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.


2 Answers 2


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 $\|x - y\|_p^p < 2\epsilon$.

  • $\begingroup$ 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! $\endgroup$ Commented Oct 29, 2021 at 13:46
  • $\begingroup$ 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! $\endgroup$ 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.

  • $\begingroup$ 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. $\endgroup$ Commented Oct 29, 2021 at 13:29
  • 1
    $\begingroup$ $s^{-1}(\{0\})$ 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. $\endgroup$
    – jakobdt
    Commented Oct 29, 2021 at 13:33
  • $\begingroup$ Thank you very much, I missed this point before! $\endgroup$ Commented Oct 29, 2021 at 13:39
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    $\begingroup$ 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, $|\sum y_n| = 0$. $\endgroup$
    – Alex Ortiz
    Commented Oct 29, 2021 at 13:41

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