Closed set in $\ell^p$

Question

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.

Answer 1

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

Answer 2

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.