon$$ for any $$k \ge \overline{k}$$. At this point, my idea was to estimate $$\sum_{n=1}^{\infty} y_n \le |\sum_{n=1}^{\infty} y_n| = |\sum_{n=1}^{\infty} (y_n - x^k_n)|$$, since by hypothesis $$x^k \in C$$. Then, using Holder inequality, I get $$|\sum_{n=1}^{\infty} 1 \cdot (y_n - x^k_n)| \le ||1||_{\ell^{\infty}} ||x^k - y ||_{\ell^1} < \epsilon$$, since the sequence $$(1)_n \in \ell^{\infty}$$. So I have that $$\sum_{n=1}^{\infty} y_n = \sum_{n=1}^{\infty} (y_n - x^k_n) = \sum_{n=1}^{\infty} y_n - \sum_{n=1}^{\infty} x^k_n < \epsilon$$ and I can conclude that $$\sum_{n=1}^{\infty} y_n = 0$$, that is $$y \in C$$. However, I am not really sure about the final part of the reasoning, since I have no information on the signs of each $$y_n$$.

The other implication intuitively may follow from the fact that $$(1)_n \notin \ell^q$$ for every $$1 < q < \infty$$ and thus, assuming $$C$$ is closed, I cannot apply Holder inequality. But I have difficulties in formalizing this reasoning.

If anyone could give me some hint on how to approach this problem, it would be greatly appreciated.

If $$p > 1$$ it is straightfoward to find sequences $$x = (x_n)$$ with $$\|x\|_1$$ arbitrarily large (even infinite) yet with $$\|x\|_p$$ arbitrarily small. For instance for $$M$$ large you could take $$x_n = 0$$ for $$n < M$$ and $$x_n = \frac 1n$$ for $$n \ge M$$. You can exploit this idea to show that the set $$S$$ is in fact dense in $$\ell^p$$ and in particular cannot be closed. I'll try to provide a sketch.

Fix a sequence $$x \in \ell^p$$ and let $$\epsilon > 0$$. Choose $$N$$ large enough to guarantee that $$\displaystyle \sum_{n=N+1}^\infty|x_n|^p < \epsilon$$.

Define a sequence $$y$$ that is eventually zero by $$y_n = x_n$$ if $$n \le N$$, and for a finite number of terms $$n \ge N + 1$$ select $$y_n$$ so that you have $$\displaystyle \sum_{n \ge N+1} y_n = - \sum_{n \le N} y_n$$ and $$\displaystyle \sum_{n = N+1}^\infty |y_n|^p < \epsilon$$. The details might get messy but the idea follows from the first paragraph above.

The density of $$S$$ in $$\ell^p$$ follows from the fact that $$y \in S$$ and $$\|x - y\|_p^p < 2\epsilon$$.

• From the density of $C$ in $\ell^p$ it follows that $\overline{C} = \ell^p$. However, by hypothesis $C$ is closed, hence $C = \ell^p$ which is absurd (as you suggested, I can take $x_n = \frac{1}{n} \in \ell^p \setminus C$). Thank you very much! Commented Oct 29, 2021 at 13:46
• I apologize since it seems like I can only accept one answer as a solution to the problem, even though yours has been extremely useful! Commented Oct 29, 2021 at 13:51

For the implication '$$p=1\Rightarrow C$$ is closed' there is a somewhat simpler solution. Consider the map $$s\colon\ell^1\to\mathbb{R}$$ defined as $$s((x_n))=\sum_{n=1}^\infty x_n$$ for $$(x_n)\in\ell^1$$. Note that $$s$$ is continuous since $$\lvert s((x_n))-s((y_n))\rvert=\lvert\sum_{n=1}^\infty x_n-\sum_{n=1}^\infty y_n\rvert\leq\sum_{n=1}^\infty\lvert x_n-y_n\rvert=\lVert(x_n)-(y_n)\rVert$$ for any $$(x_n),(y_n)\in\ell^1$$. My point is then that $$C=s^{-1}(\{0\})$$ is closed since it is nothing more than the preimage of a closed set for a continuous function.

• I thought about something of this kind, but shouldn't I prove that $s$ is at least injective to conclude that $s^{-1}$ exists? For instance, $s(x_n) = 0$ does not imply $x_n = 0$ (it suffices to take any $x_n \in C \setminus \{0\}$). I agree with you that if $s^{-1}$ exists and if it is continuous, then the thesis easily follows from your reasoning. Commented Oct 29, 2021 at 13:29
• $s^{-1}(\{0\})$ denotes the preimage of $\{0\}$ under $s$. This is the set of sequences $(x_n)\in\ell^1$ with $s((x_n))=0$. This is defined for any function - not only for injective functions. Commented Oct 29, 2021 at 13:33
• Thank you very much, I missed this point before! Commented Oct 29, 2021 at 13:39
• A slight modification that doesn't rely on the notion of continuity is to write $|\sum y_n| \le |\sum y_n-x_n| + |\sum x_n| \le \sum|y_n-x_n| < \epsilon$ for every $\epsilon$, so in particular, $|\sum y_n| = 0$. Commented Oct 29, 2021 at 13:41